problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
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class | __index_level_0__ int64 0 742k |
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285. $\int_{0}^{1}\left(4 x^{3}+1\right)^{5} x^{2} d x$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
285. $\int_{0}^{1}\left(4 x^{3}+1\right)^{5} x^{2} d x$. | Solution. $\int_{0}^{1}\left(4 x^{3}+1\right)^{5} x^{2} d x=\left|\begin{array}{c}t=4 x^{3}+1, t_{1}=4 \cdot 0^{3}+1=1, \\ d t=12 x^{2} d x, t_{2}=4 \cdot 1^{3}+1=5 \\ x^{2} d x=\frac{1}{12} d t\end{array}\right|=$
$$
=\frac{1}{12} \int_{1}^{5} t^{5} d t=\left.\frac{1}{12} \cdot \frac{t^{6}}{6}\right|_{1} ^{5}=\frac{1... | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,175 | |
291. $\int_{0}^{1}\left(e^{x}-1\right)^{4} e^{x} d x$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
291. $\int_{0}^{1}\left(e^{x}-1\right)^{4} e^{x} d x$. | Solution. $\int_{0}^{1}\left(e^{x}-1\right)^{4} e^{x} d x=\left|\begin{array}{l}e^{x}-1=t, t_{1}=0, \\ e^{x} d x=d t ; t_{2}=e-1\end{array}\right|=\int_{0}^{e-1} t^{4} d t=\left.\frac{1}{5} t^{5}\right|_{0} ^{e-1}=$ $=\frac{1}{5}(e-1)^{5}$. | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,176 | |
294. $\int_{1}^{e} \frac{3 \ln ^{2} x d x}{x}$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
294. $\int_{1}^{e} \frac{3 \ln ^{2} x \, dx}{x}$. | Solution. $\int_{1}^{e} \frac{3 \ln ^{2} x d x}{x}=\left|\begin{array}{l}\ln x=t, t_{1}=\ln 1=0, \\ \frac{d x}{x}=d t ; t_{2}=\ln e=1\end{array}\right|=3 \int_{0}^{1} t^{2} d t=\left.t^{3}\right|_{0} ^{1}=1$. | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,177 |
299. $\int_{\pi / 4}^{\pi / 2} \sin ^{3} z \cos z d z$
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
299. $\int_{\pi / 4}^{\pi / 2} \sin ^{3} z \cos z d z$ | Solution. $\int_{\pi / 4}^{\pi / 2} \sin ^{3} z \cos z d z=\left|\begin{array}{l}\sin z=u, u_{1}=\sqrt{2} / 2, \\ \cos z d z=d u ; u_{2}=1\end{array}\right|=\int_{\sqrt{2} / 2}^{1} u^{3} d u=$ $=\left.\frac{u^{4}}{4}\right|_{\sqrt{2} / 2} ^{1}=\frac{1}{4}-\frac{1}{16}=\frac{3}{16}$. | Calculus | proof | Yes | Yes | olympiads | false | 32,178 | |
307. $y^{2}=9 x, x=16, x=25$ and $y=0$ (Fig. 150). | Solution. For any $x \in [16,25]$, the function $y=\sqrt{9 x}$ takes positive values; therefore, to calculate the area of the given curvilinear trapezoid, we should use formula (1):
$$
S=\int_{16}^{25} \sqrt{9} x d x=\int_{16}^{25} 3 x^{1 / 2} d x=\left.3 \frac{x^{3 / 2}}{3 / 2}\right|_{16} ^{25}=\left.2 x \sqrt{x}\ri... | 122 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,179 |
312. $x y=6$ and $x+y-7=0$. | Solution. Let's draw the figure whose area we need to determine (Fig. 157). We will find the abscissas of the points of intersection of the equilateral hyperbola and the line by solving the system of equations $\left\{\begin{array}{l}x y=6, \\ x+y-7=0 .\end{array}\right.$
.
Construct the line $x-2 y+4=0: \quad y=0, x=-4, A(-4 ; 0) ; x=0$, $y=2, \quad B(0 ; 2)$. Construct the line $x+y-5=0: y=0, x=5, C(5 ; 0)$; $x=0, y=5, D(0 ; 5)$.
$2^{6}$. Find the intersection point of the lines by solving the system
$$
\left\{\begin{array}{l}... | 13.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,181 |
320. $y=4x-x^2$, $y=0$ and $x=5$. | The parabola $y=4 x-x^{2}$ intersects the x-axis at points $x=0$ and $x=4$. The figure whose area we need to find is marked in color in Fig. 167. Let $S_{1}$ and $S_{2}$ be the areas of the parts of this figure corresponding to the segments $[0,4]$ and $[4, 5]$, and let $S$ be the required area; then $S=S_{1}+S_{2}$. U... | 13 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,183 |
324. $y=\frac{1}{2} x^{3}, x=-2, x=4$ and $y=0$. | Solution. The graph of the function $y=\frac{1}{2} x^{3}$ lies below the $O x$ axis on the interval $[-2,0]$, and above the $O x$ axis on the interval $[0,4]$ (Fig. 171). Therefore,
$S=\left|\int_{-2}^{0} \frac{1}{2} x^{3} d x\right|+\int_{0}^{4} \frac{1}{2} x^{3} d x=\left|\frac{1}{8} x^{4}\right|_{-2}^{0}+\left.\fra... | 34 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,184 |
326. $y=-x^{2}+5$ and $y=x+3$. | Solution. Let's find the abscissas of the points of intersection of the parabola $y = -x^{2}+5$ and the line $y=x+3$. For this, we solve the system $\left\{\begin{array}{l}y=-x^{2}+5, \\ y=x+3,\end{array}\right.$ from which $x_{1}=-2, x_{2}=1$.
Let's find the area $S_{1}$ of the figure bounded by the parabola $y=-x^{2... | 4.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,185 |
334. Calculate the area of the figure bounded by the curves $y^{2}=4 x$ and $x^{2}=4 y$ | Solution. In Fig. 182, the area of the figure we need to calculate is shown. As can be seen from the figure, the area of the figure OBM AO can be represented as the difference between the areas of the figures OBMPO and OAMPO, where $MP$ is the perpendicular dropped from point $M$ to the $Ox$ axis.
Let's find the coord... | \frac{16}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,187 |
337. $y=x^{2}+1, x=-2, x=2$ and $y=0$ (Fig. 185). | Solution. \( S = \int_{-2}^{2} \left(x^{2} + 1\right) dx = 2 \int_{0}^{2} \left(x^{2} + 1\right) dx = 2 \left( \frac{x^{3}}{3} + x \right) \bigg|_{\sigma}^{2} = 9 \frac{1}{3} \) (sq. units).
.
We determine the values of the function at the partition points: $f\left(x_{0}\right)=f(1)=2$; $f\left(x_{1}\right)=f(1.5)=3.25 ; f\left(x_{2}\right)=f(2)=5 ; f\le... | 8.75 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,189 |
345. Find the value of the integral $\int_{0}^{1} x d x$, considering it as the limit of integral sums. | Solution. Divide the interval $[0,1]$ (Fig. 194) into $n$ equal parts; in this case, the length of each part is $1 / n$. Draw lines through the division points parallel to the ordinate axis. Then each strip, for example, $A B C D$, will represent a trapezoid. For each trapezoid, construct a rectangle $A B E D$. If $A$ ... | \frac{1}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,190 |
348. Using the rectangle method, compute $\int_{0}^{\pi / 4} \cos x d x$ | Solution. Divide the interval of integration into 5 parts (Fig. 196). Then $n=5 ; \quad b-a=\pi / 4 ; \quad \Delta x=(b-a) / n=\pi / 20 \approx 0.1571$. The corresponding values of the integrand function will be found using tables: $y_{0}=\cos 0^{\circ}=1 ; y_{1}=\cos (\pi / 20)=\cos 9^{\circ}==0.9877 ; y_{2}=$ $=\cos ... | 0.7288 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,192 |
356. Calculate the integral $\int_{0}^{5} \frac{d x}{\sqrt{x+4}}$ using the trapezoidal rule, dividing the interval of integration into 5 equal parts. | Solution. Here $n=5 ; \Delta x=1 ; x_{0}=0 ; x_{1}=1 ; x_{2}=2 ; x_{3}=3 ; x_{4}=4$; $x_{5}=5$. Next, we find $y_{0}=\frac{1}{\sqrt{4}}=\frac{1}{2} ; y_{1}=\frac{1}{\sqrt{1+4}} \approx 0.447 ; y_{2}=\frac{1}{\sqrt{2+4}} \approx$ $\approx 0.409 ; y_{3}=\frac{1}{\sqrt{3+4}} \approx 0.377 ; y_{4}=\frac{1}{\sqrt{4+4}} \app... | 2.002 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,194 |
357. Calculate the integral $\int_{2}^{12} \frac{d x}{x}$ using the trapezoidal rule, dividing the interval of integration into 10 equal parts. Estimate the error. | Solution. According to the condition, $a=2, b=12, n=10, \Delta x=1$. We will calculate the values of the integrand function $y=1 / x$ at the corresponding division points: $x_{0}=2, \quad y_{0}=1 / 2=0.5 ; \quad x_{1}=3, y_{1}=1 / 3 \approx 0.3333 ; \quad x_{2}=4, y_{2}=$
$=1 / 4=0.25 ; x_{3}=5, y_{3}=1 / 5=0.2 ; x_{4}... | 1.812 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,195 |
358. Calculate the integral $\int_{1}^{9} \sqrt{6 x-5} d x$ using the Newton-Leibniz formula and approximate formulas for rectangles and trapezoids, dividing the interval of integration into 8 equal parts. Estimate the error of the results. | Solution. According to the Newton-Leibniz formula, we find
$$
I=\int_{1}^{9} \sqrt{6 x-5} d x=\frac{1}{6} \int_{1}^{9}(6 x-5)^{1 / 2} d(6 x-5)=\left.\frac{1}{9}(6 x-5)^{3 / 2}\right|_{1} ^{9}=\frac{7^{3}-1}{9}=38
$$
Since the interval $[1,9]$ is divided into 8 equal parts, then $\Delta x=1$. We find the values $y_{i}... | 38 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,196 |
362. Calculate the integral $\int_{1}^{4} x^{2} d x$ using Simpson's formula. | Solution. Divide the interval of integration into 10 equal parts. Then $(b-a) / 3 n=3 / 30=1 / 10=0.1$. Substituting into the integrand function $y=x^{2}$ the values of the argument $x_{0}=1, x_{1}=1.3 ; x_{2}=1.6, \ldots$, $x_{10}=4$, we find the corresponding values of the ordinates: $y_{0}=1 ; y_{1}=1.69$; $y_{2}=2.... | 21 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,197 |
363. The velocity of a material point is given by the formula $v=\left(4 t^{3}-2 t+1\right)$ m/s. Find the distance traveled by the point in the first 4 s from the start of the motion. | Solution. According to formula (1), we have
$$
s=\int_{0}^{4}\left(4 t^{3}-2 t+1\right) d t=\left.\left(t^{4}-t^{2}+t\right)\right|_{0} ^{4}=256-16+4=244(\text { m })
$$
Thus, in 4 s, the point has traveled 244 m. | 244 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,198 |
370. The velocity of a body is given by the equation $v=$ $=\left(12 t-3 t^{2}\right) \mathrm{m} / \mathrm{s}$. Determine the distance traveled by the body from the start of the motion until it stops. | Solution. The speed of the body is zero at the beginning of its motion and at the moment of stopping. To find the moment of stopping, we set the speed to zero and solve the equation with respect to $t$; we get $12 t-3 t^{2}=0 ; 3 t(4-t)=0 ; t_{1}=0, t_{2}=4$. Therefore,
$$
s=\int_{0}^{4}\left(12 t-3 t^{2}\right) d t=\... | 32 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,199 |
371. A body is thrown vertically upward with a velocity that varies according to the law \( v = (29.4 - 9.8 t) \) m/s. Find the maximum height of ascent. | Solution. Let's find the time during which the body was rising: $29.4 - 9.8 t = 0$ (at the moment of maximum ascent, the velocity is zero); $t = 3$ (s). Therefore,
$$
s=\int_{0}^{3}(29.4 - 9.8 t) \, dt = 9.8 \left(3 t - \frac{t^{2}}{2}\right) \bigg|_{0}^{3} = 44.1 \quad (\mathrm{m})
$$ | 44.1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,200 |
378. Two bodies simultaneously leave one point: one with a speed of $v_{1}=5 t \mathrm{~m} / \mathrm{c}$, the other with a speed of $v_{2}=3 t^{2} \mathrm{~m} / \mathrm{c}$. At what distance from each other will they be after $20 \mathrm{c}$, if they move in a straight line in the same direction? | Solution. $s_{1}=\int_{0}^{20} 5 t d t=5 \cdot \frac{t^{2}}{2} \left\lvert\, \frac{20}{\bar{\sigma}} 5 \cdot 200=1000(\mathrm{m})\right. ;$
$$
s_{2}=\int_{0}^{20} 3 t^{2} d t=\left.t^{3}\right|_{0} ^{20}=8000(\mathrm{m}) ; s=s_{2}-s_{1}=8000-1000=7000(\mathrm{m})
$$ | 7000(\mathrm{}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,201 |
380. Two bodies simultaneously started linear motion from a certain point in the same direction with speeds $v_{1}=$ $=\left(6 t^{2}+4 t\right) \mathrm{m} /$ s and $v_{2}=4 t \mathrm{m} / \mathrm{s}$. After how many seconds will the distance between them be 250 m? | Solution. Let $t_{1}$ be the moment of the meeting. Then
$$
s_{1}=\int_{0}^{t_{1}}\left(6 t^{2}+4 t\right) d t=\left.\left(2 t^{3}+2 t^{2}\right)\right|_{0} ^{t_{1}}=2 t_{1}^{3}+2 t_{1}^{2} ; \quad s_{2}=\int_{0}^{t_{1}} 4 t d t=2 t_{1}^{2}
$$
Since $s_{1}-s_{2}=250$, we get the equation $2 t_{1}^{3}+2 t_{1}^{2}-2 t_... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,202 |
382. A force of $60 \mathrm{H}$ stretches a spring by 2 cm. The original length of the spring is $14 \mathrm{~cm}$. How much work is required to stretch it to 20 cm? | Solution. We have $k=\frac{60}{0.02}=3000(\mathrm{H} /$ m) and, consequently, $F=$ $=3000 x$. Since the spring needs to be stretched by 0.06 (m), then
$$
A=\int_{0}^{0.06} 3000 x d x=\left.1500 x^{2}\right|_{0}^{0.06}=1500 \cdot 0.06^{2}=5.4 \quad(\text{J})
$$ | 5.4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,204 |
385. When a spring is stretched by 5 cm, the work expended is 29.43 J. By how much will the spring stretch if the work expended is 9.81 J? | Solution. Here $A_{1}=29.4$ J, $x_{0}=0, x_{1}=0.05$ m. Since $A_{1}=$ $=k \int_{x_{0}}^{x_{1}} x d x$, we obtain the equation $29.43=k \int_{0}^{0.05} x d x \quad$ or $\quad 29.43=k \times$ $\times\left.\frac{x^{2}}{2}\right|_{0} ^{0.05}$, from which $k=\frac{29.43 \cdot 2}{0.0025}=23544$.
Next, from the equation $A_... | 0.029 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,205 |
389. Determine the force of water pressure on the gate wall, which is $20 \mathrm{~m}$ long and $5 \mathrm{m}$ high (assuming the gate is completely filled with water). | Solution. Here $y=f(x)=20, a=0, b=5(m), \gamma=1000$ kg $/ \mathbf{~}^{3}$. Using formula (5), we find
$$
P=9810 \int_{0}^{5} 20 x d x=\left.9810 \cdot 20 \frac{x^{2}}{2}\right|_{0} ^{5}=9810 \cdot 10 \cdot 25=2,45 \cdot 10^{6}
$$ | 2,45\cdot10^{6} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 32,206 |
392. A rectangular plate is immersed in water, positioned vertically. Its horizontal side is 1 m, the vertical side is $2 \mathrm{m}$. The top side is at a depth of 0.5 m. Determine the force of water pressure on the plate. | Solution. Here $y=1, \quad a=0.5, \quad b=2+0.5=2.5$ (m), $\gamma=$ $=1000 \mathrm{kr} / \mathrm{m}^{3}$. Therefore,
$$
\begin{gathered}
P=9810 \int_{a}^{b} x y d x=9810 \int_{a}^{b} x d x=\left.9810 \frac{x^{2}}{2}\right|_{0.5} ^{2.5}=9810 \frac{2.5^{2}-0.5^{2}}{2}= \\
=29430 \text { (N) }
\end{gathered}
$$ | 29430 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,207 |
395. A plate in the form of a triangle, the base of which is $4 \mathrm{~cm}$, and the height is $3 \mathrm{~cm}$, is vertically submerged in water. Find the force of water pressure on this plate, if its vertex lies on the water surface (Fig. 200). | Solution. Again, we will use formula (5). Here $a=0$, $b=3 \mathrm{~cm}=0.03 \mathrm{~m}$, $|A C|=4 \mathrm{~cm}=0.04 \mathrm{~m}$, $\gamma=1000 \mathrm{kr} / \mathrm{m}^{3}$.
From the similarity of triangles $D B E$ and $A B C$ (Fig. 200), we have $\frac{|D F|}{|A C|}=$ $=\frac{|B K|}{|B E|}$, or $\frac{y}{0.04}=\fra... | 0.117\quad(\mathrm{N}) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 32,208 |
400. Calculate the force of water pressure on a dam that has the shape of a trapezoid, where the upper base, coinciding with the water surface, is $10 \mathrm{~m}$ long, the lower base is $20 \mathrm{~m}$, and the height is $3 \mathrm{m}$. | Solution. Using formula (6), we find
$$
P=9810 \frac{(10+2 \cdot 20) 3^{2}}{6}=9810 \cdot \frac{450}{6}=735750(\mathrm{H})
$$ | 735750 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 32,209 |
403. Determine the force of water pressure on a vertical parabolic segment, the base of which is $4 \mathrm{m}$ and is located on the water surface, while the vertex is at a depth of $4 \mathrm{m}$ (Fig. 204). | Solution. We have $|B A|=2 x=4$ (m). Point $A$ in the chosen coordinate system has coordinates $(2 ; 4)$. The equation of the parabola relative to this system is $y = a x^{2}$ or $4=a \cdot 2^{2}$, from which $a=1$, i.e., $y=x^{2}$.
Consider an elementary area $d S$ at a distance $y$ from the origin. The length of thi... | 167424 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,210 |
404. A cylindrical glass is filled with mercury. Calculate the force of pressure of the mercury on the side surface of the glass, if its height is 0.1 m, and the radius of the base is 0.04 m. The density of mercury is $13600 \mathrm{kg} / \mathrm{m}^{3}$. | Solution. Let's calculate the area of the circular strip:
$$
\Delta S=2 \pi r d x=0.08 \pi d x
$$
The elementary force of pressure is
$$
\Delta P=9.81 \cdot 136,00 \cdot 0.08 \pi d x=10673 \pi d x
$$
Therefore,
$$
P=\int_{0}^{0.1} 10673 \pi x d x=\left.10673 \pi \frac{x^{2}}{2}\right|_{0} ^{0.1}=53.37 \pi \approx ... | 167.6(\mathrm{N}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,211 |
1. Determine which of the following equations are differential: a) $y^{\prime}+3 x=0$; b) $y^{2}+x^{2}=5$; c) $y=e^{x}$; d) $y^{\prime} y-x=0$; e) $y=\ln |x|+C$; f) $2 d y+3 x d x=0$. | Solution. Equations b), c), d) are not differential, as they do not contain the derivative of the unknown function or differentials of the argument and the unknown function; equations a), e), f) are differential. | ),e),f) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,212 |
6. Given the equation $x y^{\prime}=y-1$. Which of the functions $y=3 x+1$ or $y=C x+1$ is its solution? | Solution. Each of the given functions is a solution to the given equation. Indeed:
if $y=3 x+1$, then $y^{\prime}=3$, from which $x \cdot 3=(3 x+1)-1$, i.e., $3 x=3 x$;
if $y=C x+1$, then $y^{\prime}=C$, from which $x C=(C x+1)-1$, i.e., $C x=C x$. | proof | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,213 |
7. Check if the function $y=x^{2}+x+C$ is a solution to the differential equation $d y=(2 x+1) d x$. | Solution. We find $y^{\prime}=2 x+1$, from which $d y=(2 x+1) d x$. Substituting the found expression for $d y$ into the left part of the given equation, we get $(2 x+1) d x \equiv(2 x+1) d x$, i.e., the given function is a solution to this equation. | proof | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,214 |
11. Check that the function $y=C e^{-x^{2}}$ is a solution to the differential equation $\frac{d y}{y}+2 x d x=0$. | Solution. We have $\frac{C e^{-x^{2}}(-2 x)}{C e^{-x^{2}}} d x+2 x d x=0$, i.e. $-2 x d x+2 x \times$ $\times d x \equiv 0$. | proof | Calculus | proof | Yes | Yes | olympiads | false | 32,215 |
15. Knowing that the function $y=C x+1$ is the general solution of the equation $x y^{\prime}=y-1$, determine its particular solution if $y(1)=5$. | Solution. Substituting the given initial conditions \(x=1, y=5\) into the general solution \(y=Cx+1\), we get \(5=C \cdot 1+1\), from which \(C=4\).
Now, substituting the value \(C=4\) into the general solution, we find the particular solution \(y=4x+1\).
Thus, when solving a differential equation, the general soluti... | 4x+1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,216 |
17. Solve the equation $y^{\prime}=x+3$. | Solution. It is required to find the function $y(x)$, the derivative of which is equal to $x+3$, i.e., to find the antiderivative of the function $x+3$. According to the rules of integration, we obtain the general solution of the equation in the form $y=\frac{x^{2}}{2}+3 x+C$, where $C$ is an arbitrary constant. | \frac{x^{2}}{2}+3x+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,217 |
18. Find the solution $y(x)$ of the differential equation $y^{\prime}=$ $=\cos x$, satisfying the condition $y(0)=1$. | Solution. The general solution of this equation is found by integration: $y=\sin x+C$. By substituting the initial conditions and determining $C=1$, we find the particular solution (solution of the Cauchy problem): $y=\sin x+1$.
The problems considered above on finding the equation of a curve for which the slope of th... | \sinx+1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,218 |
27. Solve the equation $x d x+y d y=0$. | Solution. Here the variables are separated. Integrating, we get
$$
\int x d x+\int y d y=C ; \frac{x^{2}}{2}+\frac{y^{2}}{2}=C ; x^{2}+y^{2}=2 C .
$$
Since $C$ is arbitrary, we can denote $2 C$ by $C^{2}$, considering that the left-hand side of the last equation is positive. Then this equation will take the form $x^{... | x^{2}+y^{2}=C^{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,220 |
28. Solve the equation $2 y d y=3 x^{2} d x$. | Solution. Here $\varphi(y)=2 y, f(x)=3 x^{2}$. Integrating both sides of the equation, we have
$$
\int 2 y d y=\int 3 x^{2} d x, y^{2}=x^{3}+C
$$
We obtained the general solution of the differential equation. This solution can be written in explicit form: $y=\sqrt{x^{3}+C}$. | \sqrt{x^{3}+C} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,221 |
31. Find the particular solution of the differential equation $d y=\left(x^{2}-1\right) d x$, if $y=4$ when $x=1$. | Solution. We have $\int d y=\int\left(x^{2}-1\right) d x ; y=\frac{x^{3}}{3}-x+C ; \quad 4=\frac{1}{3}-$ $-1+C$, from which $C=\frac{14}{3}$. Therefore, we obtain the answer: $y=\frac{x^{3}}{3}-x+\frac{14}{3}$. | \frac{x^{3}}{3}-x+\frac{14}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,222 |
33. Solve the equation $\frac{d y}{y+1}=\frac{d x}{x-1}$. | Solution. Here the variables are separated. Integrating, we have
$$
\int \frac{d y}{y+1}=\int \frac{d x}{x-1} ; \ln (y+1)=\ln (x-1)+C
$$
The arbitrary constant $C$ can be denoted as $\ln C$; then $\ln (y+1)=\ln (x-1)+\ln C$. Representing the sum of logarithms on the right side of the equation as the logarithm of a pr... | y+1=C(x-1) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,223 |
40. Solve the equation $x\left(y^{2}-1\right) d x+y\left(x^{2}+1\right) d y=0$. | Solution. Dividing all terms of the equation by the product $(x^{2}+1)\left(y^{2}-1\right)$, we get $\frac{x d x}{x^{2}+1}+\frac{y d y}{y^{2}-1}=0$. Now the variables are separated; integrating, we find
$$
\int \frac{x d x}{x^{2}+1}+\int \frac{y d y}{y^{2}-1}=C_{1} ; \frac{1}{2} \ln \left(x^{2}+1\right)+\frac{1}{2} \l... | (x^{2}+1)(y^{2}-1)=C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,224 |
42. Integrate the differential equation
$$
\left(1+x^{2}\right) d y-2 x y d x=0
$$ | Solution. The given equation is an equation with separable variables. By dividing both sides of the equation by the product $y\left(1+x^{2}\right)$, we obtain an equation with separated variables:
$$
\frac{d y}{y}-\frac{2 x d x}{1+x^{2}}=0
$$
Integrating this equation, we find
$$
\ln |y|-\ln \left(1+x^{2}\right)=\ln... | C(1+x^{2}) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,225 |
43. Find the general solution of the equation $1+y^{\prime}+y+x y^{\prime}=0$. | Solution. $1^{0}$. Note $y^{\prime}$ as $\frac{d y}{d x}$ :
$$
1+\frac{d y}{d x}+y+x \frac{d y}{d x}=0
$$
$2^{0}$. Multiply all terms of the equation by $d x$ :
$$
d x+d y+y d x+x d y=0
$$
Group all terms containing $d y$ and $d x$, and write the obtained expressions in different parts of the equation:
$$
(1+x) d ... | \frac{C}{1+x}-1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,226 |
52. Find the particular solution of the equation $2 y d x=(1+x) d y$, if $y=4$ when $x=1$. | Solution. Separating variables:
$$
\frac{2 d x}{1+x}=\frac{d y}{y}
$$
Integrating, we get
$$
\int \frac{2 d x}{1+x}=\int \frac{d y}{y}, 2 \ln (1+x)=\ln y+C, \text { or } \ln (1+x)^{2}=\ln y+\ln C
$$
(here $C$ is replaced by $\ln C$). Exponentiating, we find $(1+x)^{2}=C y$ - the general integral of the given differ... | (1+x)^2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,227 |
53. Find the particular solution of the differential equation $y^{\prime}=2+y$, if $y=3$ when $x=0$. | Solution. Replacing $y^{\prime}$ with $\frac{d y}{d x}$, and then multiplying all terms by $d x$, we get
$$
d y=2 d x+y d x, \text{ i.e., } d y=(2+y) d x
$$
Dividing both sides of the equation by $2+y$ and integrating:
$$
\frac{d y}{2+y}=d x ; \int \frac{d y}{2+y}=\int d x ; \ln (2+y)=x+\ln C
$$
Express $x$ in term... | 5e^{x}-2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,228 |
57. Find the particular solution of the equation $y^{\prime}+y \operatorname{tg} x=0$, if $y=2$ when $x=0$. | Solution. $1^{0} . \frac{d y}{d x}+y \tan x=0$.
$2^{0} . d y=-y \tan x d x$.
$3^{0} . \frac{d y}{y}=-\tan x d x$
$4^{0} \cdot \int \frac{d y}{y}=-\int \tan x d x ; \ln |y|=\ln |\cos x|+\ln |C|, y=C \cos x$.
$5^{0}$. When $x=0, y=2$ we have $2=C \cdot 1, C=2$, i.e., $y=2 \cos x$. | 2\cosx | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,229 |
60. Find the equation of the line passing through the point $(1 ; 3)$ and having a tangent with a slope equal to $2 x-3$ (see problem 19). | Solution. Using the condition, we form the differential equation:
$$
y^{\prime}=2 x-3 ; \frac{d y}{d x}=2 x-3 ; d y=(2 x-3) d x
$$
Let's find the general solution of this equation:
$$
\int d y=\int(2 x-3) d x ; y=x^{2}-3 x+C
$$
Substituting the initial data $x=1, y=3$ into the general solution, we get $C=5$. Theref... | x^{2}-3x+5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,230 |
65. The speed of a body coming out of a state of rest is equal to $5 t^{2} \mathrm{m} / \mathrm{s}$ after $t$ seconds. Determine the distance the body will travel in 3 s (see problem 20). | Solution. Using the condition, we form the differential equation: $\frac{d s}{d t}=5 t^{2}$, since velocity $v=\frac{d s}{d t}$.
Let's find the general solution of this equation:
$$
d s=5 t^{2} d t ; \int d s=5 \int t^{2} d t ; s=\frac{5}{3} t^{3}+C
$$
Now, let's find the particular solution of this equation. The in... | 45 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,231 |
67. A material point moves in such a way that its speed is proportional to the distance traveled. At the initial moment, the point was 1 m away from the origin, and after 2 s - at a distance of $e$ m. Find the law of motion of the material point (see problem 21). | Solution. Let the speed of the material point be denoted by $v$. As is known, speed is the derivative of the path with respect to time, i.e., $v=\frac{d s}{d t}$. According to the problem, the speed is proportional to the distance traveled, i.e., $v=\frac{d s}{d t}=k s$, where $k$ is the proportionality constant.
Thus... | e^{/2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,232 |
74. According to Newton's law, the cooling rate of a body in air is proportional to the difference between the body's temperature $T$ and the air temperature $T_{0}$. Determine the law of temperature change of the body over time, if the experiment is conducted at $T_{0}=20^{\circ} \mathrm{C}$, and the body cooled from ... | Solution. The cooling rate of a body (the rate of change of its temperature) is equal to the derivative $\frac{d T}{d t}$, and thus Newton's law can be expressed by the equation
$$
\frac{d T}{d t}=k(T-20)
$$
where $k-$ is the proportionality coefficient.
Solving the obtained differential equation with separable vari... | 20+80(\frac{1}{2})^{/20} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,233 |
78. Given the equations:
a) $y^{\prime}+\frac{2 y}{x+1}=(x+1)^{3}$
b) $y^{\prime \prime}+2 x y=0$;
c) $\frac{d y}{d x}+x y^{2}=(x-3)^{2}$.
Which of them are linear first-order equations, and which are not, and why? | Solution. Equation a) is a linear first-order equation, since $y$ and $y^{\prime}$ enter to the first power, and $p=\frac{2}{x+1}, q=(x+$ $+1)^{3}$ are functions of one variable $x$. Equations b) and c) are not linear, as they contain the second derivative and $y^{2}$, respectively. | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,234 |
81. Solve the equation $y^{\prime}+y \tan x=\cos ^{2} x$. | Solution. This is a linear equation, as it has the form $y^{\prime}+p y=q$, where $p=\operatorname{tg} x, q=\cos ^{2} x$. Let $y=u v$; then $y^{\prime}=u^{\prime} v+$ $+v^{\prime} u$
$$
u^{\prime} v+v^{\prime} u-u v \operatorname{tg} x=\cos ^{2} x
$$
or
$$
u^{\prime} v+\left(v^{\prime}+v \operatorname{tg} x\right) u... | (\sinx+C)\cosx | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,235 |
82. Solve the equation $\frac{d y}{d x} \cos x+y \sin x=1$. | Solution. $1^{0}$. Dividing all terms of the equation by $\cos x$, we get
$$
\frac{d y}{d x}+y \operatorname{tg} x=\frac{1}{\cos x}
$$
$2^{0}$. Let $y=u v ; y^{\prime}=\frac{d y}{d x}=u^{\prime} v+v^{\prime} u$.
Substituting the expressions for $y$ and $\frac{d y}{d x}$ into the equation, we have
$$
u^{\prime} v+v^... | \sinx+C\cosx | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,236 |
102. Solve the equation $y^{\prime}+2 y+3=0$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | Solution. $\frac{d y}{d x}=-(2 y+3) ; \frac{d y}{2 y+3}=-d x ; \int \frac{d y}{2 y+3}=-\int d x ;$ $\ln |2 y+3|=-x+\ln C ; \ln \frac{2 y+3}{C}=-x$,
$$
\frac{2 y+3}{C}=e^{-x} ; 2 y+3=C e^{-x}, y=C e^{-x}-\frac{3}{2}
$$ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,238 | |
106. Solve the equation $y^{\prime}\left(y^{2}-x\right)=y$. | Solution. This equation is not linear with respect to the unknown function $y$, since it cannot be brought to the form $y^{\prime} + p(x) y = q(x)$. However, if $x$ is considered a function and $y$-- an argument, then this equation will be linear with respect to the unknown function $x$, since it can be brought to the ... | (\frac{y^{3}}{3}+C)\frac{1}{y} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,240 |
110. Find the general solution of the equation $y^{\prime \prime}=4 x$. | Solution. We have
$$
\begin{gathered}
\frac{d y^{\prime}}{d x}=4 x ; d y^{\prime}=4 x d x ; y^{\prime}=4 \int x d x=2 x^{2}+C_{1} \\
\frac{d y}{d x}=2 x^{2}+C_{1} ; d y=\left(2 x^{2}+C_{1}\right) d x \\
y=\int\left(2 x^{2}+C_{1}\right) d x=2 \int x^{2} d x+C_{1} \int d x=\frac{2}{3} x^{3}+C_{1} x+C_{2}
\end{gathered}
... | \frac{2}{3}x^{3}+C_{1}x+C_{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,241 |
111. Find the general solution of the equation $y^{\prime \prime}=\sin 2 x$. | Solution. Multiply both sides of the equation by $d x$ and then integrate:
$$
d y^{\prime}=\sin 2 x d x ; \int d y^{\prime}=\int \sin 2 x d x ; y^{\prime}=-\frac{1}{2} \cos 2 x+C_{\mathrm{t}}
$$
Multiply both sides of the last equation by $d x$ and integrate:
$$
d y=-\frac{1}{2} \cos 2 x d x+C_{1} d x ; \int d y=-\f... | -\frac{1}{4}\sin2x+C_{1}x+C_{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,242 |
120. Find the solution of the equation $y^{\prime \prime}=2$, satisfying the following initial conditions: $y=1$ and $y^{\prime}=2$ when $x=0$. | Solution. Integrating the given equation with respect to $x$ twice in succession, we first obtain $y^{\prime}=2 x+C_{1}$, and then $y=x^{2}+C_{1} x+C_{2}$. Using the initial conditions now, we have
$$
\left\{\begin{array}{l}
y^{\prime}=2 x+C_{1} \\
y=x^{2}+C_{1} x+C_{2};
\end{array} \quad \left\{\begin{array}{l}
2=2 \... | (x+1)^{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,243 |
121. Solve the Cauchy problem for the equation $y^{\prime \prime}=1+x+x^{2}+$ $+x^{3}$, if $y=1$ and $y^{\prime}=1$ when $x=0$. | Solution. $1^{0}$. $\frac{d y}{d x}=\int\left(1+x+x^{2}+x^{3}\right) d x=x+\frac{x^{2}}{2}+\frac{x^{3}}{3}+\frac{x^{4}}{4}+C_{1}$.
$2^{0} \cdot y=\int\left(x+\frac{x^{2}}{2}+\frac{x^{3}}{3}+\frac{x^{4}}{4}+C_{1}\right) d x=\frac{x^{2}}{2}+\frac{x^{3}}{6}+\frac{x^{4}}{12}+\frac{x^{5}}{20}+C_{1} x+C_{2}$.
$3^{0}$. Subst... | \frac{x^{2}}{2}+\frac{x^{3}}{6}+\frac{x^{4}}{12}+\frac{x^{5}}{20}+x+1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,244 |
129. A body moves in a straight line with acceleration $a=6 t-4$. At $t=0$, the initial path $s_{0}=0$, initial velocity $v_{0}=4$. Find the velocity and the distance traveled as functions of time. | Solution. According to the condition, we have $s^{\prime \prime}=6 t-4$. Integrating both sides of this equation, we get $\frac{d s}{d t}=3 t^{2}-4 t+C_{1}$, i.e., the velocity is expressed as a function of time.
Integrating both sides of the last equation, we express the path as a function of time:
$$
s=t^{3}-2 t^{2... | v=3^{2}-4,=^{3}-2^{2}+4 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,245 |
134. Solve the equation $y^{\prime \prime}-3 y^{\prime}+2 y=0$. | Solution. Substitute the function $y=e^{k x}$ and its derivatives $y^{\prime}=$ $=k e^{k x}$ and $y^{\prime \prime}=k^{2} e^{k x}$ into the given equation:
$$
k^{2} e^{k x}-3 k e^{k x}+2 e^{k x}=0
$$
Factor out $e^{k x}$: $e^{k x}\left(k^{2}-3 k+2\right)=0$. Since $e^{k x} \neq 0$, then $k^{2}-3 k+2=0$.
Solve the re... | C_{1}e^{x}+C_{2}e^{2x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,246 |
136. Solve the equation $y^{\prime \prime}+2 y^{\prime}-8 y=0$. | Solution. Let's form the characteristic equation $k^{2}+2 k$ $-8=0$. Here $D=p^{2}-4 q=2^{2}-4 \cdot(-8)=4+32=36>0$. Therefore, the characteristic equation has two distinct real roots. Let's determine them: $k_{1}=-4, k_{2}=2$.
We find the particular solutions of the given differential equation: $y_{1}=e^{-4 x}, y_{2}... | C_{1}e^{-4x}+C_{2}e^{2x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,247 |
140. Find the particular solution of the equations $y^{\prime \prime}-2 y^{\prime}-3 y=0$, if $y=8$ and $y^{\prime}=0$ when $x=0$. | Solution. We have $k^{2}-2 k-3=0, D>0, k_{1}=-1, k_{2}=3$; hence, $y_{1}=e^{-x}, y_{2}=e^{3 x}$ are particular linearly independent solutions of the given equation. Its general solution is of the form
$$
y=C_{1} y_{1}+C_{2} y_{2}=C_{1} e^{-x}+C_{2} e^{3 x}
$$
Next, we find
$$
y^{\prime}=-C_{1} e^{-x}+3 C_{2} e^{3 x}... | 6e^{-x}+2e^{3x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,248 |
142. Find the linearly independent particular solutions and the general solution of the equation $y^{\prime \prime}-4 y^{\prime}+4 y$. | Solution. The characteristic equation $k^{2}-4 k+4=0$ has equal roots: $k_{1}=k_{2}=2$.
One particular solution is $y_{1}=e^{2 x}$. We will show that $y_{2}=x e^{2 x}$. We have $y^{\prime}=e^{2 x}+2 x e^{2 x}, y^{\prime}=2 e^{2 x}+2 e^{2 x}+4 x e^{2 x}=4 e^{2 x}+4 x e^{2 x}$, from which
$$
4 e^{2 x}+4 x e^{2 x}-4\lef... | (C_1+C_2x)e^{2x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,249 |
143. Find the general solution of the equation $y^{\prime \prime}-6 y^{\prime}+9 y=0$. | Solution. The characteristic equation $k^{2}-6 k+9=0$ has equal roots: $k_{1}=k_{2}=3$. The particular linearly independent solutions of this equation are the functions $y_{1}=e^{3 x}, y_{2}=x e^{3 x}$, and the general solution is given by
$$
y=C_{1} e^{3 x}+C_{2} x e^{3 x}=\left(C_{1}+C_{2} x\right) e^{3 x}
$$ | (C_{1}+C_{2}x)e^{3x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,250 |
146. Find the particular solution of the equation $y^{\prime \prime}-2 y^{\prime}+y=0$, if $y(0)=4, y^{\prime}(0)=2$. | Solution. The characteristic equation $k^{2}-2 k+1=0$ has equal real roots: $k_{2}=k_{1}=1$. Therefore, the general solution of the differential equation is the function $y=e^{x} \times$ $x\left(C_{1}+C_{2} x\right)$.
Next, substitute the initial conditions $y_{(0)}=4$ and $y_{(0)}^{\prime}=2$ into the equations $y=e^... | e^{x}(4-2x) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,251 |
149. Find the general solution of the equation $y^{\prime \prime}-4 y^{\prime}+13 y=0$. | Solution. Let's form the characteristic equation and find its roots: $k^{2}-4 k+13=0 ; k_{1}=2+3 i, k_{2}=2-3 i$.
Write down the general solution of the equation:
$$
y=e^{2 x}\left(C_{1} \cos 3 x+C_{2} \sin 3 x\right)
$$ | e^{2x}(C_{1}\cos3x+C_{2}\sin3x) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,252 |
152. Find the particular solution of the equation $y^{\prime \prime}-2 y^{\prime}+50 y=0$, if $y=1$ and $y^{\prime}=1$ when $x=0$. | Solution. The characteristic equation $k^{2}-2 k+50=0$ has roots $k_{1,2}=1 \pm \sqrt{1-50}=1 \pm 7 i$. Therefore, the general solution of the equation is
$$
y=e^{x}\left(C_{1} \cos 7 x+C_{2} \sin 7 x\right)
$$
Next, we find
$$
y^{\prime}=e^{x}\left(C_{1} \cos 7 x+C_{2} \sin 7 x\right)+e^{x}\left(-7 C_{1} \sin 7 x+7... | e^{x}(\cos7x-\frac{2}{7}\sin7x) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,253 |
189. Find the integral curve of the differential equation $y^{\prime \prime}+2 y^{\prime}+2 y=0$, passing through the point $(0 ; 1)$ and tangent to the line $y=x+1$ at this point. | Solution. The roots of the characteristic equation $k^{2}+2 k+2=$ $=0$ are complex conjugates: $-1 \pm i$. The equation of the family of integral curves of the given differential equation is written in the form
$$
y=e^{-x}\left(C_{1} \cos x+C_{2} \sin x\right)
$$
To find the equation of the desired integral curve, su... | e^{-x}(\cosx+2\sinx) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,254 |
191. The acceleration of a material point moving in a straight line depends on time and is expressed by the formula $a(t)=$ $=6 t-2$. Find the law of motion, if at the initial moment of time $t=0$ the velocity $v=1 \mathrm{m} / \mathrm{s}$, and the path $s=0$. | Solution. $1^{\text {" }}$. The law of motion of a point is expressed by the function $s(t)$. Then $v(t)=\frac{d s}{d t}$ - the velocity of the point; $a(t)=\frac{d v}{d t}=\frac{d^{2} s}{d t^{2}}$ - the acceleration of the motion. According to the condition, we form the differential equation
$$
\frac{d^{2} s}{d t^{2}... | ^{3}-^{2}+ | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,255 |
192. A metal ball, the temperature of which at the beginning of the experiment was $12^{\circ} \mathrm{C}$, is cooled by a stream of water having a temperature of $0^{\circ}$. After 8 minutes, the ball cooled down to $9^{\circ}$. Assuming the cooling rate is proportional to the difference between the temperature of the... | Solution. $1^{\circ}$. Let $T$ denote the temperature of the sphere, and $t-$ the time elapsed since the start of the experiment. Then the cooling rate of the sphere is the derivative $\frac{d T}{d t}\left(T^{\prime}\right)$. According to the condition, $T^{\prime}=k(T-0)=k T$, where $k$ is the proportionality coeffici... | 15 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,256 |
193. A vessel with a capacity of 100 liters is filled with brine containing 10 kg of dissolved salt. In 1 minute, 3 liters of water flow into it, and the same amount of the mixture is pumped into another vessel of the same capacity, initially filled with water, from which the excess liquid is drained. At what moment in... | Let at time $t$ (min) the first vessel contains $x$ (kg) of salt, and let in the subsequent small time interval $d t$ the amount of salt in this vessel decreases by $d x$.
During the time $d t$, $3 d t$ (l) of brine will flow out of the vessel. The concentration of the brine (the amount of salt in one liter of solutio... | 33\frac{1}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,257 |
195. In a motorboat with an initial speed of $v_{0}=$ $=5 \mathrm{m} / \mathrm{s}$, the engine was turned off. During movement, the boat experiences water resistance, the force of which is proportional to the square of the boat's speed, with a proportionality coefficient of $m / 50$, where $m$ is the mass of the boat. ... | Solution. $1^{0}$. According to the condition, we can take the function as the path $s$, and the argument as time $t$. Using Newton's second law $F=m a$, we obtain the equation
$$
m s_{(t)}^{\prime \prime}=-\frac{m}{50}\left(s_{(t)}^{\prime}\right)^{2}, \text { or } s_{(t)}^{\prime \prime} \doteq-\frac{1}{50}\left(s_{... | 10 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,258 |
196. A material point of mass $m$ is attracted to a fixed point $O$ with a force proportional to the mass $m$ and the distance $x$ from point $O$; the coefficient of proportionality is $\omega^{2}$. Find the law of motion. | Solution. $1^{0}$. From mechanics, it is known that if a material point of mass $m$ moves in a straight line under the action of a force $F$, then $F=m a$, i.e., $F=m x^{\prime \prime}$, where $t$ is time, and $x^{\prime \prime}$ is the acceleration caused by the force $F$.
Considering that the attractive force $F$ is... | R\sin(\omega+\alpha) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,259 |
197. A weight supported by a spiral spring is raised to a distance $b$ and then released. It begins to fall, with its acceleration determined by the equation $g=-p^{2} s$, where $p$ is a constant, and $s$ is the distance from the equilibrium position. Find the equation of motion. | Solution. Let's write down the differential equation
$$
s^{\prime \prime}=-p s, \text { or } s^{\prime \prime}+p^{2} s=0 .
$$
We form the characteristic equation and find its roots: $k^{2}+p=0, k= \pm p i, a=0, b=p$. Therefore,
$$
s=e^{0}(A \sin p t+B \cos p t), \text { i.e. } s=A \sin p t+B \cos p t
$$
To determin... | b\cosp | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,260 |
198. A $\mathrm{K}$ source with an electromotive force (emf) equal to $e(t)$ is connected to a circuit consisting of a coil with inductance $L$, an ohmic resistance $R$, and a capacitance $C$ connected in series (Fig. 206). Find the current $i$ in the circuit as a function of time $t$, if at the initial moment the curr... | Solution. $1^{0}$. According to Kirchhoff's law, the electromotive force in the circuit is equal to the sum of the voltage drops across the inductance, resistance, and capacitance:
$$
e(t)=U_{L}+U_{R}+U_{c}
$$
which are related to the current by the relations
$$
U_{L}=L \frac{d i}{d t} ; U_{R}=R i ; U_{c}=\frac{1}{C... | \frac{E}{L\omega_{1}}\cdote^{-\delta}\cdot\sin\omega_{1} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 32,261 |
1. Calculate: a) $3!$; b) $7!-5!$; c) $\frac{7!+5!}{6!}$. | Solution. a) $3!=1 \cdot 2 \cdot 3=6$.
b) Since $7!=1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7$ and $5!=1 \cdot 2 \cdot 3 \cdot 4 \cdot 5$, we can factor out 5! Then we get
$$
5!(6 \cdot 7-1)=5! \cdot 41=1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 41=120 \cdot 41=4920
$$
c) $\frac{5!(6 \cdot 7+1)}{6!}=\frac{1 \cd... | 6,4920,\frac{43}{6} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 32,262 |
2. Simplify: a) $\frac{(n+1)!}{n!}$; b) $\frac{(n+1)!}{(n-1)!}$; c) $\frac{1}{(n+1)!}+\frac{1}{n!}$. | Solution. a) Given that $(n+1)!=1 \cdot 2 \cdot 3 \ldots n(n+1)$, and $n!=1 \cdot 2 \cdot 3 \ldots n$, we can simplify the fraction; $\frac{(n+1)!}{n!}=n+1$.
6) Since $(n+1)!=1 \cdot 2 \cdot 3 \ldots(n-1) n(n+1)$, after simplification we get $\frac{(n+1)!}{(n-1)!}=n(n+1)$.
c) We have $(n+1)!=1 \cdot 2 \cdot 3 \ldots ... | )n+1;b)n(n+1);)\frac{n+2}{(n+1)!} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,263 |
17. Calculate: a) $A_{6}^{3} ;$ b) $\frac{A_{15}^{3}+A_{5}^{4}}{A_{15}^{5}}$. | Solution. a) $A_{6}^{3}=6 \cdot 5 \cdot 4=120$.
6) Since $A_{15}^{3}=15 \cdot 14 \cdot 13, A_{15}^{4}=15 \cdot 14 \cdot 13 \cdot 12, A_{15}^{5}=15 \cdot 14 \cdot 13 \cdot 12 \times$ $\times 11$, then
$$
\frac{A_{15}^{3}+A_{15}^{4}}{A_{15}^{5}}=\frac{15 \cdot 14 \cdot 13+15 \cdot 14 \cdot 13 \cdot 12}{15 \cdot 14 \cdo... | \frac{13}{132} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 32,264 |
18. How many two-digit numbers can be formed from the five digits $1,2,3,4,5$ if no digit is repeated? | Solution. Since two-digit numbers differ from each other either by the digits themselves or by their order, the desired number is equal to the number of arrangements of two out of five elements: \(A_{5}^{2}=5 \cdot 4=20\). Therefore, 20 different two-digit numbers can be formed.
When finding the number of arrangements... | 20 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 32,265 |
19. Calculate in factorial form $A_{6}^{3}$. | Solution. $A_{6}^{3}=\frac{6!}{(6-3)!}=6 \cdot 5 \cdot 4=120$.
20-25. Calculate in any way: | 120 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 32,266 |
30. Calculate: a) $C_{8}^{3}$; b) $C_{10}^{8}$. | Solution. a) Applying formula (6) for $m=8, n=3$, we find
$$
C_{8}^{3}=-\frac{8!}{(8-3)!3!}=-\frac{8!}{5!3!}=\frac{5!6 \cdot 7 \cdot 8}{5!3 \cdot 2 \cdot 1}=56
$$
b) $C_{10}^{\beta}=\frac{10!}{(10-8)!\cdot 8!}=\frac{10!}{2!\cdot 8!}=\frac{8!9 \cdot 10}{1 \cdot 2 \cdot 8!}=45$.
Let's note the main property of the num... | 56 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 32,267 |
4. A die is rolled. Find the probability that: a) an even number of points will fall (event $A$); b) a number of points divisible by 3 will fall (event $B$); c) any number of points except 5 will fall (event $C$). | Solution. a) On the faces of a die, there are three even numbers (2, 4, and 6), i.e., the number of favorable outcomes $m=3$. The number of all possible outcomes is 6 (any number of points from 1 to 6 can fall). Therefore, $P(A)=3 / 6=1 / 2$.
b) Here, there are two numbers that are multiples of three: 3 and 6. Thus, $... | P(A)=\frac{1}{2},P(B)=\frac{1}{3},P(C)=\frac{5}{6} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 32,268 |
44. Given 5 points, no three of which lie on the same line. Find the probability that, by randomly selecting two points, a student will get the required line. | Solution. Let event $A$ be the selection of the desired line. The number of all possible outcomes is equal to the number of lines passing through the given five points. Since a line is determined by a pair of points and the order of the points within this pair does not matter, each pair must differ by at least one poin... | \frac{1}{10} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 32,269 |
49. There are 100 lottery tickets. It is known that 5 tickets have a prize of 20 rubles, 10 tickets have a prize of 15 rubles, 15 tickets have a prize of 10 rubles, 25 tickets have a prize of 2 rubles, and the rest have nothing. Find the probability that the purchased ticket will have a prize of at least 10 rubles. | Solution. Let $A, B$ and $C$ be the events that the purchased ticket wins 20, 15, and 10 rubles, respectively. Since events $A, B$ and $C$ are mutually exclusive, then
$$
P(A+B+C)=P(A)+P(B)+P(C)=\frac{.5}{100}+\frac{10}{100}+\frac{15}{100}=0.3
$$ | 0.3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 32,270 |
51. In a box, there are 250 light bulbs, of which 100 are 100W, 50 are 60W, 50 are 25W, and 50 are 15W. Calculate the probability that the power of any randomly selected light bulb will not exceed 60W. | Solution. Let $A$ be the event that the power of the bulb is $60 \mathrm{Wt}$, $B$-- $25 \mathrm{Wt}$, $C$-- $15 \mathrm{Wt}$, $D$-- $100$ Wt. Events $A, B, C$ and $D$ form a complete system, as they are all mutually exclusive and one of them will definitely occur in the given trial (selection of a bulb). The probabili... | \frac{3}{5} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 32,271 |
65. In the first urn, there are 6 black and 4 white balls, in the second - 5 black and 7 white balls. One ball is drawn from each urn. What is the probability that both balls will be white? | Solution. Let $A_{1}$ be the event that a white ball is drawn from the first urn; $A_{2}$ be the event that a white ball is drawn from the second urn. Obviously, events $A_{1}$ and $A_{2}$ are independent. Since $P\left(A_{1}\right)=4 / 10=2 / 5, P\left(A_{2}\right)=7 / 12$, then by formula (5) we find
$$
P\left(A_{1}... | \frac{7}{30} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 32,273 |
66. The device consists of two elements operating independently. The probability of the first element failing is 0.2; the probability of the second element failing is 0.3. Find the probability that: a) both elements will fail; b) both elements will work. | Solution. Let event $A$ be the failure of the first element, and event $B$ be the failure of the second element. These events are independent (by condition).
a) The simultaneous occurrence of $A$ and $B$ is the event $A B$. Therefore, $P(A B)=0.2 \cdot 0.3=0.06$.
6) If the first element is working, then the event $\b... | 0.06 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,274 |
67. Each exam ticket includes two theoretical questions and one problem. In total, 28 tickets have been prepared. Calculate the probability that a student, randomly drawing a ticket, will answer all questions, if the student has prepared 50 theoretical questions and 22 problems: | Solution. The full answer to the ticket consists of the product of two events: the student will simultaneously answer two questions (event $A$) and solve a problem (event $B$). Let's calculate the probabilities of these events.
The number of all possible combinations of 2 questions out of 56 is
$$
C_{56}^{2}=\frac{56... | 0.625 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 32,275 |
75. There are two identical urns. The first contains 2 black and 3 white balls, the second - 2 black and 1 white ball. First, an urn is chosen at random, and then one ball is drawn at random from it. What is the probability that a white ball will be selected? | Solution. Let event $A$ consist in the extraction of a white ball from a randomly chosen urn, and $H_{1}$ and $H_{2}$ be the hypotheses that it belongs to the first or second urn, respectively. Then the probability $P\left(H_{1}\right)=$ $=P\left(H_{2}\right)=\frac{1}{2}$, the probability that a white ball belongs to t... | \frac{7}{15} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 32,276 |
79. A shooter makes three attempts. The success (hitting the target) and failure (miss) of each of them are independent of the outcomes of the other attempts, and the probability of successful completion of each attempt is constant and equal to p. Find the probability of successful completion of two attempts out of thr... | Solution. Let $A_{1}, A_{2}$ and $A_{3}$ be the success in the first, second, and third attempts, respectively. Then, two successful attempts correspond to the following events: $A_{1} A_{2} \bar{A}_{3}$ - the first two attempts are successful, the third is not; $A_{1} \bar{A}_{2} A_{3}$ - the first and third attempts ... | 3p^{2}(1-p) | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 32,277 |
80. A coin is flipped 10 times. What is the probability that heads will appear exactly three times? | Solution. Let $A_{10.3}$ be the event that a coin lands heads up three times out of ten tosses. In this case, the probability of landing heads is $\frac{1}{2}$, i.e., $p=\frac{1}{2}$. Then $q=1-\frac{1}{2}=$ $=\frac{1}{2}$. Substituting these values into the Bernoulli formula, we get
$$
\begin{aligned}
P\left(A_{10.3}... | \frac{15}{128} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 32,278 |
81. The probability that a lamp will remain defective after 1000 hours of operation is 0.2. What is the probability that out of five lamps, at least three will remain operational after 1000 hours of operation? | Solution. We will consider the burning of each lamp for 1000 hours as a separate experiment. Then we can say that 5 experiments have been conducted. We are interested in the events "3 out of 5 lamps are burning", "4 out of 5 lamps are burning", and "5 out of 5 lamps are burning", i.e., we can find the probability of ea... | 0.0579 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 32,279 |
87. Two items worth 5 rubles each and one item worth 30 rubles are being raffled. Formulate the distribution law of winnings for a person who bought one ticket out of 50. | Solution. The random variable $X$ represents the winnings and can take three values: 0, 5, and 30 rubles. The first outcome is favorable in 47 cases, the second in two cases, and the third in one case. Let's find their probabilities:
$$
P\left(x_{1}\right)=47 / 50=0.94 ; P\left(x_{2}\right)=2 / 50=0.04 ; P\left(x_{3}\... | \begin{pmatrix} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 32,280 |
95. Find the mathematical expectation of the number of points that fall when throwing a dice. | Solution. The random variable $X$ of the number of points takes the values $1,2,3,4,5,6$. We will construct its distribution law:
| Values $x_{i}$ | 1 | 2 | 3 | 4 | 5 | 6 |
| :--- | :---: | :---: | :---: | :---: | :---: | :---: |
| Probabilities $p_{l}$ | $\frac{1}{6}$ | $\frac{1}{6}$ | $\frac{1}{6}$ | $\frac{1}{6}$ |... | 3.5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 32,281 |
96. Let $X$ be the number of points that fall when a die is rolled once. Find the variance of the random variable $X$. | Solution. The distribution law of the random variable $X$ and its mathematical expectation $M(X)=3.5$ were found in the previous example. Let's find the deviations for $x_{1}, x_{2}, \ldots, x_{6}$:
$$
\begin{aligned}
x_{1}^{0}=1-3.5 ; x_{2}^{0}=2-3.5 ; x_{3}^{0} & =3-3.5 ; x_{9}^{0}=4-3.5 ; x_{5}^{0}=5-3.5 ; \\
x_{6}... | \frac{35}{12} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 32,282 |
1. Two dice are thrown. Find the probability that the sum of the points on the faces that come up is even, and that a six appears on at least one of the dice. | The problem is solved. On the emerged face of the "first" die, one point, two points, ..., six points can appear. Similar six elementary outcomes are possible when throwing the "second" die. Each outcome of throwing the "first" die can combine with each outcome of throwing the "second". Thus, the total number of possib... | \frac{5}{36} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 32,283 |
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