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Find the solution of the initial value problem\n\n\[ \n{y}^{\prime \prime } + y = 0,\;y\left( 0\right) \text{ and }{y}^{\prime }\left( 0\right) = 1.\n\] | We know that \( y = \sin x, y = \cos x \), and more generally \( y = {c}_{1}\sin x + {c}_{2}\cos x \) for any constants \( {c}_{1} \) and \( {c}_{2} \), are all solutions of the differential equation.\n\nAlso, \( y = \sin x \) clearly satisfies the initial conditions, because \( \sin 0 = 0 \) and \( \cos 0 = 1 \) . By ... | No |
Example 2. Solve\n\n\[ \n{y}^{\prime \prime } + {y}^{\prime } = 0\text{.} \n\] | By inspection we see that \( {y}_{1} = 1 \) and \( {y}_{2} = {e}^{-x} \) are solutions. It is obvious that neither function is a constant multiple of the other, so (assuming the theorem stated above, but not yet proved) we conclude that\n\n\[ \ny = {c}_{1} + {c}_{2}{e}^{-x} \n\]\n\nis the general solution. | No |
Example 3. Solve\n\n\[ \n{x}^{2}{y}^{\prime \prime } + {2x}{y}^{\prime } - {2y} = 0. \n\] | Since differentiating a power pushes down the exponent by one unit, the form of this equation suggests that we look for possible solutions of the type \( y = {x}^{\prime \prime } \) . On substituting this in the differential equation and dividing by the common factor \( {x}^{\prime \prime } \), we obtain the quadratic ... | Yes |
Lemma 1. If \( {y}_{1}\left( x\right) \) and \( {y}_{2}\left( x\right) \) are any two solutions of equation (1) on \( \left\lbrack {a, b}\right\rbrack \), then their Wronskian \( W = W\left( {{y}_{1},{y}_{2}}\right) \) is either identically zero or never zero on \( \left\lbrack {a, b}\right\rbrack \) . | Proof. We begin by observing that\n\n\[ \n{W}^{\prime } = {y}_{1}{y}_{2}^{\prime \prime } + {y}_{1}^{\prime }{y}_{2}^{\prime } - {y}_{2}{y}_{1}^{\prime \prime } - {y}_{2}^{\prime }{y}_{1}^{\prime } \n\]\n\n\[ \n= {y}_{1}{y}_{2}^{\prime \prime } - {y}_{2}{y}_{1}^{\prime \prime }\text{.} \n\]\n\nNext, since \( {y}_{1} \)... | Yes |
Lemma 2. If \( {y}_{1}\left( x\right) \) and \( {y}_{2}\left( x\right) \) are two solutions of equation (1) on \( \left\lbrack {a, b}\right\rbrack \), then they are linearly dependent on this interval if and only if their Wronskian \( W\left( {{y}_{1},{y}_{2}}\right) = {y}_{1}{y}_{2}^{\prime } - {y}_{2}{y}_{1}^{\prime ... | Proof. We begin by assuming that \( {y}_{1} \) and \( {y}_{2} \) are linearly dependent, and we show as a consequence of this that \( {y}_{1}{y}_{2}^{\prime } - {y}_{2}{y}_{1}^{\prime } = 0 \) . First, if either function is identically zero on \( \left\lbrack {a, b}\right\rbrack \), then the conclusion is clear. We may... | Yes |
Show that \( y = {c}_{1}\sin x + {c}_{2}\cos x \) is the general solution of \( {y}^{\prime \prime } + y = 0 \) on any interval, and find the particular solution for which \( y\left( 0\right) = 2 \) and \( {y}^{\prime }\left( 0\right) = 3 \) . | The fact that \( {y}_{1} = \sin x \) and \( {y}_{2} = \cos x \) are solutions is easily verified by substitution. Their linear independence on any interval \( \left\lbrack {a, b}\right\rbrack \) follows from the observation that \( {y}_{1}/{y}_{2} = \tan x \) is not constant, and also from the fact that their Wronskian... | Yes |
Example 1. \( {y}_{1} = x \) is a solution of \( {x}^{2}{y}^{\prime \prime } + x{y}^{\prime } - y = 0 \) which is simple enough to be discovered by inspection. Find the general solution. | We begin by writing the given equation in the form of (1):\n\n\[ \n{y}^{\prime \prime } + \frac{1}{x}{y}^{\prime } - \frac{1}{{x}^{2}}y = 0.\n\]\n\nSince \( P\left( x\right) = 1/x \), a second linearly independent solution is given by \( {y}_{2} = v{y}_{1} \), where\n\n\[ \nv = \int \frac{1}{{x}^{2}}{e}^{-\int \left( {... | Yes |
Find a particular solution of\n\n\[ \n{y}^{\prime \prime } + y = \sin x. \n\] | The reduced homogeneous equation \( {y}^{\prime \prime } + y = 0 \) has \( y = {c}_{1}\sin x + {c}_{2}\cos x \) as its general solution, so it is useless to take \( {y}_{p} = A\sin x + B\cos x \) as a trial solution for the complete equation (15). We therefore try \( {y}_{p} = x(A\sin x + \) \( B\cos x) \) . This yield... | Yes |
Example 2. Find the general solution of\n\n\[ \n{y}^{\prime \prime } - {y}^{\prime } - {2y} = 4{x}^{2} \n\]\n\n(19) | The reduced homogeneous equation \( {y}^{\prime \prime } - {y}^{\prime } - {2y} = 0 \) has \( {m}^{2} - m - 2 = 0 \) or \( \left( {m - 2}\right) \left( {m + 1}\right) = 0 \) as its auxiliary equation, so the general solution of the reduced equation is \( {y}_{g} = {c}_{1}{e}^{2x} + {c}_{2}{e}^{-x} \). \n\nSince the rig... | Yes |
Example 1. Find a particular solution of \( {y}^{\prime \prime } + y = \csc x \) . | The corresponding homogeneous equation \( {y}^{\prime \prime } + y = 0 \) has \( y\left( x\right) = {c}_{1}\sin x + {c}_{2}\cos x \) as its general solution, so \( {y}_{1} = \sin x,{y}_{1}^{\prime } = \cos x,{y}_{2} = \cos x \), and \( {y}_{2}^{\prime } = - \sin x \) . The Wronskian of \( {y}_{1} \) and \( {y}_{2} \) i... | Yes |
The differential equation\n\n\[ \n{y}^{\left( 4\right) } - 5{y}^{\prime \prime } + {4y} = 0 \n\] | has auxiliary equation\n\n\[ \n{r}^{4} - 5{r}^{2} + 4 = \left( {{r}^{2} - 1}\right) \left( {{r}^{2} - 4}\right) = \left( {r - 1}\right) \left( {r + 1}\right) \left( {r - 2}\right) \left( {r + 2}\right) = 0. \n\]\n\nIts general solution is therefore\n\n\[ \ny = {c}_{1}{e}^{x} + {c}_{2}{e}^{-x} + {c}_{3}{e}^{2x} + {c}_{4... | Yes |
The equation\n\n\[ \n{y}^{\left( 4\right) } - 8{y}^{\prime \prime } + {16y} = 0 \n\] | has auxiliary equation\n\n\[ \n{r}^{4} - 8{r}^{2} + {16} = {\left( {r}^{2} - 4\right) }^{2} = {\left( r - 2\right) }^{2}{\left( r + 2\right) }^{2} = 0, \n\]\n\nso the general solution is\n\n\[ \ny = \left( {{c}_{1} + {c}_{2}x}\right) {e}^{2x} + \left( {{c}_{3} + {c}_{4}x}\right) {e}^{-{2x}}. \n\] | Yes |
Example 3. The equation\n\n\[ \n{y}^{\left( 4\right) } - 2{y}^{\prime \prime \prime } + 2{y}^{\prime \prime } - 2{y}^{\prime } + y = 0 \n\] | has auxiliary equation\n\n\[ \n{r}^{4} - 2{r}^{3} + 2{r}^{2} - {2r} + 1 = 0, \n\]\n\nor, after factoring, \( {}^{12} \)\n\n\[ \n{\left( r - 1\right) }^{2}\left( {{r}^{2} + 1}\right) = 0. \n\]\n\nThe general solution is therefore\n\n\[ \ny = \left( {{c}_{1} + {c}_{2}x}\right) {e}^{x} + {c}_{3}\cos x + {c}_{4}\sin x. \n\... | Yes |
Example 4. Coupled harmonic oscillators. Linear equations of order \( n > 2 \) arise most often in physics by eliminating variables from simultaneous systems of second order equations. We can see an example of this by linking together two simple harmonic oscillators of the kind discussed at the beginning of Section 20.... | We can now obtain a single fourth order equation for \( {x}_{1} \) by solving the first equation for \( {x}_{2} \) and substituting in the second equation (Problem 17). | No |
Find a particular solution of the differential equation \( {y}^{\prime \prime \prime } + 2{y}^{\prime \prime } - {y}^{\prime } = 3{x}^{2} - {2x} + 1 \) . | Our experience in Section 18 suggests that we take a trial solution of the form\n\n\[ y = x\left( {{a}_{0} + {a}_{1}x + {a}_{2}{x}^{2}}\right) \]\n\n\[ = {a}_{0}x + {a}_{1}{x}^{2} + {a}_{2}{x}^{3}. \]\n\nSince \( {y}^{\prime } = {a}_{0} + 2{a}_{1}x + 3{a}_{2}{x}^{2},{y}^{\prime \prime } = 2{a}_{1} + 6{a}_{2}x \), and \... | Yes |
Find a particular solution of \( {y}^{\prime \prime } - 3{y}^{\prime } + {2y} = x{e}^{x} \) . | Solution. We have \( \left( {{D}^{2} - {3D} + 2}\right) y = x{e}^{x} \), so\n\n\[ \left( {D - 1}\right) \left( {D - 2}\right) y = x{e}^{x}\;\text{ and }\;y = \frac{1}{D - 1}\frac{1}{D - 2}x{e}^{x}. \]\n\nBy (9) and an integration by parts, we obtain\n\n\[ \frac{1}{D - 2}x{e}^{x} = {e}^{2x}\int {e}^{-{2x}}x{e}^{x}{dx} =... | Yes |
Find a particular solution of \( {y}^{\prime \prime } - y = {e}^{-x} \) . | Solution. We have \( \left( {{D}^{2} - 1}\right) y = {e}^{-x} \), so\n\n\[ \left( {D - 1}\right) \left( {D + 1}\right) y = {e}^{-x},\;y = \frac{1}{D - 1}\frac{1}{D + 1}{e}^{-x}, \]\n\n\[ \frac{1}{D + 1}{e}^{-x} = {e}^{-x}\int {e}^{x}{e}^{-x}{dx} = x{e}^{-x}, \]\n\n\[ y = \frac{1}{D - 1}x{e}^{-x} = {e}^{x}\int {e}^{-x}x... | Yes |
Solve the problem in Example 1 by this method. | Solution. We have\n\n\[ y = \frac{1}{\left( {D - 1}\right) \left( {D - 2}\right) }x{e}^{x} = \left\lbrack {\frac{1}{D - 2} - \frac{1}{D - 1}}\right\rbrack x{e}^{x} \]\n\n\[ = \frac{1}{D - 2}x{e}^{x} - \frac{1}{D - 1}x{e}^{x} \]\n\n\[ = {e}^{2x}\int {e}^{-{2x}}x{e}^{x}{dx} - {e}^{x}\int {e}^{-x}x{e}^{x}{dx} \]\n\n\[ = -... | Yes |
Example 4. Solve the problem in Example 2 by this method. | Solution. We have\n\n\[ y = \frac{1}{\left( {D - 1}\right) \left( {D + 1}\right) }{e}^{-x} = \frac{1}{2}\left\lbrack {\frac{1}{D - 1} - \frac{1}{D + 1}}\right\rbrack {e}^{-x} \]\n\n\[ = \frac{1}{2}{e}^{x}\int {e}^{-x}{e}^{-x}{dx} - \frac{1}{2}{e}^{-x}\int {e}^{x}{e}^{-x}{dx} \]\n\n\[ = - \frac{1}{4}{e}^{-x} - \frac{1}{... | Yes |
Example 5. Find a particular solution of \( {y}^{\prime \prime \prime } - 2{y}^{\prime \prime } + y = {x}^{4} + {2x} + 5 \) . | Solution. We have \( \left( {{D}^{3} - 2{D}^{2} + 1}\right) y = {x}^{4} + {2x} + 5 \), so\n\n\[ y = \frac{1}{1 - 2{D}^{2} + {D}^{3}}\left( {{x}^{4} + {2x} + 5}\right) .\n\]\n\nBy ordinary long division we find that\n\n\[ \frac{1}{1 - 2{D}^{2} + {D}^{3}} = 1 + 2{D}^{2} - {D}^{3} + 4{D}^{4} - 4{D}^{5} + \cdots ,\n\]\n\ns... | Yes |
Example 6. Find a particular solution of \( {y}^{\prime \prime \prime } + {y}^{\prime \prime } + {y}^{\prime } + y = {x}^{5} - 2{x}^{2} + x \) . | Solution. We have \( \left( {{D}^{3} + {D}^{2} + D + 1}\right) y = {x}^{5} - 2{x}^{2} + x \), so\n\n\[ y = \frac{1}{1 + D + {D}^{2} + {D}^{3}}\left( {{x}^{5} - 2{x}^{3} + x}\right) \]\n\n\[ = \frac{1}{1 - {D}^{4}}\left( {1 - D}\right) \left( {{x}^{5} - 2{x}^{2} + x}\right) \]\n\n\[ = \frac{1}{1 - {D}^{4}}\left\lbrack {... | Yes |
Example 7. Solve the problem in Example 1 by this method. | Solution. We have \( \left( {{D}^{2} - {3D} + 2}\right) y = x{e}^{x} \), so\n\n\[ y = \frac{1}{{D}^{2} - {3D} + 2}x{e}^{x} = {e}^{x}\frac{1}{{\left( D + 1\right) }^{2} - 3\left( {D + 1}\right) + 2}x \]\n\n\[ = {e}^{x}\frac{1}{{D}^{2} - D}x = - {e}^{x}\frac{1}{D}\frac{1}{1 - D}x \]\n\n\[ = - {e}^{x}\left( {\frac{1}{D} +... | Yes |
Find the Fourier series of the function \( f\left( x\right) = x, - \pi \leq x \leq \pi \) . | First, by (3) we have\n\n\[ \n{a}_{0} = \frac{1}{\pi }{\int }_{-\pi }^{\pi }{xdx} = {\left. \frac{1}{\pi } \cdot \frac{{x}^{2}}{2}\right\rbrack }_{-\pi }^{\pi } = 0.\n\]\n\nIf \( n \geq 1 \), then we find \( {a}_{n} \) by using (7) and integrating by parts with \( u = x \) , \( {dv} = \cos {nxdx} \)\n\n\[ \n{a}_{n} = \... | Yes |
Find the Fourier series of the function defined by\n\n\[ f\left( x\right) = 0, - \pi \leq x < 0 \]\n\n\[ f\left( x\right) = \pi ,0 \leq x \leq \pi . \] | By (3), (7) and (9) we have\n\n\[ {a}_{0} = \frac{1}{\pi }\left\lbrack {{\int }_{-\pi }^{0}{0dx} + {\int }_{0}^{\pi }{\pi dx}}\right\rbrack = \pi \]\n\n\[ {a}_{n} = \frac{1}{\pi }{\int }_{0}^{\pi }\pi \cos {nxdx} = 0, n \geq 1 \]\n\n\[ {b}_{n} = \frac{1}{\pi }{\int }_{0}^{\pi }\pi \sin {nxdx} = \frac{1}{n}\left( {1 - \... | Yes |
Find the Fourier series of the function defined by\n\n\[ f\left( x\right) = - \frac{\pi }{2},\; - \pi \leq x < 0 \]\n\n\[ f\left( x\right) = \frac{\pi }{2},\;0 \leq x \leq \pi . \] | This is the function in Example 2 minus the constant \( \pi /2 \) . Its Fourier series can therefore be obtained by subtracting \( \pi /2 \) from the series (13), which gives\n\n\[ f\left( x\right) = 2\left( {\sin x + \frac{\sin {3x}}{3} + \frac{\sin {5x}}{5} + \cdots }\right) . \] | No |
Find the Fourier series of the function defined by\n\n\[ f\left( x\right) = - \frac{\pi }{2} - \frac{1}{2}x,\; - \pi \leq x < 0; \]\n\n\[ f\left( x\right) = \frac{\pi }{2} - \frac{1}{2}x,\;0 \leq x \leq \pi . \] | This is the function defined in Example 3 minus one-half the function in Example 1. The Fourier series can therefore be obtained by subtracting one-half the series (11) term by term the series (14):\n\n\[ f\left( x\right) = 2\left( {\sin x + \frac{\sin {3x}}{3} + \frac{\sin {5x}}{5} + \cdots }\right) \]\n\n\[ - \left( ... | Yes |
(a) First, we briefly consider the function \( f\left( x\right) = x \) on the interval \( - \pi \leq x \leq \pi \) . Since this is an odd function, its Fourier series is automatically a sine series, and therefore it is not necessary to bother calculating the cosine coefficients. We found in Section 33 that the Fourier ... | \[ x = 2\left( {\sin x - \frac{\sin {2x}}{2} + \frac{\sin {3x}}{3} - \cdots }\right) ,\] | Yes |
Example 2. Find the sine series, and also the cosine series, for the function \( f\left( x\right) = \cos x,0 \leq x \leq \pi \) . | For the sine series, (6) gives\n\n\[ \n{a}_{n} = 0\;\text{ and }\;{b}_{n} = \frac{2}{\pi }{\int }_{0}^{\pi }\cos x\sin {nxdx}.\n\]\n\nFor \( n = 1 \) we have \( {b}_{1} = 0 \), and for \( n > 1 \) a short calculation yields\n\n\[ \n{b}_{n} = \frac{2n}{\pi }\left\lbrack \frac{1 + {\left( -1\right) }^{n}}{{n}^{2} - 1}\ri... | Yes |
Expand \( f\\left( x\\right) \) in a Fourier series on the interval \( - 2 \\leq x < 2 \) if \( f\\left( x\\right) = 0 \) for \( - 2 \\leq x < 0 \) and \( f\\left( x\\right) = 1 \) for \( 0 \\leq x < 2 \). | Here we introduce \( t \) by writing\n\n\\[ \n\\frac{t}{\\pi } = \\frac{x}{2},\\;\\text{ so }\\;t = \\frac{\\pi x}{2}\\;\\text{ and }\\;x = \\frac{2t}{\\pi }.\n\\]\n\nThen \( g\\left( t\\right) = 0 \) for \( - \\pi \\leq t < 0 \) and \( g\\left( t\\right) = 1 \) for \( 0 \\leq t < \\pi \), and we have\n\n\\[ \n{a}_{0} ... | Yes |
Use the differential equation \( {y}^{\prime \prime } + {\lambda y} = 0 \), or equivalently \( {y}^{\prime \prime } = - {\lambda y} \), to show that the trigonometric sequence (2) is orthogonal on \( \left\lbrack {-\pi ,\pi }\right\rbrack \) . | Let \( m \) and \( n \) be positive integers. If \( {y}_{m} = \sin {mx} \) or \( \cos {mx} \) and \( {y}_{n} = \sin {nx} \) or \( \cos {nx} \), then\n\n\[ \n{y}_{m}^{\prime \prime } = - {m}^{2}{y}_{m}\;\text{ and }\;{y}_{n}^{\prime \prime } = - {n}^{2}{y}_{n} \n\]\n\nIf the first equation is multiplied by \( {y}_{n} \)... | Yes |
For each positive integer \( n \), the \( n \) th partial sum of the Fourier series of \( f \), namely,\n\n\[ \mathop{\sum }\limits_{{k = 1}}^{n}{a}_{k}{\phi }_{k} = {a}_{1}{\phi }_{1} + \cdots + {a}_{n}{\phi }_{n} \]\n\ngives a smaller mean square error \( {E}_{n} = {\int }_{a}^{b}{\left( f - {p}_{n}\right) }^{2}{dx} ... | Formula (6) tells us that we always have \( {E}_{n} \geq 0 \), because the integrand in (6), being a square, is nonnegative. Since \( {E}_{n} \geq 0 \) for all choices of the \( {b}_{k} \), it is clear that the minimum value of \( {E}_{n} \) (which arises when \( {b}_{k} = {a}_{k} \) ) is also \( \geq 0 \) . Therefore ... | Yes |
Theorem 2. If the numbers \( {a}_{n} = {\int }_{a}^{b}f{\phi }_{n}{dx} \) are the Fourier coefficients of \( f \) with respect to the orthonormal sequence \( \left\{ {\phi }_{n}\right\} \), then the series \( \sum {a}_{n}^{2} \) converges and satisfies Bessel's inequality, | \[ \mathop{\sum }\limits_{{n = 1}}^{\infty }{a}_{n}^{2} \leq {\int }_{a}^{b}{\left\lbrack f\left( x\right) \right\rbrack }^{2}{dx} \] | Yes |
Theorem 5. If \( f\left( x\right) \) is any function defined and integrable on \( \left\lbrack {-\pi ,\pi }\right\rbrack \), then \( f\left( x\right) \) is represented by its ordinary Fourier series in the sense of mean convergence,\n\n\[ f\left( x\right) = \frac{1}{2}{a}_{0} + \mathop{\sum }\limits_{{n = 1}}^{\infty }... | \( {}^{16} \) The basic tools for the proof we have in mind are two major theorems of classical analysis, Fejer's summability theorem and the Weierstrass approximation theorem. | No |
Suppose that the thin rod discussed above is first immersed in boiling water so that its temperature is \( {100}^{ \circ }\mathrm{C} \) throughout, and then removed from the water at time \( t = 0 \) with its ends immediately put in ice so that these ends are kept at temperature \( {0}^{ \circ }\mathrm{C} \) . Find the... | This is the special case of the above discussion in which the initial temperature distribution is given by the constant function\n\n\[ f\left( x\right) = {100},0 < x < \pi \text{.} \]\n\nWe must therefore find the sine series of this function, which we can either calculate from scratch by using (15) or obtain in some o... | Yes |
Example 2. Find the steady-state temperature of the thin rod discussed above if the fixed temperatures at the ends \( x = 0 \) and \( x = \pi \) are \( {w}_{1} \) and \( {w}_{2} \) , respectively. | Solution. \ | No |
Lemma 1. Let \( y\left( x\right) \) and \( z\left( x\right) \) be nontrivial solutions of \[ {y}^{\prime \prime } + q\left( x\right) y = 0 \] and \[ {z}^{\prime \prime } + r\left( x\right) z = 0, \] where \( q\left( x\right) \) and \( r\left( x\right) \) are positive continuous functions such that \( q\left( x\right) >... | Proof. By the Sturm comparison theorem (Theorem 25-B), \( y\left( x\right) \) has at least one zero in each of the open intervals \( \left( {{b}_{0},{b}_{1}}\right) ,\left( {{b}_{1},{b}_{2}}\right) ,\ldots ,\left( {{b}_{n - 1},{b}_{n}}\right) \), and both statements follow at once from this. | Yes |
Lemma 2. Let \( q\left( x\right) \) be a positive continuous function that satisfies the inequalities\n\n\[ 0 < {m}^{2} < q\left( x\right) < {M}^{2} \]\n\non a closed interval \( \left\lbrack {a, b}\right\rbrack \) . If \( y\left( x\right) \) is a nontrivial solution of \( {y}^{\prime \prime } + q\left( x\right) y = 0 ... | Proof. To prove (1), we begin by comparing the given equation with \( {z}^{\prime \prime } + {m}^{2}z = 0 \) . A nontrivial solution of this that vanishes at \( {x}_{1} \) is \( z\left( x\right) = \sin m\left( {x - {x}_{1}}\right) \) . Since the next zero of \( z\left( x\right) \) is \( {x}_{1} + \pi /m \), and Theorem... | Yes |
Lemma 3. Let \( q\left( x\right) \) be a positive continuous function and consider the differential equation\n\n\[ \n{y}^{\prime \prime } + {\lambda q}\left( x\right) y = 0 \n\]\n\non a closed interval \( \left\lbrack {a, b}\right\rbrack \) . For each \( \lambda \), let \( {\mathrm{y}}_{\lambda }\left( x\right) \) be t... | Proof. It is clear by Theorem 24-B that \( {y}_{\lambda }\left( x\right) \) has no zeros to the right of \( a \) when \( \lambda \leq 0 \) . Our plan is to watch the oscillation behavior of \( {y}_{\lambda }\left( x\right) \) as \( \lambda \) increases from 0 . We begin with the observation that by the continuity of \(... | Yes |
Example 1. Find the solution of\n\n\[ \n{y}^{\prime \prime } + {4y} = {4x} \]\n\n(6)\n\nthat satisfies the initial conditions \( y\left( 0\right) = 1 \) and \( {y}^{\prime }\left( 0\right) = 5 \) . | When \( L \) is applied to both sides of (6), we get\n\n\[ \nL\left\lbrack {y}^{\prime \prime }\right\rbrack + {4L}\left\lbrack y\right\rbrack = {4L}\left\lbrack x\right\rbrack \]\n\n(7)\n\nIf we recall that \( L\left( x\right) = 1/{p}^{2} \), and use (4) and the initial conditions, then (7) becomes\n\n\[ \n{p}^{2}L\le... | Yes |
\[ L\left\lbrack {\sin {bx}}\right\rbrack = \frac{b}{{p}^{2} + {b}^{2}} \] | \[ L\left\lbrack {{e}^{ax}\sin {bx}}\right\rbrack = \frac{b}{{\left( p - a\right) }^{2} + {b}^{2}}. \] | No |
[{L}^{-1}\left\lbrack \frac{1}{{p}^{2}}\right\rbrack = x] | [{L}^{-1}\left\lbrack \frac{1}{{\left( p - a\right) }^{2}}\right\rbrack = {e}^{ax}x] | No |
Since \( L\left\lbrack {\sin {ax}}\right\rbrack = a/\left( {{p}^{2} + {a}^{2}}\right) \), we have | \[ L\left\lbrack {x\sin {ax}}\right\rbrack = - \frac{d}{dp}\left( \frac{a}{{p}^{2} + {a}^{2}}\right) = \frac{2ap}{{\left( {p}^{2} + {a}^{2}\right) }^{2}}. \] | Yes |
We know from Section 49 that \( L\left\lbrack {x}^{-1/2}\right\rbrack = \sqrt{\pi /p} \), so | \[ L\left\lbrack {x}^{1/2}\right\rbrack = L\left\lbrack {x\left( {x}^{-1/2}\right) }\right\rbrack = - \frac{d}{dp}\left( \sqrt{\frac{\pi }{p}}\right) = \frac{1}{2p}\sqrt{\frac{\pi }{p}}. \] | Yes |
Bessel's equation of order zero is\n\n\[ x{y}^{\prime \prime } + {y}^{\prime } + {xy} = 0. \] | It is known to have a single solution \( y\left( x\right) \) with the property that \( y\left( 0\right) = 1 \) . To find this solution, we apply \( L \) to (8) and use (5) and (7), which gives\n\n\[ - \frac{d}{dp}\left\lbrack {{p}^{2}Y - p}\right\rbrack + {pY} - 1 - \frac{dY}{dp} = 0 \]\n\nor\n\n\[ \left( {{p}^{2} + 1}... | Yes |
Example 4. Since \( L\left\lbrack {\sin x}\right\rbrack = 1/\left( {{p}^{2} + 1}\right) \) ,(12) gives | \[ {\left. {\int }_{0}^{\infty }\frac{\sin x}{x}dx = {\int }_{0}^{\infty }\frac{dp}{{p}^{2} + 1} = {\tan }^{-1}p\right\rbrack }_{0}^{\infty } = \frac{\pi }{2}. \] | Yes |
The integral equation\n\n\[ y\left( x\right) = {x}^{3} + {\int }_{0}^{x}\sin \left( {x - t}\right) y\left( t\right) {dt} \] | is of this type, and by applying \( L \) we get\n\n\[ L\left\lbrack {y\left( x\right) }\right\rbrack = L\left\lbrack {x}^{3}\right\rbrack + L\left\lbrack {\sin x}\right\rbrack L\left\lbrack {y\left( x\right) }\right\rbrack .\n\]\nSolving for \( L\left\lbrack {y\left( x\right) }\right\rbrack \) yields\n\n\[ L\left\lbrac... | Yes |
Use formula (13) to solve \( {y}^{\prime \prime } + {y}^{\prime } - {6y} = 2{e}^{3t} \), where \( y\left( 0\right) = {y}^{\prime }\left( 0\right) = 0 \) . | Here we have\n\n\[ L\left\lbrack {A\left( t\right) }\right\rbrack = \frac{1}{p\left( {{p}^{2} + p - 6}\right) } \]\n\nso by partial fractions and inversion we find that\n\n\[ A\left( t\right) = - \frac{1}{6} + \frac{1}{15}{e}^{-{3t}} + \frac{1}{10}{e}^{2t}. \]\n\nSince \( f\left( t\right) = 2{e}^{3t},{f}^{\prime }\left... | Yes |
Consider again the equation \( {y}^{\prime \prime } + {y}^{\prime } - {6y} = 2{e}^{3t} \) solved in Example 1. | We have\n\n\[ h\left( t\right) = {L}^{-1}\left\lbrack \frac{1}{\left( {p + 3}\right) \left( {p - 2}\right) }\right\rbrack = \frac{1}{5}\left( {{e}^{2t} - {e}^{-{3t}}}\right) ,\]\n\nso that\n\n\[ y\left( t\right) = {\int }_{0}^{t}\frac{1}{5}\left\lbrack {{e}^{2\left( {t - \tau }\right) } - {e}^{-3\left( {t - \tau }\righ... | Yes |
Example 1. In the case of the system \[ \left\{ \begin{array}{l} \frac{dx}{dt} = x + y \\ \frac{dy}{dt} = {4x} - {2y} \end{array}\right. \] | The auxiliary equation here is \[ {m}^{2} + m - 6 = 0\;\text{ or }\;\left( {m + 3}\right) \left( {m - 2}\right) = 0, \] so \( {m}_{1} \) and \( {m}_{2} \) are -3 and 2 . With \( m = - 3 \) ,(9) becomes \[ {4A} + B = 0 \] \[ {4A} + B = 0\text{.} \] A simple nontrivial solution of this system is \( A = 1, B = - 4 \), so ... | Yes |
In the case of the system\n\n\\[ \n\\left\\{ \\begin{array}{l} \\frac{dx}{dt} = {3x} - {4y} \\\\ \\frac{dy}{dt} = x - y, \\end{array}\\right.\n\\]\n\n(21) | The auxiliary equation is\n\n\\[ \n{m}^{2} - {2m} + 1 = 0\\;\\text{ or }\\;{\\left( m - 1\\right) }^{2} = 0,\n\\]\n\nwhich has equal real roots 1 and 1 . With \\( m = 1 \\) ,(22) becomes\n\n\\[ \n{2A} - {4B} = 0\n\\]\n\n\\[ \nA - {2B} = 0\\text{.}\n\\]\n\nA simple nontrivial solution of this system is \\( A = 2, B = 1 ... | Yes |
Consider the system\n\n\[ \left\{ \begin{array}{l} \frac{dx}{dt} = x \\ \frac{dy}{dt} = - x + {2y}. \end{array}\right. \] | It is clear that the origin is the only critical point, and the general solution can be found quite easily by the methods of Section 56:\n\n\[ \left\{ \begin{array}{l} x = {c}_{1}{e}^{t} \\ y = {c}_{1}{e}^{t} + {c}_{2}{e}^{2t} \end{array}\right. \]\n\nWhen \( {c}_{1} = 0 \), we have \( x = 0 \) and \( y = {c}_{2}{e}^{2... | Yes |
The system\n\n\[ \left\{ \begin{array}{l} \frac{dx}{dt} = - y \\ \frac{dy}{dt} = x \end{array}\right. \] | has the origin as its only critical point, and its general solution is\n\n\[ \left\{ \begin{array}{l} x = - {c}_{1}\sin t + {c}_{2}\cos t \\ y = {c}_{1}\cos t + {c}_{2}\sin t. \end{array}\right. \] | Yes |
If \( a \) is an arbitrary constant, then the system\n\n\[ \left\{ \begin{array}{l} \frac{dx}{dt} = {ax} - y \\ \frac{dy}{dt} = x + {ay} \end{array}\right. \]\n\nhas the origin as its only critical point (why?). | The differential equation of the paths,\n\n\[ \frac{dy}{dx} = \frac{x + {ay}}{{ax} - y} \]\n\nis most easily solved by introducing polar coordinates \( r \) and \( \theta \) defined by \( x = r\cos \theta \) and \( y = r\sin \theta \) . Since\n\n\[ {r}^{2} = {x}^{2} + {y}^{2}\;\text{ and }\;\theta = {\tan }^{-1}\frac{y... | No |
Consider the equation of motion of a mass \( m \) attached to a spring:\n\n\[ m\frac{{d}^{2}x}{d{t}^{2}} + c\frac{dx}{dt} + {kx} = 0. \] | Here \( c \geq 0 \) is a constant representing the viscosity of the medium through which the mass moves, and \( k > 0 \) is the spring constant. The autonomous system equivalent to (5) is\n\n\[ \left\{ \begin{array}{l} \frac{dx}{dt} = y \\ \frac{dy}{dt} = - \frac{k}{m}x - \frac{c}{m}y, \end{array}\right. \]\n\nand its ... | Yes |
The system\n\n\\[ \n\\left\\{ \\begin{array}{l} \\frac{dx}{dt} = - {2xy} \\\\ \\frac{dy}{dt} = {x}^{2} - {y}^{3} \\end{array}\\right.\n\\]\n\n(8)\n\nhas \\( \\left( {0,0}\\right) \\) as an isolated critical point. Let us try to prove stability by constructing a Liapunov function of the form \\( E\\left( {x, y}\\right) ... | It is clear that\n\n\\[ \n\\frac{\\partial E}{\\partial x}F + \\frac{\\partial E}{\\partial y}G = 2{\\max }^{{2m} - 1}\\left( {-{2xy}}\\right) + {2nb}{y}^{{2n} - 1}\\left( {{x}^{2} - {y}^{3}}\\right)\n\\]\n\n\\[ \n= \\left( {-4{\\max }^{2m}y + {2nb}{x}^{2}{y}^{{2n} - 1}}\\right) - {2nb}{y}^{{2n} + 2}.\n\\]\n\nWe wish t... | Yes |
Example 1. In the case of the system\n\n\\[ \n\\left\\{ \\begin{array}{l} \\frac{dx}{dt} = - {2x} + {3y} + {xy} \\\\ \\frac{dy}{dt} = - x + y - {2x}{y}^{2} \\end{array}\\right.\n\\]\n\n(7) | we have\n\n\\[ \n\\left| \\begin{array}{ll} {a}_{1} & {b}_{1} \\\\ {a}_{2} & {b}_{2} \\end{array}\\right| = \\left| \\begin{array}{ll} - 2 & 3 \\\\ - 1 & 1 \\end{array}\\right| = 1 \\neq 0,\n\\]\n\nso (5) is satisfied. Furthermore, by using polar coordinates we see that\n\n\\[ \n\\frac{\\left| f\\left( x, y\\right) \\r... | Yes |
Example 2. We know from Section 58 that the equation of motion for the damped vibrations of a pendulum is\n\n\[ \n\\frac{{d}^{2}x}{d{t}^{2}} + \\frac{c}{m}\\frac{dx}{dt} + \\frac{g}{a}\\sin x = 0 \n\]\n\nwhere \( c \) is a positive constant. The equivalent nonlinear system is\n\n\[ \n\\left\\{ \\begin{array}{l} \\frac{... | Let us now write (14) in the form\n\n\[ \n\\left\\{ \\begin{array}{l} \\frac{dx}{dt} = y \\\\ \\frac{dy}{dt} = - \\frac{g}{a}x - \\frac{c}{m}y + \\frac{g}{a}\\left( {x - \\sin x}\\right) . \\end{array}\\right. \n\]\n\n(15)\n\nIt is easy to see that\n\n\[ \n\\frac{x - \\sin x}{\\sqrt{{x}^{2} + {y}^{2}}} \\rightarrow 0 \... | Yes |
We saw in Section 58 that the equation of motion of an undamped pendulum is\n\n\\[ \n\\frac{{d}^{2}x}{d{t}^{2}} + k\\sin x = 0 \n\\]\n\n(10)\n\nwhere \\( k \\) is a positive constant. Since this equation is of the form (2), it can be interpreted as describing the undamped rectilinear motion of a unit mass under the inf... | and by separating variables and integrating, we see that the equation of the family of paths is\n\n\\[ \n\\frac{1}{2}{y}^{2} + \\left( {k - k\\cos x}\\right) = E. \n\\]\n\nThis is evidently of the form (8), where \\( m = 1 \\) and\n\n\\[ \nV\\left( x\\right) = {\\int }_{0}^{x}f\\left( x\\right) {dx} = k - k\\cos x \n\\... | Yes |
To find the shortest curve joining two points \( \left( {{x}_{1},{y}_{1}}\right) \) and \( \left( {{x}_{2},{y}_{2}}\right) \) -which we know intuitively to be a straight line-we must minimize the arc length integral\n\n\[ I = {\int }_{{x}_{1}}^{{x}_{2}}\sqrt{1 + {\left( {y}^{\prime }\right) }^{2}}{dx} \] | The variables \( x \) and \( y \) are missing from \( f\left( {y}^{\prime }\right) = \sqrt{1 + {\left( {y}^{\prime }\right) }^{2}} \), so this problem falls under Case A. Since\n\n\[ {f}_{{y}^{\prime }{y}^{\prime }} = \frac{{\partial }^{2}f}{\partial {y}^{\prime 2}} = \frac{1}{{\left\lbrack 1 + {\left( {y}^{\prime }\ri... | Yes |
To find the curve joining the points \( \left( {{x}_{1},{y}_{1}}\right) \) and \( \left( {{x}_{2},{y}_{2}}\right) \) that yields a surface of revolution of minimum area when revolved about the \( x \) -axis, we must minimize\n\n\[ I = {\int }_{{x}_{1}}^{{x}_{2}}{2\pi y}\sqrt{1 + {\left( {y}^{\prime }\right) }^{2}}{dx} ... | The variable \( x \) is missing from \( f\left( {y,{y}^{\prime }}\right) = {2\pi y}\sqrt{1 + {\left( {y}^{\prime }\right) }^{2}} \), so Case \( \mathrm{C} \) tells us that Euler's equation becomes\n\n\[ \frac{y{\left( {y}^{\prime }\right) }^{2}}{\sqrt{1 + {\left( {y}^{\prime }\right) }^{2}}} - y\sqrt{1 + {\left( {y}^{\... | Yes |
To find the curve of quickest descent in Figure 99, we must minimize\n\n\\[ I = {\\int }_{{x}_{1}}^{{x}_{2}}\\frac{\\sqrt{1 + {\\left( {y}^{\\prime }\\right) }^{2}}}{\\sqrt{2gy}}{dx} \\]\n\nAgain the variable \\( x \\) is missing from the function \\( f\\left( {y,{y}^{\\prime }}\\right) = \\sqrt{1 + {\\left( {y}^{\\pri... | so by Case C, Euler's equation becomes\n\n\\[ \\frac{{\\left( {y}^{\\prime }\\right) }^{2}}{\\sqrt{y}\\sqrt{1 + {\\left( {y}^{\\prime }\\right) }^{2}}} - \\frac{\\sqrt{1 + {\\left( {y}^{\\prime }\\right) }^{2}}}{\\sqrt{y}} = {c}_{1} \\]\n\nThis reduces to\n\n\\[ y\\left\\lbrack {1 + {\\left( {y}^{\\prime }\\right) }^{2... | Yes |
We shall find the curve of fixed length \( L \) that joins the points \( \\left( {0,0}\\right) \) and \( \\left( {1,0}\\right) \), lies above the \( x \) -axis, and encloses the maximum area between itself and the \( x \) -axis. This is a restricted version of the original isoperimetric problem in which part of the cur... | Here we have \( F = y + \\lambda \\sqrt{1 + {\\left( {y}^{\\prime }\\right) }^{2}} \), so Euler’s equation is\n\n\[ \n\\frac{d}{dx}\\left( \\frac{\\lambda {y}^{\\prime }}{\\sqrt{1 + {\\left( {y}^{\\prime }\\right) }^{2}}}\\right) - 1 = 0 \n\]\n\n(18)\n\nor, after carrying out the differentiation,\n\n\[ \n\\frac{{y}^{\\... | Yes |
In Example 1 it is clearly necessary to have \( L > 1 \) . Also, if \( L > \pi /2 \) the circular arc determined by (20) will not define \( y > 0 \) as a single-valued function of \( x \) . We can avoid these artificial issues by considering curves in parametric form \( x = x\left( t\right) \) and \( y = y\left( t\righ... | Here we have\n\n\[ \nF = \frac{1}{2}\left( {x\dot{y} + y\dot{x}}\right) + \lambda \sqrt{{\dot{x}}^{2} + {\dot{y}}^{2}} \n\]\n\nso the Euler equations (17) are\n\n\[ \n\frac{d}{dt}\left( {-\frac{1}{2}y + \frac{\lambda \dot{x}}{\sqrt{{\dot{x}}^{2} + {\dot{y}}^{2}}}}\right) - \frac{1}{2}\dot{y} = 0 \n\]\n\nand\n\n\[ \n\fr... | Yes |
Example 3. If we choose the surface (21) to be the sphere \( {x}^{2} + {y}^{2} + {z}^{2} = {a}^{2} \) | then \( G\left( {x, y, z}\right) - {x}^{z} + {y}^{2} + {z}^{2} - {a}^{2} \) and (28) is\n\n\[ \frac{f\ddot{x} - \dot{x}\dot{f}}{{2x}{f}^{2}} = \frac{f\ddot{y} - \dot{y}\dot{f}}{{2y}{f}^{2}} = \frac{f\ddot{z} - \dot{z}\dot{f}}{{2z}{f}^{2}} \]\n\nwhich can be rewritten in the form\n\n\[ \frac{x\ddot{y} - y\ddot{x}}{x\dot... | Yes |
If a particle of mass \( m \) is constrained to move on a given surface \( G\left( {x, y, z}\right) = 0 \), and if no force acts on it, then it glides along a geodesic. | To establish this, we begin by observing that since no force is present we have \( V = 0 \), so the Lagrangian \( L = T - V \) reduces to \( T \) where\n\n\[ T = \frac{1}{2}m\left\lbrack {{\left( \frac{dx}{dt}\right) }^{2} + {\left( \frac{dy}{dt}\right) }^{2} + {\left( \frac{dz}{dt}\right) }^{2}}\right\rbrack .\n\]\n\n... | Yes |
If a particle of mass \( m \) moves in a plane under the influence of a gravitational force of magnitude \( {km}/{r}^{2} \) directed toward the origin, then it is natural to choose polar coordinates as the generalized coordinates: \( {q}_{1} = r \) and \( {q}_{2} = \theta \) . It is easy to see that \( T = \left( {m/2}... | Since \( L \) does not depend explicitly on \( \theta \), equation (6) shows that \( \partial L/\partial \dot{\theta } = m{r}^{2}\dot{\theta } \) is constant, so \[ {r}^{2}\frac{d\theta }{dt} = h \] (7) for some constant \( h \) assumed to be positive. We next observe that (5) can easily be written in the form \[ \frac... | Yes |
In its simplest form, the problem of minimal surfaces was first proposed by Euler as follows: to find the surface of smallest area bounded by a given closed curve in space. If we assume that this curve projects down to a closed curve \( C \) surrounding a region \( R \) in the \( {xy} \) - plane, and also that the surf... | Euler’s equation (11) for this integral is \[ \frac{\partial }{\partial x}\left( \frac{{z}_{x}}{\sqrt{1 + {z}_{x}^{2} + {z}_{y}^{2}}}\right) + \frac{\partial }{\partial y}\left( \frac{{z}_{y}}{\sqrt{1 + {z}_{x}^{2} + {z}_{y}^{2}}}\right) = 0, \] which can be written in the form \[ {z}_{xx}\left( {1 + {z}_{y}^{2}}\right... | Yes |
In Section 40 we obtained the one-dimensional wave equation from Newton's second law of motion. In this example we deduce it from Hamilton's principle with the aid of equation (11). Assume the following: a string of constant linear mass density \( m \) is stretched with a tension \( T \) and fastened to the \( x \) -ax... | When the string is displaced, an element of length \( {dx} \) is stretched to a length \( {ds} \), where\n\n\[ {ds} = \sqrt{1 + {y}_{x}^{2}}{dx} \cong \left( {1 + \frac{1}{2}{y}_{x}^{2}}\right) {dx}. \]\n\nThis approximation results from expanding \( \sqrt{1 + {y}_{x}^{2}} = {\left( 1 + {y}_{x}^{2}\right) }^{1/2} \) in... | Yes |
A certain population has \( P\left( t\right) \) individuals at time \( t \), and its rate of growth is proportional to its size (i.e., \( {\mathrm{P}}^{\prime }\left( \mathrm{t}\right) = \mathrm{{aP}}\left( \mathrm{t}\right) \), for some constant \( \mathrm{a} > 0 \) ). Find \( \mathrm{P}\left( \mathrm{t}\right) \) in ... | The equation \( {\mathrm{P}}^{\prime }\left( \mathrm{t}\right) = \mathrm{{aP}}\left( \mathrm{t}\right) \) is separable, since we can write it in the form\n\n\[ \frac{\mathrm{{dP}}}{\mathrm{P}} = \mathrm{a}\mathrm{{dt}} \]\n\nIntegrating, we obtain (assuming \( \mathrm{P} > 0 \) ) \( \log \left( \mathrm{P}\right) = \mat... | Yes |
A particle is carried along by a fluid flow in the xy-plane. Suppose that the velocity of the fluid at the arbitrary point \( \left( {x, y}\right) \) is \( {2y}\mathbf{i} + {4x}\mathbf{j} \) (i.e., the direction and magnitude of the fluid flow varies from point to point). Find the path traced out by the particle, if it... | The slope of the path of a particle at \( \left( {x, y}\right) \) is the ratio \( {4x}/{2y}\; \) (assuming that \( y \neq 0) \) of the components of the fluid velocity vector at \( \left( {x, y}\right) \) . Assuming that the path is the graph of a function \( y \) of \( x \), we then obtain the ODE \( {y}^{\prime }\lef... | Yes |
Example 3. Sclve \( \left( {1 + {x}^{2}}\right) {y}^{\prime } + {2xy} = 3{x}^{2} \) . | Solution. First, put the equation into the standard form (6), namely\n\n\[ \n{y}^{\prime } + \left\lbrack {{2x}/\left( {1 + {x}^{2}}\right) }\right\rbrack y = 3{x}^{2}/\left( {1 + {x}^{2}}\right) .\n\]\n\n(14)\n\nThe integrating factor for equation (14) (cf. (10)) is\n\n\[ \nm\left( x\right) = \exp \left\lbrack {\int \... | Yes |
Consider two identical cans, A and B. Assume that syrup will leak out of either can at a rate which is proportional to the volume \( V \) of the syrup in the can, say \( {V}^{\prime }\left( t\right) = - {kV}\left( t\right) \) , where \( k > 0 \), due to the leakage. Suppose that the initial volume of syrup in can A is ... | Solution. The rate of change of \( {\mathrm{V}}_{\mathrm{B}}\left( \mathrm{t}\right) \) is \[ {\mathrm{V}}_{\mathrm{B}}^{\prime }\left( \mathrm{t}\right) = - \mathrm{k}{\mathrm{V}}_{\mathrm{B}}\left( \mathrm{t}\right) - {\mathrm{V}}_{\mathrm{A}}^{\prime }\left( \mathrm{t}\right) . \] (16) Since \( {\mathrm{V}}_{\mathrm... | Yes |
Suppose that tank \( A \) contains salt water with 4 pounds of salt per 100 gallons. Tank B is initially filled with 100 gallons of pure water. Over a period of one hour, the water in tank \( B \) is drained at the rate of \( 3 \) gallons per minute. The water in tank \( A \) flows into tank \( B \) at the rate of 5 ga... | Let \( S\left( t\right) \) denote the number of \( p \) ounds of salt in tank \( B \) at time \( t. \) At time \( t, \) tank \( B \) loses salt (via draining) at the rate of 3 gallons per minute, times the amount of salt per gallon in tank \( B \), namely, \( {3S}\left( t\right) /\left( {{100} + \left( {5 - 3}\right) t... | Yes |
An object of mass m is attached to a spring which lies along the x-axis, as shown in Figure 2 below. With Hooke's law in effect, when the object is displaced to the position \( x \), the spring exerts a force \( - \mathrm{{kx}} \) (toward the origin, since the constant \( \mathrm{k} \) is positive) on the object. Let \... | Solution. Since the total force on the object is \( \mathrm{F}\left( \mathrm{t}\right) = - \mathrm{{kx}}\left( \mathrm{t}\right) - {\mathrm{{bx}}}^{\prime }\left( \mathrm{t}\right) \), Newton’s second law yields the ODE\n\n\[ m{x}^{\prime \prime }\left( t\right) + b{x}^{\prime }\left( t\right) + {kx}\left( t\right) = 0... | Yes |
Suppose that a battery of voltage \( \mathrm{V} \), a resistor of resistance \( \mathrm{R} \), a coil of inductance \( \mathrm{L} \) and a capacitor of capacitance \( \mathrm{C} \) are placed in series as shown below in Figure 4. We wish to find the most general expression for the current \( i\left( t\right) \) in this... | Solution. Kirchhoff’s second law asserts that the sum of the voltage drops across the elements of any closed loop in a circuit must be zero. At time t, the voltage drop across the resistor is R times the current i(t). The voltage drop across the coil is L \( {i}^{\prime }\left( t\right) \) . (This drop is due to the fa... | Yes |
Find a particular solution of \( \left( {40}\right) \) with \( {abc} \neq 0 \), in the case when \( f\left( t\right) = A\cos \left( {\omega t}\right) \) or \( \mathrm{f}\left( \mathrm{t}\right) = \mathrm{A}\sin \left( {\omega \mathrm{t}}\right) \) for a real constant \( \omega \), by using the following approach. Deter... | Solution. Substituting the trial solution \( y\left( t\right) = C{e}^{i\omega t} \) into (40) with \( f\left( t\right) = A{e}^{i\omega t} \), we obtain\n\n\[ \n{\operatorname{Ce}}^{i\omega t}\left\lbrack {a{\left( i\omega \right) }^{2} + {bi\omega } + c}\right\rbrack = A{e}^{i\omega t}\;\text{ or }\;C \cdot \left\lbrac... | Yes |
In Example 2, calculate the position \( \left( {x\left( t\right), y\left( t\right) }\right) \) of the particle at any time \( t \), given that \( x\left( 0\right) = 1 \) and \( y\left( 0\right) = 3 \) . | Solution. The velocity vector at time \( \mathrm{t} \) is \( {\mathrm{x}}^{\prime }\left( \mathrm{t}\right) \mathrm{i} + {\mathrm{y}}^{\prime }\left( \mathrm{t}\right) \mathrm{j} \) . Thus, we have the system\n\n\[ \n{\mathrm{x}}^{\prime }\left( \mathrm{t}\right) = 2\mathrm{y}\left( \mathrm{t}\right) \n\]\n\n(49a)\n\n\... | Yes |
Show that if \( {\mathrm{A}}^{2} < 4\mathrm{{BC}} \), then the animal will eventually starve to death (after a number of diet/binge cycles), unless \( w\left( 0\right) = \mathrm{{AK}}/\mathrm{{BC}} \) and \( \mathrm{s}\left( 0\right) = \mathrm{K}/\mathrm{C} \), in which case \( \mathrm{w}\left( \mathrm{t}\right) \) and... | Solution. We have the system\n\n\[ \n{\mathrm{s}}^{\prime }\left( \mathrm{t}\right) = \mathrm{{As}}\left( \mathrm{t}\right) - \mathrm{{Bw}}\left( \mathrm{t}\right) \n\]\n\n(50a)\n\n\[ \n{\mathrm{w}}^{\prime }\left( \mathrm{t}\right) = \mathrm{{Cs}}\left( \mathrm{t}\right) - \mathrm{K}. \n\]\n\n(50b)\n\nDifferentiating ... | Yes |
Suppose that we seek a solution \( u\left( {x, y, z}\right) \) of Laplace’s equation which is spherically symmetric in the sense that \( u\left( {x, y, z}\right) \) only depends on the distance \( r = {\left| {x}^{2} + {y}^{2} + {z}^{2}\right| }^{\frac{1}{2}} \) to the origin \( \left( {0,0,0}\right) \) . In other word... | Using the chain rule, we have\n\n\[ \n{u}_{x} = \frac{df}{dr}\frac{\partial r}{\partial x} = {f}^{\prime }\left( r\right) {r}_{x},\text{ where }{r}_{x} = \frac{\partial }{\partial x}{\left\lbrack {x}^{2} + {y}^{2} + {z}^{2}\right\rbrack }^{\frac{1}{2}} = \frac{1}{2}{r}^{-1}{2x} = x{r}^{-1}. \]\n\nThen\n\n\[ \n{u}_{xx} ... | Yes |
As another instance of (16), in the case of a vibrating string with a transverse applied force density proportional to \( - \mathrm{{cu}}\left( {\mathrm{x},\mathrm{t}}\right) + \mathrm{F}\left( {\mathrm{x},\mathrm{t}}\right) \), we obtain the one-dimensional inhomogeneous Klein-Gordon equation\n\n\[ \n- {a}^{2}{u}_{xx}... | If \( F \equiv 0 \) and \( c = 0 \) ,(18) reduces to the (homogeneous) wave equation \( - {a}^{2}{u}_{xx} + {u}_{tt} = 0 \) or (12). We refer to (18) as the generalized wave equation. | No |
Find a solution of \( {u}_{xx} - {u}_{y} = {18x} + {8y} \) . | Solution. Here \( L\left\lbrack u\right\rbrack = {u}_{xx} - {u}_{y},\;{f}_{1}\left( {x, y}\right) = {6x} \) and \( {f}_{2}\left( {x, y}\right) = - {2y} \) . Note that \( {18x} + {8y} \) \( = 3{f}_{1}\left( {x, y}\right) - 4{f}_{2}\left( {x, y}\right) \), and thus \( {c}_{1} = 3 \) and \( {c}_{2} = - 4 \) . The superpos... | Yes |
Observe that \( {u}_{1}\left( {x, t}\right) = \sin \left( t\right) \cos \left( x\right) \) and \( {u}_{2}\left( {x, t}\right) = \cos \left( {3t}\right) \sin \left( {3x}\right) \) are solutions of the wave equation \( {u}_{tt} = {u}_{xx} \) . By applying the superposition principle, find infinitely many other solutions,... | Note that \( \;{u}_{tt} = {u}_{xx}\; \) can be written in the form of an homogeneous linear PDE \( {\mathrm{u}}_{\mathrm{{tt}}} - {\mathrm{u}}_{\mathrm{{xx}}} = 0 \) . According to the superposition principle, for any constants \( {\mathrm{c}}_{1} \) and \( {\mathrm{c}}_{2} \) , \[ {c}_{1}\sin \left( t\right) \cos \lef... | Yes |
Consider the nonlinear first-order PDE \( {u}_{x}{u}_{y} - u\left( {{u}_{x} + {u}_{y}}\right) + {u}^{2} = 0 \) or equivalently \( \left( {{u}_{x} - u}\right) \left( {{u}_{y} - u}\right) = 0 \) . Note that we have two solutions, namely \( {e}^{x} \) and \( {e}^{y} \) . However, show that \( {\mathrm{c}}_{1}{\mathrm{e}}^... | Solution. Defining \( \mathrm{N}\left\lbrack \mathrm{u}\right\rbrack = \left( {{\mathrm{u}}_{\mathrm{x}} - \mathrm{u}}\right) \left( {{\mathrm{u}}_{\mathrm{y}} - \mathrm{u}}\right) \), observe that for any \( {\mathrm{C}}^{1} \) functions \( \mathrm{v} \) and \( \mathrm{w} \)\n\n\[ N\left\lbrack {v + w}\right\rbrack = ... | Yes |
Imagine a soap film surface which remains after a (possibly nonplanar) loop of wire is dipped in a soap solution. Due to the surface tension of the film, it will form a surface of least area spanning the loop (i.e. a minimal surface). If the surface is the graph \( z = u\left( {x, y}\right) \) of some function \( u \),... | \[ \left( {1 + {u}_{y}^{2}}\right) {u}_{xx} + \left( {1 + {u}_{x}^{2}}\right) {u}_{yy} - 2{u}_{x}{u}_{y}{u}_{xy} = 0. \] | Yes |
Find the general solution of the first-order linear PDE for \( u = u\\left( {x, y}\\right) \)\n\n\[ \n{u}_{x}\\left( {x, y}\\right) = {2xy}\\text{, for all }\\left( {x, y}\\right) .\n\] | Solution. If we hold \( y \) fixed and integrate with respect to \( x \), we obtain\n\n\[ \nu\\left( {x, y}\\right) = {x}^{2}y + f\\left( y\\right) \]\n\nNote that the constant of integration may depend on \( y, \) and indeed any function of the form \( \\left( 4\\right) \) satisfies (3). As a technical point, recall t... | Yes |
Find the general solution of the third-order PDE\n\n\\[ \n{u}_{xyy} = 2\\sin \\left( x\\right) ,\\;u = u\\left( {x, y, z}\\right) ,\\;\\text{ for all }\\left( {x, y, z}\\right) .\n\\] | Solution. Integrating (5) once with respect to \\( y, \\) we get \\( {u}_{xy}\\left( {x, y, z}\\right) = {2y}\\sin \\left( x\\right) + f\\left( {x, z}\\right) \\) . Integrating again with respect to \\( y \\), we obtain \\( {u}_{x}\\left( {x, y, z,}\\right) = {y}^{2}\\sin \\left( x\\right) + {yf}\\left( {x, z}\\right) ... | Yes |
Find the general solution of\n\n\[ \n{\left( {u}_{xx}\right) }^{2} + {\left( {u}_{yy}\right) }^{2} = 0,\;u = u\left( {x, y}\right) ,\;\text{ for all }\left( {x, y}\right) .\n\] | Solution. A \( {C}^{2} \) function \( u\left( {x, y}\right) \) solves this equation, if and only if \( {u}_{xx} = 0 \) and \( {u}_{yy} = 0 \) . Since \( {u}_{xx} = 0,\;u\; \) must have the form \( \;u\left( {x, y}\right) = f\left( y\right) x + g\left( y\right) .\; \) However, since \( \;{u}_{yy} = 0,\;u\; \) must also ... | Yes |
Find a generic solution of the nonlinear first-order PDE\n\n\[ \n{u}_{x}\left( {x, y}\right) = {\left\lbrack u\left( x, y\right) \right\rbrack }^{2}. \n\] | Solution. By fixing \( y, \) we may regard \( \left( 9\right) \) as a first-order separable ODE, namely \( {u}^{-2}{du} = {dx}, \) assuming that \( u \neq 0 \) . Integrating, we get the solution \( - {u}^{-1} = x + g\left( y\right) \), or\n\n\[ \nu\left( {x, y}\right) = - {\left\lbrack x + g\left( y\right) \right\rbrac... | Yes |
Show that\n\n\[ \mathrm{u}\left( {\mathrm{x},\mathrm{y}}\right) = \mathrm{{xf}}\left( {2\mathrm{x} + \mathrm{y}}\right) \]\n\n(12)\n\nis a generic solution of (11), where \( \mathrm{f} \) is an arbitrary \( {\mathrm{C}}^{1} \) function. | Solution. First note that despite the involvement of both \( x \) and \( y \) in \( f\left( {{2x} + y}\right) \), the function \( f \) is still really a function of one variable, since \( f \) has only one \ | No |
Find a solution of (11) which satisfies the condition \( u\left( {1, y}\right) = {y}^{2} \) for all \( y \) . | Solution. The condition \( \;u\left( {1, y}\right) = {y}^{2}\; \) specifies the values of the solution \( \;u\left( {x, y}\right) \; \) for points \( \;\left( {x, y}\right) \) on the line \( \mathrm{x} = 1 \), parallel to the \( \mathrm{y} \) -axis. Since (12) is a generic solution, it suffices to find a function \( f ... | Yes |
Show that the wave equation \( {u}_{tt} = {c}^{2}{u}_{xx} \) has a generic solution of the form\n\n\[ u\left( {x, t}\right) = f\left( {x + {ct}}\right) + g\left( {x - {ct}}\right) ,\]\n\nwhere \( f \) and \( g \) are arbitrary \( {C}^{2} \) functions. Find a particular solution meeting the initial conditions\n\n\[ \tex... | Solution. One can directly verify that (13) is a solution of \( {\mathrm{u}}_{\mathrm{{tt}}} = {\mathrm{c}}^{2}{\mathrm{u}}_{\mathrm{{xx}}} \), as in Problem 4(d) of Section 1.2 . Since the wave equation is second-order and there are two arbitrary functions in (13), neither of which can be eliminated without losing sol... | Yes |
Find the general solution of the PDE\n\n\[ \n{u}_{xx} + {cu} = 0,\;u = u\left( {x, t}\right) \n\]\n\nin the three cases \( c > 0, c = 0 \) and \( c < 0 \) . | Solution. For fixed \( t \) ,(17) is a second-order linear ODE with constant coefficients (discussed in Section 1.1) for \( u \), regarded as a function of \( x \) . If \( c > 0 \), then for each fixed \( t \), the solution is of the form \( {c}_{1}\sin \left( {\sqrt{c} \cdot x}\right) + {c}_{2}\cos \left( {\sqrt{c} \c... | Yes |
Find the general solution \( u = u\\left( {x, y}\\right) \) of\n\n\[ \n{u}_{yy} + {u}_{y} = x\\text{. } \n\]\n\n(18) | Solution. By fixing \( x \), we can regard (18) as a linear, inhomogeneous, second-order ODE with \( y \) as the independent variable. A particular solution is \( u\\left( {x, y}\\right) = {xy} \) . The auxiliary equation for the related homogeneous equation is \( {r}^{2} + r = 0 \), which has roots 0 and -1 . Remember... | Yes |
Using separation of variables, find the product solutions of the heat equation with temperature-dependent sink, namely\n\n\[ \n{u}_{t} - {u}_{xx} = - u,\;u = u\left( {x, t}\right) \n\] | Solution. Substituting a product solution of the form \( u\left( {x, t}\right) = f\left( x\right) g\left( t\right) \) into (22), we get\n\n\[ \nf\left( x\right) {g}^{\prime }\left( t\right) - {f}^{\prime \prime }\left( x\right) g\left( t\right) = - f\left( x\right) g\left( t\right) .\n\]\n\n(23)\n\nThen we separate the... | Yes |
Find some nontrivial product solutions of the following wave equation for the amplitude \( u\left( {x, y, t}\right) \) of a transversely vibrating membrane at \( \left( {x, y}\right) \) at time \( t \)\n\n\[ \n{u}_{tt} = {u}_{xx} + {u}_{yy} \n\]\n\n(28) | Solution. Let \( u\left( {x, y, t}\right) \) be of the form \( X\left( x\right) Y\left( y\right) T\left( t\right) \) for functions \( X, Y \) and \( T \) . This notation for the function is helpful in keeping track of the variables which correspond to the functions. Substituting \( u \) into \( \left( {28}\right) \), w... | No |
Find the general solution of the PDE\n\n\[ 3{u}_{x} - 2{u}_{y} + u = x\text{,}u = u\left( {x, y}\right) \text{.} \] | Solution. The characteristic lines have slope \( - 2/3 \) . They constitute the family of lines \( {2x} + {3y} = d \) . Hence, we make the change of variables\n\n\[ \left\{ {\begin{array}{l} w = {2x} + {3y} \\ z = y \end{array}\;;\;\left\{ \begin{array}{l} x = \left( {w - {3z}}\right) /2 \\ y = z. \end{array}\right. }\... | Yes |
Solve the following PDE with the given side condition :\n\n\[ \n{u}_{x} - {u}_{y} + {2u} = 1,\;u\left( {x,0}\right) = {x}^{2}. \n\] | Solution. Here the side condition specifies the values of \( u \) at points on the \( x \) -axis. First, we find the general solution of the PDE, and then we try to meet the side condition. The characteristic lines have slope \( - 1 \) and are of the form \( x + y = d \) . Thus, we make the change of variables\n\n\[ \n... | Yes |
Attempt to solve the PDE \( {u}_{x} + 2{u}_{y} - {4u} = {e}^{x + y}\; \) with side condition \( \;u\left( {x,{2x} - 1}\right) = 0\;. | Solution. The general solution \( u\left( {x, y}\right) = - {e}^{x + y} + {e}^{4x}D\left( {{2x} - y}\right) \) of the PDE was found in Example 3 (cf. (16)). The side condition is given on the characteristic line \( y = {2x} - 1 \), and it implies that\n\n\[ 0 = u\left( {x,{2x} - 1}\right) = - {e}^{{3x} - 1} + {e}^{4x}D... | Yes |
Solve the PDE \( {u}_{x} \) + 2u \( {y}^{-{4u}} = {e}^{x + y} \), subject to the condition \( \;u\left( {x,{2x}}\right) = - {e}^{3x} + {e}^{4x}. \) | Solution. As in Example 3, the general solution of the PDE is \( \;u\left( {x, y}\right) = - {e}^{x + y} + {e}^{4x}D\left( {{2x} - y}\right) . \) The side condition (given on the characteristic line \( y = {2x} \) ) then tells us that\n\n\[ \n- {e}^{3x} + {e}^{4x} = u\left( {x,{2x}}\right) = - {e}^{3x} + {e}^{4x}D\left... | Yes |
Determine the form of the functions \( g\left( x\right) \) for which the PDE \( {u}_{x} - u = 0 \), with side condition \( u\left( {x,1}\right) = g\left( x\right) \), has a solution. | Here the characteristic lines are the horizontal lines \( y = d \) . The variable \( x \) serves as a position variable on each of these lines. Also, on the line \( y = d \), the PDE becomes the ODE\n\n\[ \frac{\partial }{\partial \mathrm{x}}\left\lbrack {\mathrm{u}\left( {\mathrm{x},\mathrm{d}}\right) }\right\rbrack -... | Yes |
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