problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
Example 13. Compute the integral
$$
\int_{-\infty}^{+\infty} \frac{e^{a x}}{1+e^{x}} d x \quad(0<a<1)
$$ | Solution. Let us choose the auxiliary function
$$
f(z)=\frac{e^{a z}}{1+e^{z}}
$$
and the contour shown in Fig. 11 (a rectangle with sides $2 R$ and $2 \pi$). Inside this contour, $f(z)$ is analytic except for the point $Z=\pi i$, which is a simple pole for it.
$$
\text { res } f(\pi i)=\left.\frac{e^{a z}}{\left(1+... | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,091 | |
Example 14. Calculate the integral
$$
I=\int_{0}^{2 \pi} \frac{d x}{(a+b \cos x)^{2}} \quad(a>b>0)
$$ | Solution. Applying the substitution $e^{i z}=z$, we obtain after simple transformations
$$
I=\frac{4}{i} \int_{C} \frac{z d z}{\left(b z^{2}+2 a z+b\right)^{2}}=\frac{4}{i} 2 \pi i \sum_{k=1}^{n} \operatorname{res} F\left(z_{k}\right)
$$
Inside the unit circle, under the condition that $a>b>0$, there is only one zero... | \frac{2\pi}{(^{2}-b^{2})^{3/2}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,092 |
Example 1. Find the residues of the logarithmic derivative of the function
$$
f(z)=\frac{\sin z}{z+1}
$$
with respect to its zeros and poles. | Solution. The given function has an infinite set of simple zeros $z=k \pi (k=0, \pm 1, \pm 2, \ldots)$ and one simple pole $z=-1$. Hence,
## Problems for Independent Solution
Find the res... | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,094 | |
Example 2. Find the logarithmic residue of the function
$$
f(z)=\frac{\operatorname{ch} z}{e^{i z}-1}
$$
with respect to the contour $C:|z|=8$. | Solution. We find the zeros $z_{k}$ of the function $f(z)$. For this, we solve the equation $\cosh z=0$ or $e^{z}+e^{-z}=0$. Writing the last equation as $e^{2 z}=-1$, we find
$2 z=\operatorname{Ln}(-1)=(2 k+1) \pi i$, so $z_{k}=\frac{2 k+1}{2} \pi i(k=0, \pm 1, \pm 2, \ldots)$ (all zeros are simple). To find the poles... | 3 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,095 |
Example 3. Find the logarithmic residue of the function
$$
f(z)=\frac{1+z^{2}}{1-\cos 2 \pi z}
$$
with respect to the circle $|z|=\pi$. | Solution. Setting $1+z^{2}=0$, we find two simple zeros of the function $f(z): a_{1}=-i, a_{2}=i$. Setting $1-\cos 2 \pi z=0$, we find the poles of the function $f(z): z_{n}=n, n=0, \pm 1, \pm 2, \ldots$. The multiplicity of the poles is $k=2$.
In the circle $|z|<\pi$, the function has two simple zeros $a_{1}=-i, a_{2... | -12 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,096 |
Example 4. Find the number of roots in the right half-plane $\operatorname{Re} z>0$ of the equation
$$
Q_{5}(z) \equiv z^{5}+z^{4}+2 z^{3}-8 z-1=0
$$ | Solution. By the argument principle, the number of zeros inside the contour $C$ is
$$
N=\frac{1}{2 \pi} \Delta_{C} \operatorname{Arg} Q_{5}(z)
$$
where the contour $C$ consists of the semicircle $C_{R}:|z|=R, \operatorname{Re} z>0$, and its diameter on the imaginary axis; the radius $R$ is taken to be so large that a... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,097 |
Example 5. Find the number of roots of the equation
$$
Q_{7}(z) \equiv z^{7}-2 z-5=0
$$
in the right half-plane. | Solution. We choose the contour $C$ as indicated in Example 4. Then $\Delta_{C_{R}} \operatorname{Arg} Q_{7}(z)=\Delta_{C_{R}} \operatorname{Arg}\left(z^{7}-2 z-5\right)=$
$$
\begin{aligned}
& =\Delta_{C_{R}} \operatorname{Arg}\left[z^{7}\left(1-\frac{2}{z^{6}}-\frac{5}{z^{7}}\right)\right]=7 \Delta_{C_{R}} \operatorn... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,098 |
Example 6. Find the number of zeros of the function
$$
F(z)=z^{8}-4 z^{5}+z^{2}-1
$$
inside the unit circle $|z|<1$. | Solution. Let us represent the function $F(z)$ as the sum of two functions $f(z)$ and $\varphi(z)$, which we choose, for example, as follows:
$$
f(z)=-4 z^{5}, \quad \varphi(z)=z^{8}+z^{2}-1
$$
Then on the circle $|z|=1$ we will have
$$
\begin{aligned}
& |f(z)|=\left|-4 z^{5}\right|=4 \\
& |\varphi(z)|=\left|z^{8}+z... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,099 |
Example 7. Determine the number of roots of the equation
$$
z^{6}-6 z+10=0
$$
inside the circle $|z|<1$. | Solution. Let, for example, $f(z)=10$ and $\varphi(z)=z^{6}-6 z$. On the circle $|z|=1$ we have
$$
|f(z)|=10, \quad|\varphi(z)|=\left|z^{6}-6 z\right| \leqslant\left|z^{6}\right|+6|z|=7
$$
Thus, in all points of the circle $|z|=1$, the inequality $|f(z)|>|\varphi(z)|$ holds. The function $f(z)=10$ has no zeros inside... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,100 |
Example 8. How many roots of the equation
$$
z^{4}-5 z+1=0
$$
lie in the annulus $1<|z|<2 ?$ | Solution. Let $N$ be the number of roots of equation (4) in the ring $1<|\varphi(z)|$, since $|f(z)|=|-5 z|=5,|\varphi(z)|=\left|z^{4}+1\right| \leqslant$ $\left|z^{4}\right|+1=2$. The function $f(z)=-5 z$ has one root in the circle $|z|<1$, and thus $N_{1}=1$.
In the circle $|z|<2$, $|f(z)|>\left|\varphi(z)\right|$, ... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,101 |
Example 9. Find the number of roots of the equation
$$
z^{2}-a e^{z}=0, \quad \text { where } \quad 0<a<e^{-1}
$$
in the unit circle $|z|<1$. | Solution. Let $f(z)=z^{2}$ and $\varphi(z)=-a e^{z}$. On the circle $|z|=1$ we have
$$
\begin{aligned}
& |f(z)|=\left|z^{2}\right|=1 \\
& |\varphi(z)|=\left|-a e^{z}\right|=a\left|e^{z}\right|=a\left|e^{x+i y}\right|=a e^{x} \leqslant a e|\varphi(z)|$, if $|z|=1$. The function $f(z)=z^{2}$ in the circle $|z|0, \quad \... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,102 |
Example 10. Find the number of roots of the equation
$$
\lambda-\boldsymbol{z}-e^{-z}=0, \quad \lambda>1
$$
in the right half-plane $\operatorname{Re} z>0$. | Solution. Consider the contour composed of the segment $[-i R, i R]$ and the right semicircle $|z|=R$. Let $f(z)=z-\lambda$ and $\varphi(z)=e^{-z}$. On the segment $[-i R, i R]$, where $z=i y$, we have
$$
\begin{aligned}
& |f(z)|=|i y-\lambda|=\sqrt{\lambda^{2}+y^{2}} \geqslant \sqrt{\lambda^{2}}=\lambda>1 \\
& |\varp... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,103 |
Example 1. In which domains $D$ are the mappings
a) $w=2z$,
b) $w=(z-2)^{2}$
conformal | Solution. a) Since the function $f(z)=2z$ is analytic and univalent in the entire complex plane $z$, and its derivative $f'(z)=2 \neq 0$, the given mapping is conformal in the entire complex plane.
b) The mapping $w=(z-2)^2$ is conformal everywhere except at the point $z=2$, where the derivative $f'(z)=2(z-2)$ is zero... | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,104 |
Example 3. Given points $z_{1}=2+3 i$ and $z_{2}=3+2 i$, symmetric with respect to the line $y=x$. Show that the function $w=e^{-i \pi / 2} z$ maps $z_{1}$ and $z_{2}$ to points $w_{1}=3-2 i$ and $w_{2}=2-3 i$, symmetric with respect to the line $v=-u$. | Solution. It is not difficult to verify that the function $w=e^{-i \pi / 2} z$ maps the line $y=x$ to the line $v=-u$. The function $w=e^{-i \pi / 2} z$ is analytic everywhere. By the principle of symmetry, the points $z_{1}=3+2 i$ and $z_{2}=2+3 i$, symmetric with respect to the line $y=x$, will be transformed into th... | proof | Algebra | proof | Yes | Yes | olympiads | false | 31,106 |
Example 4. Show that the function $w=e^{\pi z / n}$ maps the strip $0 < \operatorname{Im} z < n$ onto the upper half-plane $\operatorname{Im} w > 0$. | Solution. We will traverse the boundary of the region $D$ in such a way that the region $D$ remains to the left. Since
$$
w=u+i v=e^{\pi(x+i y) / h}=e^{\pi x / h} e^{i \pi y / h}
$$
then, when the point $z$ traverses the real axis $O x$ from $x=-\infty$ to $x=+\infty$ (with $y=0$), the corresponding point $w=e^{\pi /... | proof | Algebra | proof | Yes | Yes | olympiads | false | 31,107 |
Example 5. Show that the linear mapping $w=a z+b$ is completely determined if we require that two distinct points $z_{1}$ and $z_{2}$ are mapped respectively to arbitrarily given, but distinct points $w_{1}$ and $w_{2}$. | Solution. Indeed, the mapping $w=a z+b$ will be realized if the values of the parameters $a$ and $b$ are known. Let us show that our conditions allow us to uniquely determine these parameters. Suppose that for $z=z_{1}$ we get $w=w_{1}$, i.e., $w_{1}=a z_{1}+b$, and for $z=z_{2}$ we get $w_{2}=a z_{2}+b$.
From these e... | proof | Algebra | proof | Yes | Yes | olympiads | false | 31,108 |
Example 7. Find the linear function that maps the triangle with vertices at points $0,1, i$ in the $z$-plane to a similar triangle with vertices $1+i, 0,2$ in the $w$-plane. | Solution. First method. From Fig. 16, we see that $\triangle ABC$ transforms into a similar $\triangle A_{1} B_{1} C_{1}$ through the following operations:
1) rotation around the origin by an angle $\frac{5}{4} \pi$, which corresponds to the transformation
$$
w_{1}=e^{\frac{5}{4} \pi} z
$$
2) similarity transformati... | (1+i)(1-z) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,110 |
Example 8. Find the image of the circle $|z|=3$ under the mapping $w=\frac{25}{z}$. | Solution. First method. Let $z=x+i y, w=u+i v$. Then the relation $w=\frac{25}{z}$ can be rewritten as
$$
u+i \dot{v}=\frac{25}{x+i y}=\frac{25 x}{x^{2}+y^{2}}-i \frac{25 y}{x^{2}+y^{2}}
$$
from which
$$
u=\frac{25 x}{x^{2}+y^{2}}, \quad v=-\frac{25 y}{x^{2}+y^{2}}
$$
The equation of the circle $|z|=3$ in Cartesian... | u^{2}+v^{2}=(\frac{25}{3})^{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,111 |
Example 9. Find the conditions under which the fractional-linear function (6)
$$
w=\frac{a z+b}{c z+d}
$$
maps the upper half-plane $\operatorname{Im} z>0$ onto the upper half-plane $\operatorname{Im} w>0$. | Solution. Under this mapping, it is required that the boundary of the region $\operatorname{Im} z>0$ - the $0 x$ axis, traversed from left to right, is mapped to the boundary of the region $\operatorname{Im} w>0$, i.e., to the $O u$ axis, also traversed from left to right. Thus, for any real values of $z$, the values o... | ->0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,112 |
Example 10. Find the fractional-linear function that maps the points $z_{1}=1, z_{2}=i, z_{3}=-1$ to the points $w_{1}=-1, w_{2}=0, w_{3}=1$. | Solution. Using formula (7), we have
$$
\frac{w+1}{w-0} \cdot \frac{1-0}{1-(-1)}=\frac{z-1}{z-i} \cdot \frac{-1-i}{-1-1}
$$
from which $w=i \frac{i-z}{i+z}$;
Remark. If one of the points $z_{k}$ or $w_{k}(k=1,2,3)$ is infinitely distant, then in formula (7) all differences containing this point should be replaced by... | i\frac{i-z}{i+z} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,113 |
Example 11. Find a fractional-linear function that maps the point $z_{1}$ to the point $w_{1}=0$, and the point $z_{2}$ to the point $w_{2}=\infty$. | Solution. Let us take an arbitrary point $z_{3}$, different from points $z_{1}$ and $z_{2}$, and assume that it maps to a point $w_{3}$, different from points $w_{1}$ and $w_{2}$. Then, by formula (7) and with the remark in mind, we have
$$
\frac{w-0}{1} \cdot \frac{1}{w_{3}-0}=\frac{z-z_{1}}{z-z_{2}} \cdot \frac{z_{3... | K\frac{z-z_{1}}{z-z_{2}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,114 |
Example 12. Map the upper half-plane $\operatorname{Im} z>0$ onto the unit disk $|w|<1$ so that the point $z=i$ (where $\operatorname{Im} z>0$) is mapped to the center $\boldsymbol{w}=0$ of the disk. | Solution. Since the point $z_{0}$ is mapped by the sought fractional-linear function $w=w(z)$ to the center of the circle, i.e., $w\left(z_{0}\right)=0$, then the conjugate point $\bar{z}_{0}$ must be mapped to the point $w:=\infty$ (by the property of symmetry). Next, we use formula (8) and obtain
$$
w=K \frac{z-z_{0... | e^{i\alpha}\frac{z-z_{0}}{z-\bar{z}_{0}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,115 |
Example 13. Map the unit circle $|z|<1$ onto the unit circle $|\boldsymbol{w}|<1$. | Solution. Let the desired linear fractional transformation $w=w(z)$ map the point $z_{0}$, located inside the circle $|z|<1$, to the center of the circle $|w|<1$, so that $w\left(z_{0}\right)=0$. Then the point $z_{0}^{*}=\frac{1}{\bar{z}_{0}}$, symmetric to the unit circle $|z|=1$, will map to the point $\infty$, i.e.... | e^{i\alpha}\frac{z-z_{0}}{1-z\bar{z}_{0}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,116 |
Example 15. Map the sector $0<\arg z<\frac{\pi}{4}$ onto the unit disk $|w|<1$ such that the point $z_{1}=e^{i \pi / 8}$ is mapped to the center $w_{1}=0$, and the point $z_{2}=0$ is mapped to the point $w_{2}=1$. | Solution. Sector $00$ (Fig. 19, b). The point $z_{1}=e^{i \pi / 8}$ will map to the point $t_{1}=z_{1}^{4}=i$, and $z_{2}=0$ will map to the point $t_{2}=0$.
Then we map the upper half-plane $\operatorname{Im} t>0$ to the unit disk $|w|<1$ such that the point $t_{1}=i$ maps to the center of the disk (Fig. 19, c). Usin... | -\frac{z^{4}-i}{z^{4}+i} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,118 |
Example 16. Find the function that maps the upper half of the circle $|z|<1$, $\text{Im } z>0$, onto the upper half-plane $\text{Im } w>0$. | Solution. The given region represents a biangle with vertices at points $z_{1}=-1$ and $z_{2}=1$ and an angle at the vertex $\alpha=\frac{\pi}{2}$ (Fig. 20,a).
The auxiliary function $t=\frac{1+z}{1-z}$ performs a conformal mapping of this biangle onto the first quadrant of the $t$-plane (Fig. 20,b). The function $w=t... | (\frac{1+z}{1-z})^2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,119 |
Example 21. Using the Joukowsky function, find the image of the region
$$
0<|z|<1, \quad 0<\arg z<\frac{\pi}{4}
$$ | Solution. Substitute $z=r e^{i \varphi}$ into the Joukowski function
$$
w=\frac{1}{2}\left(z+\frac{1}{z}\right)
$$
and separate the real and imaginary parts; we get
$$
\left\{\begin{array}{l}
u=\frac{1}{2}\left(r+\frac{1}{r}\right) \cos \varphi \\
v=\frac{1}{2}\left(r-\frac{1}{r}\right) \sin \varphi
\end{array}\righ... | u^{2}-v^{2}>\frac{1}{2},\quadu>\frac{\sqrt{2}}{2},\quadv<0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,121 |
Example 22. Map the unit disk with a cut along the real axis from the center to the upper half-plane. | Solution. 1) Using the function $w_{1}=\sqrt{z}$, map the unit circle to the upper half-circle. In this process, the upper bank of the cut O remains in place, while the lower bank $O A^{\prime}$ will be mapped to the segment $[-1,0]$ on the $W_{1}$ plane.
2) Using the function
$$
w_{2}=\frac{w_{1}+1}{w_{1}-1}
$$
map... | (\frac{\sqrt{z}+1}{\sqrt{z}-1})^{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,122 |
Example 1. Find the function $w=f(z)$ that conformally maps the upper half-plane $\operatorname{Im} z>0$ onto the region
$$
0<\arg w<\alpha \pi, \quad \text { where } \quad 0<\alpha<2
$$
of the $w$-plane. | Solution. Since the given region is a polygon with vertices $\overline{A_{1}(w=0)}$ and $A_{2}(w=\infty)$, the solution can use the Christoffel-Schwarz integral, which defines the desired function as
$$
w=f(z)=C_{1} \int_{0}^{z}\left(\tau-a_{1}\right)^{\alpha-1} d \tau+C_{2}
$$
Assume that on the $O x$ axis of the $z... | z^{\alpha} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,125 |
Example 3. Conformally map the upper half-plane $\operatorname{Im} z>0$ onto a polygon in the $w$-plane (Fig. 36) such that
$$
w\left(A_{1}=0, A_{2}=1, A_{3}=\infty\right)
$$
corresponds to
$$
z\left(a_{1}=0, a_{2}=1, a_{3}=\infty\right)
$$ | Solution. Consider the given region of the plane $\boldsymbol{w}$ as the interior of a "triangle" with vertices $A_{1}=0, A_{2}=1, A_{3}=\infty$ and angles at these vertices
$$
\alpha_{1} \pi=\frac{3}{2} \pi, \quad \alpha_{2} \pi=\frac{\pi}{2}, \quad \alpha_{3} \pi=-\pi
$$
From this, we have
$$
\alpha_{1}=\frac{3}{2... | \frac{2}{\pi}(\arcsin\sqrt{z}-\sqrt{z-z^{2}}) | Other | math-word-problem | Yes | Yes | olympiads | false | 31,126 |
Example 1. The motion of a fluid is described by the complex potential $f(z)=z^{2}$. Find the velocity potential, the stream function, the level lines, the streamlines, the magnitude and direction of the velocity vector $\mathbf{V}$, and the projections of the velocity vector $V_{O x}$ and $V_{o y}$ on the coordinate a... | Solution. Assuming $z=x+i y$, we have
$$
f(z)=\left(x^{2}-y^{2}\right)+i 2 x y,
$$
from which the velocity potential $u(x, y)=x^{2}-y^{2}$ and the stream function $v(x, y)=2 x y$. The level lines $u(x, y)=$ const are hyperbolas $x^{2}-y^{2}=$ const. The streamlines $v(x, y)=$ const are hyperbolas $x y=$ const. The ma... | V_{x}=2x,\quadV_{y}=-2y | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,128 |
Example 2. The motion of a fluid is described by the complex potential $f(z)=\ln \operatorname{sh} \pi z$. Find the magnitude of the flow $N_{L}$ through the circle $2|z|=3$ and the circulation $\Gamma_{L}$ around it. | Solution. We find the derivative of the complex potential
$$
f^{\prime}(z)=\pi \operatorname{cth} \pi z
$$
Applying formula (2), we get
$$
\Gamma_{L}+i N_{L}=\pi \int_{|z|=3 / 2} \operatorname{cth} \pi z d z=\pi \int_{|z|=3 / 2} \frac{\operatorname{ch} \pi z}{\operatorname{sh} \pi z} d z
$$
The integrand has three ... | \Gamma_{L}=0,\quadN_{L}=6\pi^{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,129 |
Example 3. Find the complex potential $f(z)$ of the fluid flow, given the equation of equipotential lines
$$
\operatorname{ch} x \sin y + 2 x y = c
$$
where $c=$ const and $f(0)=0$. | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,130 | |
Example 1. Show that the function $\varphi(x)=\frac{1}{\left(1+x^{2}\right)^{3 / 2}}$ is a solution to the Volterra integral equation
$$
\varphi(x)=\frac{1}{1+x^{2}}-\int_{0}^{x} \frac{t}{1+x^{2}} \varphi(t) d t
$$ | Solution. Substituting the function $\frac{1}{\left(1+x^{2}\right)^{3 / 2}}$ for $\varphi(x)$ in the right-hand side of (4), we get
$$
\frac{1}{1+x^{2}}-\int_{0}^{x} \frac{t}{1+x^{2}} \frac{1}{\left(1+t^{2}\right)^{3 / 2}} d t=\frac{1}{1+x^{2}}-\left.\frac{1}{1+x^{2}}\left(-\frac{1}{\left(1+t^{2}\right)^{1 / 2}}\right... | proof | Calculus | proof | Yes | Yes | olympiads | false | 31,131 |
Example 1. Formulate the integral equation corresponding to the differential equation
$$
y^{\prime \prime}+x y^{\prime}+y=0
$$
and the initial conditions
$$
y(0)=1, \quad y^{\prime}(0)=0
$$ | Solution. Suppose
$$
\frac{d^{2} y}{d x^{2}}=\varphi(x)
$$
Then
$$
\frac{d y}{d x}=\int_{0}^{x} \varphi(t) d t+y^{\prime}(0)=\int_{0}^{x} \varphi(t) d t, \quad y=\int_{0}^{x}(x-t) \varphi(t) d t+1
$$
Substituting (9) and (10) into the given differential equation, we find
$$
\varphi(x)+\int_{0}^{x} x \varphi(t) d t... | \varphi(x)=-1-\int_{0}^{x}(2x-)\varphi() | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,132 |
Example 2. Solve the integral equation
$$
\varphi(x)=x+\int_{0}^{x} x t \varphi(t) d t
$$ | Solution. Rewrite equation (11) in the following form:
$$
\varphi(x)=x\left(1+\int_{0}^{x} t \varphi(t) d t\right)
$$
and set
$$
y(x)=1+\int_{0}^{x} t \varphi(t) d t
$$
Differentiate the last equality:
$$
y^{\prime}(x)=x \varphi(x)
$$
But since according to (12) and (13)
$$
\varphi(x)=x y(x),
$$
we obtain a dif... | \varphi(x)=xe^{x^{3}/3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,133 |
Example 1. Find the resolvent of the Volterra integral equation with the kernel $K(x, t) \equiv 1$. | Solution. We have $K_{1}(x, t)=K(x, t)=1$. Further, according to formulas (5)
\[
\begin{aligned}
& K_{2}(x, t)=\int_{t}^{x} K(x, z) K_{1}(z, t) d z=\int_{t}^{x} d z=x-t \\
& K_{3}(x, t)=\int_{t}^{x} 1 \cdot(z-t) d z=\frac{(x-t)^{2}}{2} \\
& K_{4}(x, t)=\int_{t}^{x} 1 \cdot \frac{(z-t)^{2}}{2} d z=\frac{(x-t)^{3}}{3!}
... | R(x,;\lambda)=e^{\lambda(x-)} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,134 |
Example 2. Find the resolvent of the integral equation
$$
\varphi(x)=f(x)+\int_{0}^{x}(x-t) \varphi(t) d t
$$ | Solution. In this case, $K(x, t)=x - t, \lambda=1$, hence, according to (8), $a_{1}(x)=1$, and all other $a_{k}(x)=0$.
Equation (9) in this case has the form
$$
\frac{d^{2} g(x, t ; 1)}{d x^{2}}-g(x, t ; 1)=0
$$
from which
$$
g(x, t ; 1)=g(x, t)=C_{1}(t) e^{x}+C_{2}(t) e^{-x}
$$
Conditions (10) give
$$
\left\{\be... | \operatorname{sh}(x-) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,135 |
Example 3. Using the resolvent, find the solution to the integral equation
$$
\varphi(x)=e^{x^{2}}+\int_{0}^{x} e^{x^{2}-t^{2}} \varphi(t) d t
$$ | Solution. The resolvent of the kernel $K(x, t)=\mathrm{e}^{x^{2}-t^{2}}$ for $\lambda=1$ is $R(x, t ; 1)=$ $e^{x-t} e^{\frac{x^{2}-t^{2}}{}}$ (see problem 26). According to formula (7), the solution of the given integral equation is the function
$$
\varphi(x)=e^{x^{2}}+\int_{0}^{x} e^{x-t} e^{x^{2}-t^{2}} e^{t^{2}} d ... | \varphi(x)=e^{x+x^{2}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,136 |
Example. Solve the integral equation
$$
\int_{0}^{x}(x-t) \varphi(t) d t=x^{2}
$$ | Solution. In this case, $\beta=1, \lambda=2$. Since $\lambda-\beta+k \neq 0$ ( $k=$ $0,1,2, \ldots, n)$, then by formula (13)
$$
\varphi(x)=\frac{\Gamma(3)}{\Gamma(2) \Gamma(1)} x^{2-1-1}=2
$$
$\Delta$
## Problems for Independent Solution
Solve the integral equations:
59. $\int_{0}^{x}(x-t)^{1 / 3} \varphi(t) d t=... | \varphi(x)=2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,137 |
Example 2. Show that the function $\varphi(x)=\sin \frac{\pi x}{2}$ is a solution to the Fredholm integral equation
$$
\varphi(x)-\frac{\pi^{2}}{4} \int_{0}^{1} K(x, t) \varphi(t) d t=\frac{x}{2}
$$
where the kernel is given by
$$
K(x, t)= \begin{cases}\frac{x(2-t)}{2}, & 0 \leqslant x \leqslant t \\ \frac{t(2-x)}{2... | Solution. The left-hand side of the equation can be written as
$$
\varphi(x)-\frac{\pi^{2}}{4} \int_{0}^{1} K(x, t) \varphi(t) d t=\varphi(x)-\frac{\pi^{2}}{4}\left\{\int_{0}^{x} K(x, t) \varphi(t) d t+\int_{x}^{1} K(x, t) \varphi(t) d t\right\}=
$$
$$
\begin{aligned}
& =\varphi(x)-\frac{\pi^{2}}{4}\left\{\int_{0}^{x... | proof | Calculus | proof | Yes | Yes | olympiads | false | 31,138 |
Example 1. Using Fredholm determinants, find the resolvent of the kernel $K(x, t)=x e^{t} ; a=0, b=1$. | Solution. We have $B_{0}(x, t)=x e^{t}$. Further,
$$
\begin{aligned}
& B_{1}(x, t)=\int_{0}^{1}\left|\begin{array}{ll}
x e^{t} & x e^{t_{1}} \\
t_{1} e^{t} & t_{1} e^{t_{1}}
\end{array}\right| d t_{1}=0 \\
& B_{2}(x, t)=\int_{0}^{1} \int_{0}^{1}\left|\begin{array}{lll}
x e^{t} & x e^{t_{1}} & x e^{t_{2}} \\
t_{1} e^{t... | R(x,;\lambda)=\frac{xe^{}}{1-\lambda} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,139 |
Example 2. Using formulas (8) and (9), find the resolvent of the kernel $K(x, t)=x-2 t$, where $0 \leqslant x \leqslant 1,0 \leqslant t \leqslant 1$. | Solution. We have $C_{0}=1, B_{0}(x, t)=x-2 t$. Using formula (9), we find
$$
C_{1}=\int_{0}^{1}(-s) d s=-\frac{1}{2}
$$
By formula (8), we get
$$
B_{1}(x, t)=-\frac{x-2 t}{2}-\int_{0}^{1}(x-2 s)(s-2 t) d s=-x-t+2 x t+\frac{2}{3}
$$
Further, we will have
$$
\begin{gathered}
C_{2}=\int_{0}^{1}\left(-2 s+2 s^{2}+\fr... | R(x,;\lambda)=\frac{x-2+(x+-2x-\frac{2}{3})\lambda}{1+\frac{\lambda}{2}+\frac{\lambda^{2}}{6}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,140 |
Example 1. Find the iterated kernels for the kernel $K(x, t)=x-t$, if $a=0, b=1$. | Solution. Using formulas (3), we find sequentially:
$$
\begin{aligned}
& K_{1}(x, t)=x-t, \\
& K_{2}(x, t)=\int_{0}^{1}(x-s)(s-t) d s=\frac{x+t}{2}-x t-\frac{1}{3} \\
& K_{3}(x, t)=\int_{0}^{1}(x-s)\left(\frac{s+t}{2}-s t-\frac{1}{3}\right) d s=-\frac{x-t}{12} \\
& K_{4}(x, t)=-\frac{1}{12} \int_{0}^{1}(x-s)(s-t) d s=... | \begin{aligned}& | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,141 |
Example 2. Find the iterated kernels $K_{1}(x, t)$ and $K_{2}(x, t)$, if $K(x, t)=e^{\min (x, t)}, a=0, b=1$. | Solution. By definition, we have
$$
\min \{x, t\}= \begin{cases}x, & \text { if } \quad 0 \leqslant x \leqslant t \\ t, & \text { if } t \leqslant x \leqslant 1\end{cases}
$$
therefore, the given kernel can be written as
={\begin{pmatrix}(2-)e^{x+}-\frac{1+e^{2x}}{2},&\text{if}0\leqslantx\leqslant\\(2-x)e^{x+}-\frac{1+e^{2}}{2},&\text{if}\leqslantx} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,142 |
Example 3. Find the iterated kernels $K_{1}(x, t)$ and $K_{2}(x, t)$, if $a=0$, $b=1$ n
$$
K(x, t)= \begin{cases}x+t, & \text { if } \quad 0 \leqslant x<t \\ x-t, & \text { if } t<x \leqslant 1\end{cases}
$$ | Solution. We have $K_{1}(x, t)=K(x, t)$,
$$
K_{2}(x, t)=\int_{0}^{1} K(x, s) K(s, t) d s
$$
where
$$
\begin{aligned}
& K(x, s)=\left\{\begin{array}{lll}
x+s, & \text { if } & 0 \leqslant x \leqslant s \\
x-s, & \text { if } & x > s
\end{aligned}\right.
\end{aligned}
$$
1) Let $x \leqslant s$. Then (see Fig. 2)
$$
... | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,143 |
Example 4. Find the resolvent for the kernel
$$
K(x, t)=x t+x^{2} t^{2}, \quad a=-1, \quad b=1
$$ | Solution. As shown above, the kernels $M(x, t)=x t$ and $N(x, t)=x^{2} t^{2}$ are orthogonal on $[-1,1]$ (see p. 41). Therefore, the resolvent of the kernel $K(x, t)$ is equal to the sum of the resolvents of the kernels $M(x, t)$ and $N(x, t)$. Using the results of problems 104 and 105, we find
$$
R_{K}(x, t ; \lambda... | R_{K}(x,;\lambda)=\frac{3x}{3-2\lambda}+\frac{5x^{2}^{2}}{5-2\lambda} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31,144 |
Example 1. Solve the integral equation
$$
\varphi(x)-\lambda \int_{-\pi}^{\pi}\left(x \cos t+t^{2} \sin x+\cos x \sin t\right) \varphi(t) d t=x
$$ | Solution. Let's write the equation in the following form:
$$
\varphi(x)=\lambda x \int_{-\pi}^{\pi} \varphi(t) \cos t d t+\lambda \sin x \int_{-\pi}^{\pi} t^{2} \varphi(t) d t+\lambda \cos x \int_{-\pi}^{\pi} \varphi(t) \sin t d t+x
$$
Introduce the notations:
$$
C_{1}=\int_{-\pi}^{\pi} \varphi(t) \cos t d t ; \quad... | \varphi(x)=\frac{2\lambda\pi}{1+2\lambda^{2}\pi^{2}}(\lambda\pix-4\lambda\pi\sinx+\cosx)+x | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,145 |
Example 1. Find the characteristic numbers and eigenfunctions of the integral equation
$$
\varphi(x)-\lambda \int_{0}^{\pi}\left(\cos ^{2} x \cos 2 t+\cos 3 x \cos ^{3} t\right) \varphi(t) d t=0
$$ | Solution. Imesm
$$
\varphi(x)=\lambda \cos ^{2} x \int_{0}^{\pi} \varphi(t) \cos 2 t d t+\lambda \cos 3 x \int_{0}^{\pi} \varphi(t) \cos ^{3} t d t
$$
Introducing the notations
$$
C_{1}=\int_{0}^{\pi} \varphi(t) \cos 2 t d t, \quad C_{2}=\int_{0}^{\pi} \varphi(t) \cos ^{3} t d t
$$
we will have
$$
\varphi(x)=C_{1}... | \lambda_{1}=\frac{4}{\pi},\quad\lambda_{2}=\frac{8}{\pi}\\\varphi_{1}(x)=\cos^{2}x,\quad\varphi_{2}(x)=\cos3x | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,146 |
Example 4. Find the characteristic numbers and eigenfunctions of the homogeneous equation
$$
\varphi(x)-\lambda \int_{0}^{\pi} K(x, t) \varphi(t) d t=0
$$
where
$$
K(x, t)= \begin{cases}\cos x \sin t, & 0 \leqslant x \leqslant t, \\ \cos t \sin x, & t \leqslant x \leqslant \pi\end{cases}
$$ | Solution. The given equation can be represented as
$$
\varphi(x)=\lambda \int_{0}^{x} K(x, t) \varphi(t) d t+\lambda \int_{z}^{\pi} K(x, t) \varphi(t) d t
$$
or
$$
\varphi(x)=\lambda \sin x \int_{0}^{x} \varphi(t) \cos t d t+\lambda \cos x \int_{x}^{\pi} \varphi(t) \sin t d t
$$
Differentiating both sides (15), we ... | \lambda_{n}=1-(n+\frac{1}{2})^{2},\quad\varphi_{n}(x)=\cos(n+\frac{1}{2})x | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,147 |
Example 5. Show that the integral equation with a non-symmetric kernel
$$
K(x, t)=\sin \pi x \cos \pi t, \quad 0 \leqslant x, \quad t \leqslant 1,
$$
has no characteristic numbers. | Solution. We will show that the equation
$$
\varphi(x)=\lambda \int_{0}^{1} K(x, t) \varphi(t) d t
$$
where the kernel is given by formula (21), has only the trivial solution $\varphi(x) \equiv 0$ $(\lambda \neq 0)$.
Indeed, rewrite equation (22) as
$$
\varphi(x)=\lambda \sin \pi x \int_{0}^{1} \cos \pi t \varphi(t... | proof | Calculus | proof | Yes | Yes | olympiads | false | 31,148 |
Example 6. Find the maximum
$$
|(K \varphi, \varphi)|=\left|\int_{0}^{\pi} \int_{0}^{\pi} K(x, t) \varphi(x) \varphi(t) d x d t\right|
$$
under the condition
$$
(\varphi, \varphi)=\int_{0}^{\pi} \varphi^{2}(x) d x=1
$$
if
$$
K(x, t)=\cos x \cos 2 t+\cos t \cos 2 x+1 .
$$ | Solution. Solving the homogeneous integral equation
$$
\varphi(x)=\lambda \int_{0}^{\pi}(\cos x \cos 2 t+\cos t \cos 2 x+1) \varphi(t) d t
$$
as an equation with a degenerate kernel, we find the characteristic numbers $\lambda_{1}=\frac{1}{\pi}$ and $\lambda_{2,3}= \pm \frac{2}{\pi}$ and the corresponding eigenfuncti... | 2\pi | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,149 |
Example. Solve the equation
$$
\varphi(x)-\lambda \int_{0}^{\pi}\left(\cos ^{2} x \cos 2 t+\cos ^{3} t \cos 3 x\right) \varphi(t) d t=0
$$ | Solution. The characteristic numbers of the given equation are $\lambda_{1}=\frac{4}{\pi}$, $\lambda_{2}=\frac{8}{\pi}$, and the corresponding eigenfunctions are
$$
\varphi_{1}(x)=\cos ^{2} x, \quad \varphi_{2}(x)=\cos 3 x
$$
The general solution of the equation is
$$
\begin{array}{ll}
\varphi(x)=C \cos ^{2} x, & \t... | \begin{pmatrix}\varphi(x)=C\cos^{2}x,&\text{if}\lambda=\frac{4}{\pi}\\\varphi(x)=C\cos3x,&\text{if}\lambda=\frac{8}{\pi}\\\varphi(x)=0,&\text{if}\lambda\neq\frac{4}{\pi},\lambda\ne | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,150 |
Example 1. Solve the equation
$$
\varphi(x)-\lambda \int_{0}^{1} K(x, t) \varphi(t) d t=x
$$
where
$$
K(x, t)= \begin{cases}x(t-1), & \text { if } 0 \leqslant x \leqslant t \\ t(x-1), & \text { if } t \leqslant x \leqslant 1\end{cases}
$$ | Solution. The eigenvalues and the corresponding eigenfunctions are given by
$$
\lambda_{n}=-\pi^{2} n^{2} ; \quad \varphi_{n}(x)=\sin \pi n x, \quad n=1,2, \ldots .
$$
If $\lambda \neq \lambda_{n}$, then the solution to equation (7) is
$$
\varphi(x)=x-\lambda \sum_{n=1}^{\infty} \frac{a_{n}}{\lambda+n^{2} \pi^{2}} \... | \varphi(x)=x-\frac{\lambda}{\pi}\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n(\lambda+n^{2}\pi^{2})}\sinn\pix | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,151 |
## Example 2. Solve the equation
$$
\varphi(x)-\lambda \int_{0}^{1} K(x, t) \varphi(t) d t=\cos \pi x
$$
where
$$
K(x, t)= \begin{cases}(x+1) t, & \text { if } \quad 0 \leqslant x \leqslant t \\ (t+1) x, & \text { if } t \leqslant x \leqslant 1\end{cases}
$$ | Solution. Characteristic numbers:
$$
\lambda_{0}=1, \quad \lambda_{n}=-n^{2} \pi^{2} \quad(n=1,2, \ldots)
$$
The corresponding eigenfunctions:
$$
\varphi_{0}(x)=e^{x}, \quad \varphi_{n}(x)=\sin n \pi x+n \pi \cos n \pi x \quad(n=1,2, \ldots)
$$
If $\lambda \neq 1$ and $\lambda \neq-n^{2} \pi^{2}$, then the solution... | \varphi(x)=\cos\pix+\lambda[\frac{1+e}{1+\pi^{2}}\frac{e^{x}}{\lambda-1}-\frac{\pi}{2(\lambda+\pi^{2})}(\sin\pix+\pi\cos\pix)] | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,152 |
## Example 1.
$$
\varphi(x)-\lambda \int_{0}^{1}\left(5 x^{2}-3\right) t^{2} \varphi(t) d t=e^{x}
$$ | Solution. We have
$$
\varphi(x)=C \lambda\left(5 x^{2}-3\right)+e^{x}
$$
and
$$
C=\int_{0}^{1} t^{2} \varphi(t) d t
$$
Substituting (6) into (7), we get
$$
C=C \lambda \int_{0}^{1}\left(5 t^{4}-3 t^{2}\right) d t+\int_{0}^{1} t^{2} e^{t} d t
$$
from which
$$
C=e-2 \text {. }
$$
This equation has a unique soluti... | \varphi(x)=\lambda(e-2)(5x^{2}-3)+e^{x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,154 |
## Example 2.
$$
\varphi(x)-\lambda \int_{0}^{1} \sin \ln x \varphi(t) d t=2 x
$$ | ## Solution. We have
$$
\varphi(x)=C \lambda \sin \ln x+2 x
$$
where $C=\int_{0}^{1} \varphi(t) d t$. Substituting the expression $\varphi(t)$ into the integral, we find
$$
C=C \lambda \int_{0}^{1} \sin \ln t d t+1
$$
from which
$$
C\left(1+\frac{\lambda}{2}\right)=1
$$
If $\lambda \neq-2$, then the given equatio... | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,155 | |
Example 3.
$$
\varphi(x)-\lambda \int_{0}^{\pi} \cos (x+t) \varphi(t) d t=\cos 3 x
$$ | Solution. Rewrite the equation in the form
$$
\varphi(x)-\lambda \int_{0}^{\pi}(\cos x \cos t-\sin x \sin t) \varphi(t) d t=\cos 3 x
$$
From this, we have
$$
\varphi(x)=C_{1} \lambda \cos x-C_{2} \lambda \sin x+\cos 3 x
$$
where
$$
C_{1}=\int_{0}^{\pi} \varphi(t) \cos t d t, \quad C_{2}=\int_{0}^{\pi} \varphi(t) \... | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,156 |
Example 4. For what values of the parameters $\alpha$ and $\beta$ is the integral equation
$$
\varphi(x)=\lambda \int_{0}^{1} x t^{2} \varphi(t) d t+\alpha x+\beta ?
$$
solvable? | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,157 | |
Example 1. Construct the Green's function for the homogeneous boundary value problem
\[
\begin{gathered}
y^{\mathrm{IV}}(x)=0 \\
\left\{\begin{array}{l}
y(0)=y^{\prime}(0)=0 \\
y(1)=y^{\prime}(1)=0
\end{array}\right.
\end{gathered}
\] | Solution. First, we will show that the boundary value problem (15)-(16) has only a trivial solution. Indeed, the fundamental system of solutions for equation (15) is
$$
y_{1}(x)=1, \quad y_{2}(x)=x, \quad y_{3}(x)=x^{2}, \quad y_{4}(x)=x^{3}
$$
so that its general solution has the form
$$
y(x)=A+B x+C x^{2}+D x^{3}
... | G(x,\xi)=(\frac{1}{2}x-x^{2}+\frac{1}{2}x^{3})\xi^{2}-(\frac{1}{6}-\frac{1}{2}x^{2}+\frac{1}{3}x^{3})\xi^{3}\quad\text{for}\quad\xi\leqslantx\le | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,158 |
Example 2. Construct the Green's function for the differential equation
$$
x y^{\prime \prime}+y^{\prime}=0
$$
subject to the following conditions:
$$
y(x) \text { is bounded as } x \rightarrow 0, \quad y(1)=\alpha y^{\prime}(1), \quad \alpha \neq 0 .
$$ | Solution. First, we find the general solution of equation (23) and verify that conditions (24) are satisfied only when
$$
y(x) \equiv 0
$$
Indeed, denoting $y^{\prime}(x)=z(x)$, we get $x z^{\prime}+z=0$, from which $\ln z= \ln c_{1}-\ln x, z=\frac{c_{1}}{x}$, and thus,
$$
y(x)=c_{1} \ln x+c_{2}
$$
It is clear that... | G(x,\xi)=\begin{cases}\alpha+\ln\xi,&0<x\leqslant\xi\\\alpha+\lnx,&\xi\leqslantx\leqslant1\end{cases} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,159 |
Example 3. Find the Green's function for the boundary value problem
$$
y^{\prime \prime}(x)+k^{2} y=0, \quad y(0)=y(1)=0
$$ | Solution. It is easy to verify that the solution $y_{1}(x)=\sin k x$ satisfies the boundary condition $y_{1}(0)=0$, and the solution $y_{2}(x)=\sin k(x-1)$ satisfies the condition $y_{2}(1)=0$, and that they are linearly independent. Let's find the value of the Wronskian determinant for $\sin k x$ and $\sin k(x-1)$ at ... | G(x,\xi)=\begin{cases}\frac{\sink(\xi-1)\sinkx}{k\sink},&0\leqslantx\leqslant\xi\\\frac{\sink\xi\cdot\sink(x-1)}{k\sink},&\xi\leqslantx\leqslant1\end{cases} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,160 |
Example 4. Find the influence function $G(x, y)$ for a beam supported at both ends $x=0$ and $x=1$. (Here $G(x, y)$ is the displacement parallel to the $O z$ axis of the cross-section at the point $x=y$, caused by the action of a unit load concentrated at the point $x=y$ and acting parallel to the $O z$ axis. | Solution. Let $R_{0}$ and $R_{1}$ be the unknown reactions at the support points caused by the action of a unit load at point $x=y$ (Fig. 5). Then the bending moment $M$ at point $x$ of the beam will be
$$
M=\left\{\begin{array}{lll}
-R_{0} x, & \text { if } & 0 \leqslant x \leqslant y, \\
-R_{1}(1-x), & \text { if } ... | G(x,y)=\int_{0}^{1}M(x,z)M(z,y)F(z) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,161 |
Example 1. Using the Green's function, solve the boundary value problem
$$
\begin{gathered}
y^{\prime \prime}(x)-y(x)=x \\
y(0)=y(1)=0 .
\end{gathered}
$$ | Solution. a) First, let's determine whether the Green's function exists for the corresponding homogeneous boundary value problem
$$
\begin{gathered}
r y^{\prime \prime}(x)-y(x)=0 \\
y(0)=y(1)=0
\end{gathered}
$$
Obviously, $y_{1}(x)=e^{x}, y_{2}(x)=e^{-x}$ is a fundamental system of solutions for equation (6). Theref... | y(x)=\frac{\operatorname{sh}x}{\operatorname{sh}1}-x | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,162 |
Example 2. Reduce the boundary value problem for a nonlinear differential equation to an integral equation:
$$
\begin{aligned}
& y^{\prime \prime}=f(x, y(x)) \\
& y(0)=y(1)=0
\end{aligned}
$$ | Solution. Constructing the Green's function for the problem
$$
\begin{gathered}
y^{\prime \prime}=0, \\
y(0)=y(1)=0,
\end{gathered}
$$
we find
$$
G(x, \xi)= \begin{cases}(\xi-1) x, & 0 \leqslant x \leqslant \xi \\ (x-1) \xi, & \xi \leqslant x \leqslant 1\end{cases}
$$
Considering the right-hand side of equation (10... | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,163 |
Example 1. Reduce the boundary value problem
\[
\begin{gathered}
y^{\prime \prime}+\lambda y=x \\
y(0)=y\left(\frac{\pi}{2}\right)=0
\end{gathered}
\]
to an integral equation. | Solution. First, we find the Green's function $G(x, \xi)$ for the corresponding homogeneous problem
\[
\begin{gathered}
y^{\prime \prime}(x)=0 \\
y(0)=y\left(\frac{\pi}{2}\right)=0
\end{gathered}
\]
Since the linearly independent solutions of the equation $y^{\prime \prime}(x)=0$, satisfying the conditions $y(0)=0$ a... | y(x)+\lambda\int_{0}^{\pi/2}G(x,\xi)y(\xi)\xi=\frac{1}{6}x^{3}-\frac{\pi^{2}}{24}x | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,164 |
Example 1. Solve the integral equation
$$
\varphi(x)=f(x)+\lambda \int_{-\infty}^{+\infty} e^{-|x-t|} \varphi(t) d t \quad\left(\lambda<\frac{1}{2}\right)
$$ | Solution. Let $F(\omega)$ be the Fourier transform of the function $f(x)$, and $\tilde{K}(\omega)$ be the Fourier transform of the kernel $K(x)=e^{-|x|}$. Here,
$$
\begin{aligned}
\tilde{K}(\omega)=\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{+\infty} e^{-|x|} e^{-i z \omega} d x & =\frac{1}{\sqrt{2 \pi}}\left[\int_{-\infty... | \varphi(x)=\frac{e^{-\sqrt{1-2\lambda}|x|}}{\sqrt{1-2\lambda}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,165 |
Example 3. Solve the integral equation:
$$
\int_{0}^{+\infty} \varphi(t) \sin x t \, d t=e^{-x} \quad(x>0)
$$ | Solution. The function $\sqrt{\frac{2}{\pi}} \mathrm{e}^{-z}$ is obviously the sine Fourier transform of the desired function $\varphi(t)$. Applying the inversion formula (6) of the sine Fourier transform, we will have
$$
\varphi(t)=\sqrt{\frac{2}{\pi}} \int_{0}^{+\infty} \sqrt{\frac{2}{\pi}} e^{-x} \sin x t d x=\frac... | \varphi()=\frac{2}{\pi}\frac{}{1+^{2}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,167 |
Example 4. In problems concerning the oscillations of a thin elastic plate, we arrive at the following integral equation:
$$
\psi(t)=\frac{1}{2 b t} \int_{0}^{+\infty} x f(x) \sin \frac{x^{2}}{4 b t} d x
$$
where $f(x)$ is the unknown function, and $\psi(t)$ is the known function. | Solution. Equation (8) is an integral equation of the first kind. Setting
$$
t=\frac{1}{4 b a}, \quad x^{2}=v
$$
we transform equation (8) into the form
$$
\frac{1}{\alpha} \psi\left(\frac{1}{4 b a}\right)=\int_{0}^{+\infty} f(\sqrt{v}) \sin \alpha v d v
$$
Using the inversion formula for the sine Fourier transform... | f(x)=\frac{2}{\pi}\int_{0}^{+\infty}\frac{\psi()}{}\sin\frac{x^{2}}{4} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,168 |
Example 1. Solve the integral equation
$$
\varphi(x)=\sin x+2 \int_{0}^{x} \cos (x-t) \varphi(t) d t
$$ | Solution. It is known that
$$
\sin x \risingdotseq \frac{1}{p^{2}+1}, \quad \cos x \risingdotseq \frac{p}{p^{2}+1}
$$
Let $\varphi(x) \risingdotseq \Phi(p)$. Applying the Laplace transform to both sides of the equation and using the convolution theorem, we get
$$
\Phi(p)=\frac{1}{p^{2}+1}+\frac{2 p}{p^{2}+1} \Phi(p)... | \varphi(x)=xe^{x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,169 |
Example 2. Solve the integral equation:
$$
\int_{0}^{x} \varphi(t) \varphi(x-t) d t=\frac{x^{3}}{6}
$$ | Solution. Let $\varphi(x) \risingdotseq \Phi(p)$. Applying the Laplace transform to both sides of (4), we get
$$
\Phi^{2}(p)=\frac{1}{p^{4}}
$$
from which
$$
\Phi(p)= \pm \frac{1}{p^{2}}
$$
The functions $\varphi_{1}(x)=x, \varphi_{2}(x)=-x$ are solutions to equation (4) (the solution to equation (4) is not unique)... | \varphi_{1}(x)=x,\varphi_{2}(x)=-x | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,170 |
Example 3. Solve the system of integral equations
$$
\left\{\begin{array}{l}
\varphi_{1}(x)=1-2 \int_{0}^{x} e^{2(x-t)} \varphi_{1}(t) d t+\int_{0}^{x} \varphi_{2}(t) d t \\
\varphi_{2}(x)=4 x-\int_{0}^{x} \varphi_{1}(t) d t+4 \int_{0}^{x}(x-t) \varphi_{2}(t) d t
\end{array}\right.
$$ | Solution. Transitioning to images and using the convolution theorem, we obtain
$$
\left\{\begin{array}{l}
\Phi_{1}(p)=\frac{1}{p}-\frac{2}{p-2} \Phi_{1}(p)+\frac{1}{p} \Phi_{2}(p) \\
\Phi_{2}(p)=\frac{4}{p^{2}}-\frac{1}{p} \Phi_{1}(p)+\frac{4}{p^{2}} \Phi_{2}(p)
\end{array}\right.
$$
Solving the obtained system with ... | \begin{aligned}\varphi_{1}(x)&=e^{-x}-xe^{-x}\\\varphi_{2}(x)&=\frac{8}{9}e^{2x}+\frac{1}{3}xe^{-x}-\frac{8}{9}e^{-x}\end{aligned} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,171 |
Example 4. Solve the integro-differential equation
\[
\begin{gathered}
\varphi^{\prime \prime}(x)+\int_{0}^{x} e^{2(x-t)} \varphi^{\prime}(t) d t=e^{2 x} \\
\varphi(0)=\varphi^{\prime}(0)=0
\end{gathered}
\] | Solution. Let $\varphi(x) \equiv \Phi(p)$. By (11)
$$
\varphi^{\prime}(x) \risingdotseq p \Phi(p), \quad \varphi^{\prime \prime}(x) \risingdotseq p^{2} \Phi(p)
$$
Therefore, after applying the Laplace transform, equation (10) will take the form
$$
p^{2} \Phi(p)+\frac{p}{p-2} \Phi(p)=\frac{1}{p-2}
$$
or
$$
\Phi(p) ... | \varphi(x)=xe^{x}-e^{x}+1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,172 |
Example 5. Solve the integral equation
$$
\varphi(x)=x+\int_{x}^{\infty} \mathrm{e}^{2(x-t)} \varphi(t) d t
$$ | Solution. In this case, $f(x)=x, K(x)=e^{2 x}$. Therefore,
$$
F(p)=\frac{1}{p^{2}}, \quad \overline{\mathscr{K}}(-p)=\int_{0}^{\infty} e^{-2 x} e^{p x} d x=\frac{1}{2-p}, \quad \operatorname{Re} p>1$, which is related to the inclusion or exclusion in the solution of equation (16) of the solution of the corresponding h... | \varphi(x)=2x+1+Ce^{x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,173 |
Example 6. Solve the integral equation
$$
\frac{1}{\sqrt{\pi x}} \int_{0}^{\infty} e^{-t^{2} /(4 x)} \varphi(t) d t=1
$$ | Solution. Let $\varphi(x) \risingdotseq \Phi(p)$. Applying the Laplace transform to both sides of (19), we get, according to formula (18),
$$
\frac{\Phi(\sqrt{p})}{\sqrt{p}}=\frac{1}{p}
$$
from which
$$
\frac{\Phi(p)}{p}=\frac{1}{p^{2}}, \quad \text { or } \quad \Phi(p)=\frac{1}{p} \risingdotseq 1
$$
Therefore, $\v... | \varphi(x)\equiv1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,174 |
Example 7. Solve the integral equation
$$
\varphi(x)=x e^{-x}+\lambda \int_{0}^{\infty} J_{0}(2 \sqrt{x t}) \varphi(t) d t \quad(|\lambda| \neq 1)
$$ | Solution. Let $\varphi(x) \risingdotseq \Phi(p)$. Applying the Laplace transform to both sides of (20) and taking into account the Efros theorem, we find
$$
\Phi(p)=\frac{1}{(p+1)^{2}}+\lambda \frac{1}{p} \Phi\left(\frac{1}{p}\right)
$$
Replacing $p$ with $\frac{1}{p}$, we get
$$
\Phi\left(\frac{1}{p}\right)=\frac{p... | \varphi(x)=e^{-x}(\frac{x}{1+\lambda}+\frac{\lambda}{1-\lambda^{2}}) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,175 |
## Example. Solve the integral equation
$$
\int_{0}^{x} \cos (x-t) \varphi(t) d t=x
$$ | Solution. The functions $f(x)=x, K(x, t)=\cos (x-t)$ satisfy the above formulated conditions of continuity and differentiability.
Differentiating both sides of (4) with respect to $x$, we get
$$
\varphi(x) \cos 0-\int_{0}^{x} \sin (x-t) \varphi(t) d t=1
$$
or
$$
\varphi(x)=1+\int_{0}^{x} \sin (x-t) \varphi(t) d t
$... | \varphi(x)=1+\frac{x^{2}}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,177 |
Example. Solve the integral equation
$$
\int_{0}^{x} e^{x-t} \varphi(t) d t=x
$$ | Solution. Applying the Laplace transform to both sides of (3), we get
$$
\frac{1}{p-1} \Phi(p)=\frac{1}{p^{2}}
$$
from which
$$
\Phi(p)=\frac{p-1}{p^{2}}=\frac{1}{p}-\frac{1}{p^{2}} \equiv 1-x
$$
The function $\varphi(x)=1-x$ is the solution to equation (3).
## Problems for Independent Solution
Solve the integral... | \varphi(x)=1-x | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,178 |
Example 1. Solve the Fredholm integral equation of the first kind
$$
\int_{0}^{1} K(x, t) \varphi(t) d t=\sin ^{3} \pi x
$$
where
$$
K(x, t)= \begin{cases}(1-x) t, & 0 \leqslant t \leqslant x \\ x(1-t), & x \leqslant t \leqslant 1\end{cases}
$$ | Solution. The characteristic numbers of the kernel (7)
$$
\lambda_{1}=\pi^{2}, \quad \lambda_{2}=(2 \pi)^{2}, \quad \ldots, \quad \lambda_{n}=(n \pi)^{2}, \quad \ldots
$$
and the corresponding eigenfunctions
$$
\varphi_{1}(x)=\sqrt{2} \sin \pi x, \quad \varphi_{2}(x)=\sqrt{2} \sin 2 \pi x, \quad \ldots, \quad \varph... | \varphi(x)=\frac{3\pi^{2}}{4}(\sin\pix-3\sin3\pix) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,179 |
## Example 2.
$$
\int_{0}^{1} t \varphi(t) d t=\frac{1}{3}
$$ | Solution. Applying the method of finding characteristic numbers and eigenfunctions described in $\$ 10$, we find that the characteristic number of the given kernel is $\lambda=2$, and the corresponding eigenfunction is $\psi(x)=1$.
It is clear that the "system" of eigenfunctions consisting of only one function $\psi(x... | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,180 |
Example 3. Solve the integral equation
$$
\int_{-1}^{1} \frac{\varphi(t) d t}{\sqrt{1+x^{2}-2 x t}}=x+1
$$ | ## Solution. Function
$$
G(x, t)=\frac{1}{\sqrt{1+x^{2}-2 x t}}
$$
is the generating function for the Legendre polynomials $P_{n}(t)$:
$$
G(x, t)=\sum_{n=0}^{\infty} P_{n}(t) x^{n}
$$
We seek the solution of equation (16) in the form
$$
\varphi(x)=\sum_{i=0}^{\infty} a_{i} P_{i}(x)
$$
Substituting (17) and (18) i... | \varphi(x)=\frac{1}{2}+\frac{3}{2}x | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,181 |
Example 4. Solve the integral equation
$$
\int_{0}^{\infty} \frac{e^{-x t /(1-x)}}{1-x} e^{-t} \varphi(t) d t=1-x, \quad|x|<1
$$ | Solution. The function
$$
G(x, t)=\frac{e^{-x t /(1-x)}}{1-x}
$$
is the generating function for the Chebyshev-Laguerre polynomials $L_{n}(t)$:
$$
G(x, t)=\sum_{n=0}^{\infty} L_{n}(t) x^{n}
$$
We seek the solution of equation (19) in the form
$$
\varphi(t)=\sum_{k=0}^{\infty} a_{k} L_{k}(t)
$$
Substituting (20) an... | \varphi()= | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,182 |
Example 5. Solve the equation
$$
x^{2}=\frac{2}{\pi} \int_{0}^{\pi / 2} \varphi(x \sin \theta) d \theta
$$ | Solution. Equation (34) is a Schlömilch equation, where $f(x)=x^{2}$, and thus, $f(0)=0$. We find the derivative: $f^{\prime}(x)=2 x$. Applying formula (33), we find
$$
\varphi(x)=x \int_{0}^{\pi / 2} 2(x \sin \psi) d \psi=-\left.2 x^{2} \cos \psi\right|_{\psi=0} ^{\phi=\pi / 2}=2 x^{2}
$$
Answer: $\varphi(x)=2 x^{2}... | \varphi(x)=2x^{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,183 |
Example. Solve the equation
$$
\varphi(x)=\sin x+\int_{0}^{1}(1-x \cos x t) \varphi(t) d t
$$
by replacing its kernel with a degenerate one. | Solution. Expanding the kernel $K(x, t)=1-x \cos x t$ into a series, we obtain
$$
K(x, t)=1-x+\frac{x^{3} t^{2}}{2}-\frac{x^{3} t^{4}}{24}+\ldots
$$
Taking the first three terms of the expansion (3) as the degenerate kernel $L(x, t)$:
$$
L(x, t)=1-x+\frac{x^{3} t^{2}}{2}
$$
we will solve the new equation
$$
\widet... | \tilde{\varphi}(x)=1.0031(1-x)+0.1674x^{3}+\sinx | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,184 |
Example. Find an approximate solution to the integral equation
$$
\varphi(x)+\int_{0}^{1} x\left(e^{x t}-1\right) \varphi(t) d t=e^{x}-x
$$ | Solution. Let's take three points on the interval $[0,1]$: $x_{1}=0, x_{2}=0.5, x_{3}=1$ and substitute $x=0, x=0.5, x=1$ into equation (5). Then we get respectively
$$
\left\{\begin{array}{l}
\varphi(0)=1 \\
\varphi(0.5)+0.5 \int_{0}^{1}\left(e^{0.5 t}-1\right) \varphi(t) d t=e^{0.5}-0.5, \\
\varphi(1)+\int_{0}^{1}\l... | \varphi(x)\equiv1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,185 |
Example 1. Solve the integral equation by the method of successive approximations
$$
\varphi(x)=1+\int_{0}^{x} \varphi(t) d t
$$
taking $\varphi_{0}(x) \equiv 0$. | Solution. Since $\varphi_{0}(x) \equiv 0$, then $\varphi_{1}(x)=1$. Further,
\[
\begin{aligned}
& \varphi_{2}(x)=1+\int_{0}^{x} 1 \cdot d t=1+x \\
& \varphi_{3}(x)=1+\int_{0}^{x}(1+t) d t=1+x+\frac{x^{2}}{2} \\
& \varphi_{4}(x)=1+\int_{0}^{x}\left(1+t+\frac{t^{2}}{2}\right) d t=1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}
\e... | \varphi(x)=e^x | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,186 |
Example 2. Solve the integral equation by the method of successive approximations
$$
\varphi(x)=\int_{0}^{x} \frac{1+\varphi^{2}(t)}{1+t^{2}} d t
$$
taking as the zeroth approximation: 1) $\varphi_{0}(x)=0$; 2) $\varphi_{0}(x)=x$. | Solution. 1) Let $\varphi_{0}(x)=0$. Then
\[
\begin{aligned}
& \varphi_{1}(x)=\int_{0}^{x} \frac{d t}{1+t^{2}}=\operatorname{arctg} x \\
& \varphi_{2}(x)=\int_{0}^{x} \frac{1+\operatorname{arctg}^{2} t}{1+t^{2}} d t=\operatorname{arctg} x+\frac{1}{3} \operatorname{arctg}^{3} x \\
& \varphi_{3}(x)=\int_{0}^{x} \frac{1+... | \varphi(x)=x | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,187 |
Example 3. Solve the equation by the method of successive approximations
$$
\varphi(x)=\int_{0}^{1} x t^{2} \varphi(t) d t+1
$$
and estimate the error of the approximate solution. | Solution. As the zero approximation, we take $\varphi_{0}(x) \equiv 1$. Then
$$
\begin{aligned}
& \varphi_{1}(x)=\int_{0}^{1} x t^{2} \cdot 1 d t+1=1+\frac{x}{3} \\
& \varphi_{2}(x)=\int_{0}^{1} x t^{2}\left(1+\frac{t}{3}\right) d t+1=1+\frac{x}{3}\left(1+\frac{1}{4}\right) \\
& \varphi_{3}(x)=\int_{0}^{1} x t^{2}\lef... | \varphi(x)=1+\frac{4}{9}x | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,188 |
Example. Solve the equation by the Bubnov-Galerkin method
$$
\varphi(x)=x+\int_{-1}^{1} x t \varphi(t) d t
$$ | Solution. As a complete system of functions on $[-1,1]$, we choose the system of Legendre polynomials $P_{n}(x)(n=0,1,2, \ldots)$. We will seek the approximate solution $\varphi_{n}(x)$ of equation (4) in the form
$$
\varphi_{3}(x)=a_{1} \cdot 1+a_{2} x+a_{3} \frac{3 x^{2}-1}{2}
$$
Substituting $\varphi_{3}(x)$ for $... | \varphi_{3}(x)=3x | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,189 |
Example 1. Find the approximate value of the smallest characteristic number of the kernel by the Ritz method
$$
K(x, t)=x t ; \quad a=0, b=1
$$ | Solution. As the coordinate system of functions $\psi_{n}(x)$, we choose the system of Legendre polynomials: $\psi_{n}(x)=P_{n}(2 x-1)$. In formula (1), we limit ourselves to two terms, so that
$$
\varphi_{2}(x)=a_{1} \cdot P_{0}(2 x-1)+a_{2} \cdot P_{1}(2 x-1) .
$$
Noting that
$$
\psi_{1} \equiv P_{0}(2 x-1)=1 ; \q... | 3 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,190 |
Example 2. Find the first characteristic number of the kernel by the method of traces
$$
K(x, t)=\left\{\begin{array}{ll}
t, & x \geqslant t, \\
x, & x \leqslant t,
\end{array} \quad a=0, b=1\right.
$$ | Solution. Since the kernel $K(x, t)$ is symmetric, it is sufficient to find $K_{2}(x, t)$ only for $t<x$.
We have
$$
\begin{aligned}
& K_{2}(x, t)=\int_{0}^{1} K(x, z) K(z, t) d z= \\
& =\int_{0}^{t} z^{2} d z+\int_{t}^{x} z t d z+\int_{x}^{1} x t d z=x t-\frac{x^{2} t}{2}-\frac{t^{3}}{6}
\end{aligned}
$$
Next, usin... | 2.48 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,191 |
Example 3. Using the Kellogg method, calculate the smallest characteristic number of the kernel $K(x, t)=x^{2} t^{2}, 0 \leqslant x, t \leqslant 1$. | Solution. Let $\omega(x)=x$. Then
$$
\begin{aligned}
& \omega_{1}(x)=\int_{0}^{1} x^{2} t^{2} t d t=\frac{x^{2}}{4} \\
& \omega_{2}(x)=\int_{0}^{1} x^{2} t^{4} \frac{1}{4} d t=\frac{1}{4} x^{2} \cdot \frac{1}{5} \\
& \omega_{3}(x)=\int_{0}^{1} \frac{1}{4 \cdot 5} x^{2} t^{4} d t=\frac{1}{4 \cdot 5^{2}} x^{2} \\
& \ldo... | 5 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,192 |
Example 4. Find the critical force for a rod having the shape of a truncated cone with base radii $r_{0}$ and $r_{1}=r_{0}(1+q), q>0$ (Fig. 8).
Fig. 8 | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,193 | |
Example 1. Find $\int(2 x-5)^{23} d x$. | Solution. From formula 2 of the table, taking into account $u=2 x-5$, it follows that
$$
\int(2 x-5)^{23} d x=\frac{1}{2} \cdot \frac{(2 x-5)^{24}}{24}+C=\frac{(2 x-5)^{24}}{48}+C
$$ | \frac{(2x-5)^{24}}{48}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,194 |
Example 2. Find $\int 3^{7 x-1 / 9} d x$. | Solution. From formula 4 of the table, when $u=7 x-1 / 9$, we get
$$
\int 3^{7 x-1 / 9} d x=\frac{1}{7 \cdot \ln 3} 3^{7 x-1 / 9}+C
$$
Note. In the future, to ensure the continuity of integration, auxiliary transformations, notations, and remarks will be enclosed in curly braces throughout the solution. | \frac{1}{7\cdot\ln3}3^{7x-1/9}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,195 |
Example 3. Find $\int \frac{d x}{\sqrt{4+x+x^{2}}}$. | Solution. In the expression under the root, we will complete the square to apply formula 13 for \(u=x+1/2\).
\[
\begin{gathered}
\int \frac{d x}{\sqrt{4+x+x^{2}}}=\int \frac{d x}{\sqrt{\frac{15}{4}+\left(\frac{1}{4}+x+x^{2}\right)}}=\int \frac{d x}{\sqrt{\left(x+\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{15}}{2}\right)^... | \frac{1}{\sqrt{2}}\arcsin\frac{\sqrt{2}(x-1)}{\sqrt{5}}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,196 |
Example 6. Find $\int \frac{d x}{2 x^{2}-x-1}$ | Solution. The form of the denominator of the fraction (it has two real roots) suggests the use of formula 14. Therefore,
$$
\begin{aligned}
& \int \frac{d x}{2 x^{2}-x-1}=\int \frac{d x}{2\left(x^{2}-\frac{1}{2} x-\frac{1}{2}\right)}=\frac{1}{2} \int \frac{d x}{\left(x-\frac{1}{4}\right)^{2}-\frac{9}{16}}= \\
& \quad=... | \frac{1}{3}\ln|\frac{2(x-1)}{2x+1}|+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,197 |
Example 7. Find $\int\left(\frac{3}{\sqrt{x}}+1\right)^{2} \cdot x d x$. | Solution. In the integrand, we will square, expand the brackets, and apply the linearity property. We sequentially obtain
$$
\begin{aligned}
& \int\left(\frac{3}{\sqrt{x}}+1\right)^{2} \cdot x d x=\int\left(\frac{9}{x}+\frac{6}{\sqrt{x}}+1\right) x d x= \\
& =9 \int d x+6 \int \sqrt{x} d x+\int x d x= \\
& =\left\{\be... | 9x+4\sqrt{x^{3}}+\frac{x^{2}}{2}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,198 |
Example 8. Find $\int\left(x^{2}+2\right)(\sqrt{x}-3) d x$ | Solution. We will expand the brackets and apply the linearity property, i.e., integrate term by term, factoring out numerical coefficients from the integral. In the intermediate integrals, we will use fractional exponents, and express the answer in radicals (roots). We have:
$$
\begin{aligned}
& \int\left(x^{2}+2\righ... | \frac{2}{7}\sqrt{x^{7}}-x^{3}+\frac{4}{3}\sqrt{x^{3}}-6x+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,199 |
Example 11. Find $\int \operatorname{tg}^{2} x d x$. | Solution. Using the formula $1+\operatorname{tg}^{2} x=\frac{1}{\cos ^{2} x}$, we obtain the standard integrals:
$$
\int \operatorname{tg}^{2} x d x=\int\left(\frac{1}{\cos ^{2} x}-1\right) d x=\operatorname{tg} x-x+C
$$ | \operatorname{tg}x-x+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,200 |
Example 12. Find $\int\left(2^{3 x}-1\right)^{2} \cdot 4^{x} d x$. | $$
\begin{aligned}
& \int\left(2^{3 x}-1\right)^{2} \cdot 4^{x} d x=\int\left(2^{6 x}-2 \cdot 2^{3 x}+1\right) \cdot 2^{2 x} d x= \\
&=\int\left(2^{8 x}-2^{5 x+1}+2^{2 x}\right) d x=\frac{1}{8} \cdot \frac{2^{8 x}}{\ln 2}-\frac{1}{5} \cdot \frac{2^{5 x+1}}{\ln 2}+\frac{1}{2} \cdot \frac{2^{2 x}}{\ln 2}+C= \\
&=\left(2^... | (2^{8x-3}-\frac{1}{5}\cdot2^{5x+1}+2^{2x-1})\cdot\frac{1}{\ln2}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 31,201 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.