Math 308 Week in Review 8

Directions: Full worked solutions for Week in Review 8 (Differential Equations). Click Solution under any problem to reveal the worked answer. Use the Contents panel or the Prev / Up / Next buttons to move between sections, and Search to find terms on this page.

Reference

Impulse & Convolution Reference

Impulse function and convolution theorem Laplace transform reference
$f(t)$	$\mathcal{L}\{f(t)\}$ or rule
$\delta(t-c)$	$e^{-cs}$
$f(t)\,\delta(t-c)$	$e^{-cs}\,f(c)$ (sifting property)
$(f*g)(t) = \displaystyle\int_0^t f(\tau)\,g(t-\tau)\,d\tau$	$F(s)\,G(s)$ (Convolution Theorem)
$\mathcal{L}^{-1}\{F(s)\,G(s)\}$	$(f*g)(t)$
Transfer function $H(s)$; impulse response $h = \mathcal{L}^{-1}\{H\}$	$y(t) = (h*f)(t)$ when ICs $= 0$

Power Series Reference

Power series differentiation rules and key facts
Expression	Series form
$y(x)$	$\displaystyle\sum_{n=0}^{\infty} a_n x^n$
$y'(x)$	$\displaystyle\sum_{n=1}^{\infty} n\,a_n x^{n-1} = \sum_{n=0}^{\infty}(n+1)a_{n+1}x^n$
$y''(x)$	$\displaystyle\sum_{n=2}^{\infty} n(n-1)\,a_n x^{n-2} = \sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2}x^n$
$x\,y'(x)$	$\displaystyle\sum_{n=0}^{\infty} n\,a_n x^n$
$x\,y(x)$	$\displaystyle\sum_{n=0}^{\infty} a_n x^{n+1} = \sum_{n=1}^{\infty} a_{n-1}x^n$

Section 6.5: Impulse Functions

1.

Apply the sifting property $\mathcal{L}\{f(t)\,\delta(t-c)\} = e^{-cs}\,f(c)$ to find each Laplace transform.

(a) $\displaystyle f(t) = 7\,\delta(t - 4)$
(b) $\displaystyle f(t) = t^2\,\delta(t - 3)$
(c) $\displaystyle f(t) = \sin(\pi t)\,\delta\!\left(t - \tfrac{1}{2}\right)$
(d) $\displaystyle f(t) = (2t+1)\,\delta(t-2) + \cos(3t)\,\delta(t - \pi)$

SolutionSolution

Evaluate the coefficient function at $t = c$, then multiply by $e^{-cs}$.

(a) Constant $f(t) = 7$ at $t = 4$: $f(4) = 7$.

\[\mathcal{L}\{7\,\delta(t-4)\} = 7e^{-4s}\]

(b) $f(t) = t^2$ at $t = 3$: $f(3) = 9$.

\[\mathcal{L}\{t^2\,\delta(t-3)\} = 9e^{-3s}\]

(c) $f(t) = \sin(\pi t)$ at $t = \tfrac{1}{2}$: $\sin\!\left(\tfrac{\pi}{2}\right) = 1$.

\[\mathcal{L}\!\left\{\sin(\pi t)\,\delta\!\left(t-\tfrac{1}{2}\right)\right\} = e^{-s/2}\]

(d) Two terms. For $(2t+1)$ at $t=2$: $2(2)+1 = 5$. For $\cos(3t)$ at $t=\pi$: $\cos(3\pi) = -1$.

\[\mathcal{L}\{(2t+1)\,\delta(t-2) + \cos(3t)\,\delta(t-\pi)\} = 5e^{-2s} + (-1)e^{-\pi s} = 5e^{-2s} - e^{-\pi s}\]

\[\boxed{(a)\;7e^{-4s};\quad(b)\;9e^{-3s};\quad(c)\;e^{-s/2};\quad(d)\;5e^{-2s}-e^{-\pi s}}\]

2.

Use the Laplace transform to solve the initial value problem.

\[ y'' + 2y' + 5y = \delta(t - \pi), \quad y(0) = 0, \quad y'(0) = 2. \]

Take the Laplace transform of both sides and solve for $Y(s)$. Complete the square in the denominator.
Find $y(t)$ by identifying each term as a known inverse transform.
Write $y(t)$ explicitly as a piecewise function on $[0, \pi)$ and $[\pi, \infty)$.
Sketch a rough graph of $y(t)$. For $0 \leq t < \pi$ the system oscillates freely from the initial velocity; describe what changes in the oscillation immediately after $t = \pi$ when the impulse arrives.

SolutionSolution

(a) Take $\mathcal{L}$ with $y(0) = 0$, $y'(0) = 2$:

\[(s^2Y - 2) + 2(sY) + 5Y = e^{-\pi s}\]

\[Y(s^2 + 2s + 5) = 2 + e^{-\pi s}\]

Complete the square: $s^2 + 2s + 5 = (s+1)^2 + 4$.

\[\boxed{Y(s) = \dfrac{2}{(s+1)^2+4} + \dfrac{e^{-\pi s}}{(s+1)^2+4}}\]

(b) Match each term to the standard form $\mathcal{L}\{e^{-at}\sin(\omega t)\} = \dfrac{\omega}{(s+a)^2+\omega^2}$. Here $a = 1$, $\omega = 2$: $\mathcal{L}^{-1}\!\left\{\dfrac{2}{(s+1)^2+4}\right\} = e^{-t}\sin(2t)$.

For the shifted term with $c = \pi$: $\mathcal{L}^{-1}\!\left\{\dfrac{e^{-\pi s}}{(s+1)^2+4}\right\} = \dfrac{1}{2}\,u_\pi(t)\,e^{-(t-\pi)}\sin(2(t-\pi))$.

\[\boxed{y(t) = e^{-t}\sin(2t) + \dfrac{1}{2}\,u_\pi(t)\,e^{-(t-\pi)}\sin(2(t-\pi))}\]

(c)

\[y(t) = \begin{cases} e^{-t}\sin(2t), & 0 \leq t < \pi \\[8pt] e^{-t}\sin(2t) + \dfrac{1}{2}\,e^{-(t-\pi)}\sin(2(t-\pi)), & t \geq \pi \end{cases}\]

(d) For $0 \leq t < \pi$: the response is a pure decaying oscillation at $\omega_d = 2$ rad/s driven by the initial velocity $y'(0) = 2$, with amplitude envelope $e^{-t}$. At $t = \pi$ the amplitude of the first term has decayed to $e^{-\pi} \approx 0.043$ — nearly zero. The impulse immediately excites a second decaying oscillation with initial amplitude $\tfrac{1}{2}$ at the same frequency $\omega_d = 2$. Note that $\sin(2(t-\pi)) = \sin(2t - 2\pi) = \sin(2t)$, so the two oscillatory terms share the same phase; the impulse effectively re-energizes the system in the same direction at about 12 times the amplitude of the dying first term, producing a dramatic jump in oscillation amplitude just after $t = \pi$. Both terms decay exponentially to zero as $t \to \infty$.

Rough sketch of $y(t)$; dashed line marks impulse at $t=\pi$

Section 6.6: The Convolution Integral

3.

Compute each convolution directly from the definition $(f*g)(t) = \int_0^t f(\tau)\,g(t-\tau)\,d\tau$.

(a) $f(t) = t,\quad g(t) = 1$
(b) $f(t) = e^{t},\quad g(t) = e^{3t}$
(c) $f(t) = t,\quad g(t) = \sin(2t)$

SolutionSolution

(a) $f(\tau) = \tau$, $g(t-\tau) = 1$:

\[(t*1)(t) = \int_0^t \tau\cdot 1\,d\tau = \left[\dfrac{\tau^2}{2}\right]_0^t = \dfrac{t^2}{2}\]

\[\boxed{(t*1)(t) = \dfrac{t^2}{2}}\]

(b) $f(\tau) = e^\tau$, $g(t-\tau) = e^{3(t-\tau)}$. Factor out the term not depending on $\tau$:

\[(e^t * e^{3t})(t) = \int_0^t e^\tau\,e^{3(t-\tau)}\,d\tau = e^{3t}\int_0^t e^{-2\tau}\,d\tau = e^{3t}\left[-\dfrac{e^{-2\tau}}{2}\right]_0^t = e^{3t}\cdot\dfrac{1-e^{-2t}}{2}\]

\[\boxed{(e^t * e^{3t})(t) = \dfrac{e^{3t}-e^{t}}{2}}\]

(c) By commutativity $f * g = g * f$, choose to place $t - \tau$ with the algebraic factor to avoid trigonometric identities. Set $f(\tau) = \sin(2\tau)$, $g(t - \tau) = t - \tau$:

\[(t * \sin 2t)(t) = \int_0^t \sin(2\tau)\,(t - \tau)\,d\tau = t\int_0^t \sin(2\tau)\,d\tau - \int_0^t \tau\sin(2\tau)\,d\tau\]

First integral:

\[t\int_0^t \sin(2\tau)\,d\tau = t\left[-\dfrac{\cos(2\tau)}{2}\right]_0^t = t\cdot\dfrac{1 - \cos(2t)}{2}\]

Second integral — integrate by parts with $u = \tau$, $dv = \sin(2\tau)\,d\tau$:

\[\int_0^t \tau\sin(2\tau)\,d\tau = \left[-\dfrac{\tau\cos(2\tau)}{2}\right]_0^t + \int_0^t\dfrac{\cos(2\tau)}{2}\,d\tau = -\dfrac{t\cos(2t)}{2} + \left[\dfrac{\sin(2\tau)}{4}\right]_0^t = -\dfrac{t\cos(2t)}{2} + \dfrac{\sin(2t)}{4}\]

Combine — the $\pm\tfrac{t\cos(2t)}{2}$ terms cancel exactly:

\[(t*\sin 2t) = \dfrac{t}{2} - \dfrac{t\cos(2t)}{2} - \!\left(-\dfrac{t\cos(2t)}{2} + \dfrac{\sin(2t)}{4}\right)\]

\[\boxed{(t*\sin 2t)(t) = \dfrac{t}{2} - \dfrac{\sin(2t)}{4}}\]

Verification via Convolution Theorem: $\mathcal{L}\!\left\{\dfrac{t}{2}-\dfrac{\sin 2t}{4}\right\} = \dfrac{1}{2s^2}-\dfrac{1}{2(s^2+4)} = \dfrac{(s^2+4)-s^2}{2s^2(s^2+4)} = \dfrac{4}{2s^2(s^2+4)} = \dfrac{2}{s^2(s^2+4)}$, which equals $\mathcal{L}\{t\}\cdot\mathcal{L}\{\sin(2t)\} = \dfrac{1}{s^2}\cdot\dfrac{2}{s^2+4}$. ✓

4.

For each transform below, identify $F(s)$ and $G(s)$ so that the expression equals $F(s)\,G(s)$. State the corresponding functions $f(t) = \mathcal{L}^{-1}\{F(s)\}$ and $g(t) = \mathcal{L}^{-1}\{G(s)\}$, write the inverse transform as a convolution integral, then evaluate that integral.

(a) $\displaystyle \mathcal{L}^{-1}\!\left\{\dfrac{1}{s(s^2+4)}\right\}$
(b) $\displaystyle \mathcal{L}^{-1}\!\left\{\dfrac{s}{(s^2+1)(s+2)}\right\}$
(c) $\displaystyle \mathcal{L}^{-1}\!\left\{\dfrac{1}{(s+1)(s+2)^2}\right\}$

SolutionSolution

(a) Take $F(s) = \dfrac{1}{s}$ and $G(s) = \dfrac{1}{s^2+4}$, so $f(\tau) = 1$ and $g(t) = \tfrac{1}{2}\sin(2t)$.

\[(f*g)(t) = \int_0^t 1\cdot\dfrac{1}{2}\sin(2(t-\tau))\,d\tau = \dfrac{1}{2}\left[\dfrac{\cos(2(t-\tau))}{2}\right]_0^t = \dfrac{1}{4}\bigl[\cos(0)-\cos(2t)\bigr]\]

\[\boxed{\mathcal{L}^{-1}\!\left\{\dfrac{1}{s(s^2+4)}\right\} = \dfrac{1-\cos(2t)}{4}}\]

(b) Take $F(s) = \dfrac{s}{s^2+1}$ and $G(s) = \dfrac{1}{s+2}$, so $f(\tau) = \cos\tau$ and $g(t) = e^{-2t}$.

\[(f*g)(t) = \int_0^t \cos\tau\,e^{-2(t-\tau)}\,d\tau = e^{-2t}\int_0^t e^{2\tau}\cos\tau\,d\tau\]

Integrate $\int e^{2\tau}\cos\tau\,d\tau$ by parts twice (or use the formula $\int e^{a\tau}\cos(b\tau)\,d\tau = \dfrac{e^{a\tau}(a\cos b\tau + b\sin b\tau)}{a^2+b^2}$ with $a=2$, $b=1$):

\[e^{-2t}\int_0^t e^{2\tau}\cos\tau\,d\tau = e^{-2t}\left[\dfrac{e^{2\tau}(2\cos\tau+\sin\tau)}{5}\right]_0^t = e^{-2t}\cdot\dfrac{e^{2t}(2\cos t+\sin t)-2}{5}\]

\[\boxed{\mathcal{L}^{-1}\!\left\{\dfrac{s}{(s^2+1)(s+2)}\right\} = \dfrac{2\cos t+\sin t - 2e^{-2t}}{5}}\]

Partial-fraction check: $\dfrac{s}{(s^2+1)(s+2)} = \dfrac{2s+1}{5(s^2+1)} - \dfrac{2}{5(s+2)}$, giving $\tfrac{2}{5}\cos t + \tfrac{1}{5}\sin t - \tfrac{2}{5}e^{-2t}$. ✓

(c) Take $F(s) = \dfrac{1}{s+1}$ and $G(s) = \dfrac{1}{(s+2)^2}$, so $f(\tau) = e^{-\tau}$ and $g(t) = te^{-2t}$.

\[(f*g)(t) = \int_0^t e^{-\tau}(t-\tau)e^{-2(t-\tau)}\,d\tau = e^{-2t}\int_0^t (t-\tau)e^{\tau}\,d\tau\]

Split and integrate by parts on $\int_0^t \tau e^\tau\,d\tau = te^t - e^t + 1$:

\[e^{-2t}\!\left[t\int_0^t e^\tau\,d\tau - \int_0^t\tau e^\tau\,d\tau\right] = e^{-2t}\!\left[t(e^t-1)-(te^t-e^t+1)\right] = e^{-2t}\!\left[e^t - t - 1\right]\]

\[\boxed{\mathcal{L}^{-1}\!\left\{\dfrac{1}{(s+1)(s+2)^2}\right\} = e^{-t} - (1+t)e^{-2t}}\]

Partial-fraction check: $\dfrac{1}{(s+1)(s+2)^2} = \dfrac{1}{s+1} - \dfrac{1}{s+2} - \dfrac{1}{(s+2)^2}$, giving $e^{-t} - e^{-2t} - te^{-2t}$. ✓

5.

Consider the initial value problem with an unspecified forcing function $f(t)$:

\[ y'' + 4y = f(t), \quad y(0) = 0, \quad y'(0) = 0. \]

Take the Laplace transform and write $Y(s) = H(s)\,F(s)$. Identify the transfer function $H(s)$ and find the impulse response $h(t) = \mathcal{L}^{-1}\{H(s)\}$.
Use the Convolution Theorem to express the solution $y(t)$ as a single convolution integral involving $h$ and $f$.
Now take $f(t) = \cos(t)$. Substitute into your convolution integral, evaluate it using the product-to-sum identity $\sin A \cos B = \tfrac{1}{2}[\sin(A+B)+\sin(A-B)]$, and find $y(t)$ in closed form.

SolutionSolution

(a) Take $\mathcal{L}$ with zero ICs:

\[(s^2+4)\,Y(s) = F(s) \;\Rightarrow\; Y(s) = \underbrace{\dfrac{1}{s^2+4}}_{H(s)}\,F(s)\]

The transfer function is $H(s) = \dfrac{1}{s^2+4}$. The impulse response is:

\[h(t) = \mathcal{L}^{-1}\!\left\{\dfrac{1}{s^2+4}\right\} = \dfrac{\sin(2t)}{2}\]

(b) By the Convolution Theorem $Y = H(s)F(s)$ inverts to:

\[y(t) = (h*f)(t) = \int_0^t h(t-\tau)\,f(\tau)\,d\tau = \int_0^t \dfrac{\sin(2(t-\tau))}{2}\,f(\tau)\,d\tau\]

(c) With $f(\tau) = \cos\tau$:

\[y(t) = \dfrac{1}{2}\int_0^t \sin(2(t-\tau))\cos\tau\,d\tau\]

Apply the identity $\sin(2(t-\tau))\cos\tau = \sin(2t-2\tau)\cos\tau = \tfrac{1}{2}[\sin(2t-\tau)+\sin(2t-3\tau)]$:

\[y(t) = \dfrac{1}{4}\int_0^t \left[\sin(2t-\tau)+\sin(2t-3\tau)\right]d\tau\]

Integrate (noting $\int\sin(2t-\tau)\,d\tau = \cos(2t-\tau)$ and $\int\sin(2t-3\tau)\,d\tau = \tfrac{\cos(2t-3\tau)}{3}$):

\[y(t) = \dfrac{1}{4}\left[\cos(2t-\tau)\Big|_0^t + \dfrac{\cos(2t-3\tau)}{3}\Bigg|_0^t\right] = \dfrac{1}{4}\left[(\cos t-\cos 2t)+\dfrac{\cos(-t)-\cos(2t)}{3}\right]\]

Use $\cos(-t) = \cos t$:

\[y(t) = \dfrac{1}{4}\left[(\cos t-\cos 2t)+\dfrac{\cos t-\cos 2t}{3}\right] = \dfrac{1}{4}\cdot\dfrac{4(\cos t-\cos 2t)}{3}\]

\[\boxed{y(t) = \dfrac{\cos t - \cos(2t)}{3}}\]

Verification: $\mathcal{L}\{y\} = \dfrac{1}{s^2+4}\cdot\dfrac{s}{s^2+1} = \dfrac{s}{(s^2+1)(s^2+4)}$. Partial fractions give $\dfrac{s/3}{s^2+1}-\dfrac{s/3}{s^2+4}$, inverting to $\tfrac{1}{3}\cos t-\tfrac{1}{3}\cos(2t)$. ✓

6.

An equation of the form $f(t) = g(t) + \displaystyle\int_0^t K(t-\tau)\,f(\tau)\,d\tau$, where $f$ is unknown, is called a Volterra integral equation of the second kind. The integral term is a convolution, so the Laplace transform converts the equation into a simple algebraic problem for $F(s)$.

Find $f(t)$ satisfying

\[ f(t) = 1 - \int_0^t f(\tau)\,e^{-(t-\tau)}\,d\tau. \]

Rewrite the right-hand side recognizing the integral as a convolution $f * g$ for a suitable $g(t)$. Take the Laplace transform and solve for $F(s)$.
Use partial fractions to find $f(t) = \mathcal{L}^{-1}\{F(s)\}$.
Verify that your answer satisfies the original integral equation by substituting back and confirming both sides agree.

SolutionSolution

(a) Recognize $\displaystyle\int_0^t f(\tau)\,e^{-(t-\tau)}\,d\tau = (f * e^{-t})(t)$. Take $\mathcal{L}$:

\[F(s) = \dfrac{1}{s} - F(s)\cdot\dfrac{1}{s+1}\]

\[F(s)\left(1+\dfrac{1}{s+1}\right) = \dfrac{1}{s} \;\Rightarrow\; F(s)\cdot\dfrac{s+2}{s+1} = \dfrac{1}{s}\]

\[\boxed{F(s) = \dfrac{s+1}{s(s+2)}}\]

(b) Partial fractions: $\dfrac{s+1}{s(s+2)} = \dfrac{A}{s}+\dfrac{B}{s+2}$.

At $s=0$: $A = \tfrac{1}{2}$. At $s=-2$: $B = \tfrac{-1}{-2} = \tfrac{1}{2}$.

\[\boxed{f(t) = \dfrac{1}{2}+\dfrac{1}{2}\,e^{-2t}}\]

(c) Verification. Substitute $f(\tau) = \tfrac{1}{2}+\tfrac{1}{2}e^{-2\tau}$ into the right-hand side:

\[1 - \int_0^t\!\left(\tfrac{1}{2}+\tfrac{1}{2}e^{-2\tau}\right)e^{-(t-\tau)}\,d\tau = 1 - e^{-t}\int_0^t\!\left(\tfrac{e^\tau}{2}+\tfrac{e^{-\tau}}{2}\right)d\tau\]

\[= 1 - e^{-t}\left[\dfrac{e^\tau}{2}-\dfrac{e^{-\tau}}{2}\right]_0^t = 1 - e^{-t}\cdot\dfrac{e^t - e^{-t}}{2} = 1 - \dfrac{1-e^{-2t}}{2} = \dfrac{1}{2}+\dfrac{e^{-2t}}{2} = f(t) \;\checkmark\]

Sections 5.1–5.2: Power Series Methods

7.

This question builds the technical skills needed to find power series solutions of ODEs.

Find the radius of convergence $R$ of $\displaystyle y(x) = \sum_{n=1}^{\infty} \dfrac{(-1)^n\,x^n}{n\cdot 2^n}$ using the ratio test. For which values of $x$ does the series converge absolutely?
Let $y = \displaystyle\sum_{n=0}^{\infty} a_n x^n$. Show by re-indexing (substituting $m = n-2$ in the $y''$ series and $m = n$ in the $xy'$ series) that \[y'' = \sum_{n=0}^{\infty}(n+2)(n+1)\,a_{n+2}\,x^n, \qquad xy' = \sum_{n=0}^{\infty} n\,a_n\,x^n.\]
Using part (b), write out all three series in $y'' + xy' + 2y = 0$ as a single power series in $x^n$, then apply the identity theorem to derive the recurrence relation for $a_{n+2}$. Compute $a_2$ and $a_3$ explicitly.

SolutionSolution

(a) Apply the ratio test with $a_n = \dfrac{(-1)^n x^n}{n\cdot 2^n}$:

\[\left|\dfrac{a_{n+1}}{a_n}\right| = \left|\dfrac{x^{n+1}}{(n+1)\cdot 2^{n+1}}\cdot\dfrac{n\cdot 2^n}{x^n}\right| = |x|\cdot\dfrac{n}{2(n+1)} \;\xrightarrow{n\to\infty}\; \dfrac{|x|}{2}\]

The series converges when $|x|/2 < 1$, i.e., $|x| < 2$, so $R = 2$. The series converges absolutely for all $x \in (-2, 2)$. (At the endpoints: $x=2$ gives the alternating harmonic series $\sum(-1)^n/n$, which converges conditionally but not absolutely; $x=-2$ gives the harmonic series $\sum 1/n$, which diverges.)

(b) Starting from $y'' = \displaystyle\sum_{n=2}^{\infty} n(n-1)\,a_n\,x^{n-2}$, let $m = n-2$ (so $n = m+2$, and when $n=2$, $m=0$):

\[y'' = \sum_{m=0}^{\infty}(m+2)(m+1)\,a_{m+2}\,x^m\]

Renaming the dummy index $m \to n$ gives the first formula. For $xy'$: since $y' = \displaystyle\sum_{n=1}^\infty n\,a_n\,x^{n-1}$, multiply by $x$:

\[xy' = \sum_{n=1}^{\infty} n\,a_n\,x^n = \sum_{n=0}^{\infty} n\,a_n\,x^n\]

(The $n=0$ term contributes $0\cdot a_0\cdot x^0 = 0$, so the lower limit can be extended to $n=0$.) ✓

(c) Substitute all three series into $y'' + xy' + 2y = 0$:

\[\sum_{n=0}^{\infty}(n+2)(n+1)\,a_{n+2}\,x^n + \sum_{n=0}^{\infty} n\,a_n\,x^n + 2\sum_{n=0}^{\infty} a_n\,x^n = 0\]

\[\sum_{n=0}^{\infty}\left[(n+2)(n+1)\,a_{n+2} + (n+2)\,a_n\right]x^n = 0\]

Factor $(n+2)$ and apply the identity theorem (since $n+2 \neq 0$ for $n \geq 0$):

\[\boxed{a_{n+2} = -\dfrac{a_n}{n+1}, \quad n \geq 0}\]

Explicitly: $n=0$: $a_2 = -\dfrac{a_0}{1} = -a_0$. $n=1$: $a_3 = -\dfrac{a_1}{2}$.

8.

Find the general power series solution of

\[ y'' + xy' + 2y = 0. \]

Verify that $x = 0$ is an ordinary point. What does the Existence Theorem guarantee about the radius of convergence?
Use the recurrence relation from Question 7(c) to compute $a_2$ through $a_7$ in terms of $a_0$ and $a_1$.
Write the general solution $y = a_0\,y_1(x) + a_1\,y_2(x)$ through the $x^7$ term. Confirm linear independence by evaluating $W(y_1,y_2)(0)$.
Find the particular solution satisfying $y(0) = 4$ and $y'(0) = -2$.

SolutionSolution

(a) Rewrite as $y'' + P(x)y' + Q(x)y = 0$ with $P(x) = x$ and $Q(x) = 2$. Both are polynomials, hence analytic everywhere. Therefore $x = 0$ is an ordinary point. Since there are no singular points at all, the Existence Theorem guarantees that the power series solutions centered at $x = 0$ converge for all $x$ (radius of convergence $R = \infty$).

(b) Recurrence: $a_{n+2} = -\dfrac{a_n}{n+1}$. Compute even and odd indices separately.

Even-indexed (determined by $a_0$):

\[a_2 = -a_0, \quad a_4 = -\dfrac{a_2}{3} = \dfrac{a_0}{3}, \quad a_6 = -\dfrac{a_4}{5} = -\dfrac{a_0}{15}\]

Odd-indexed (determined by $a_1$):

\[a_3 = -\dfrac{a_1}{2}, \quad a_5 = -\dfrac{a_3}{4} = \dfrac{a_1}{8}, \quad a_7 = -\dfrac{a_5}{6} = -\dfrac{a_1}{48}\]

(c) Setting $(a_0,a_1) = (1,0)$ and $(0,1)$ respectively:

\[y_1(x) = 1 - x^2 + \dfrac{x^4}{3} - \dfrac{x^6}{15} + \cdots\]

\[y_2(x) = x - \dfrac{x^3}{2} + \dfrac{x^5}{8} - \dfrac{x^7}{48} + \cdots\]

At $x = 0$: $y_1(0) = 1$, $y_1'(0) = 0$, $y_2(0) = 0$, $y_2'(0) = 1$. The Wronskian is:

\[W(y_1,y_2)(0) = y_1(0)\,y_2'(0) - y_1'(0)\,y_2(0) = (1)(1)-(0)(0) = 1 \neq 0\]

The solutions are linearly independent. ✓

(d) From $y(0) = a_0 = 4$ and $y'(0) = a_1 = -2$:

\[\boxed{y(x) = 4\,y_1(x) - 2\,y_2(x) = 4 - 2x - 4x^2 + x^3 + \dfrac{4}{3}x^4 - \dfrac{x^5}{4} - \dfrac{4}{15}x^6 + \dfrac{x^7}{24} + \cdots}\]

9.

The equation $y'' - xy = 0$ is known as the Airy equation and arises in optics and quantum mechanics. Find its general power series solution centered at $x = 0$.

Confirm that $x = 0$ is an ordinary point. Substitute $y = \displaystyle\sum_{n=0}^\infty a_n x^n$ into the equation. By matching the $x^0$ coefficient, show that $a_2 = 0$. Then align the remaining series and derive the recurrence relation $a_{n+3} = \dfrac{a_n}{(n+3)(n+2)}$ for $n \geq 0$.
Use the recurrence to compute $a_3$ through $a_8$ in terms of $a_0$ and $a_1$ (recall $a_2 = 0$, so every $a_{3k+2} = 0$).
Write the general solution $y = a_0\,y_1(x) + a_1\,y_2(x)$ where each series contains at least three nonzero terms. These are the (unnormalized) Airy functions $\mathrm{Ai}(x)$ and $\mathrm{Bi}(x)$ up to scalar multiples.

SolutionSolution

(a) Written in standard form: $y'' + 0\cdot y' + (-x)y = 0$. Here $P(x) = 0$ and $Q(x) = -x$ are both polynomials, so $x = 0$ is an ordinary point. Substituting $y = \displaystyle\sum_{n=0}^\infty a_n x^n$:

\[y'' = \sum_{n=0}^{\infty}(n+2)(n+1)\,a_{n+2}\,x^n, \qquad xy = \sum_{n=0}^{\infty} a_n\,x^{n+1} = \sum_{n=1}^{\infty} a_{n-1}\,x^n\]

So $y'' - xy = 0$ becomes:

\[2\cdot 1\cdot a_2\,x^0 + \sum_{n=1}^{\infty}\left[(n+2)(n+1)\,a_{n+2} - a_{n-1}\right]x^n = 0\]

$x^0$ coefficient: $2a_2 = 0 \;\Rightarrow\; a_2 = 0$.

$x^n$ coefficient ($n \geq 1$): $(n+2)(n+1)\,a_{n+2} = a_{n-1}$. Substitute $m = n-1$ (so $n = m+1$):

\[\boxed{a_{m+3} = \dfrac{a_m}{(m+3)(m+2)}, \quad m \geq 0}\]

(b) Applying the recurrence:

From $a_0$ (indices $0, 3, 6, \ldots$):

\[a_3 = \dfrac{a_0}{3\cdot 2} = \dfrac{a_0}{6}, \qquad a_6 = \dfrac{a_3}{6\cdot 5} = \dfrac{a_0}{180}\]

From $a_1$ (indices $1, 4, 7, \ldots$):

\[a_4 = \dfrac{a_1}{4\cdot 3} = \dfrac{a_1}{12}, \qquad a_7 = \dfrac{a_4}{7\cdot 6} = \dfrac{a_1}{504}\]

From $a_2 = 0$ (indices $2, 5, 8, \ldots$):

\[a_5 = \dfrac{a_2}{5\cdot 4} = 0, \qquad a_8 = \dfrac{a_5}{8\cdot 7} = 0\]

In general, $a_{3k+2} = 0$ for all $k \geq 0$, since every such coefficient propagates from $a_2 = 0$.

(c)

\[y_1(x) = 1 + \dfrac{x^3}{6} + \dfrac{x^6}{180} + \cdots\]

\[y_2(x) = x + \dfrac{x^4}{12} + \dfrac{x^7}{504} + \cdots\]

\[\boxed{y(x) = a_0\!\left(1 + \dfrac{x^3}{6} + \dfrac{x^6}{180}+\cdots\right) + a_1\!\left(x + \dfrac{x^4}{12} + \dfrac{x^7}{504}+\cdots\right)}\]

Linear independence follows from $W(y_1, y_2)(0) = y_1(0)y_2'(0) - y_1'(0)y_2(0) = (1)(1)-(0)(0) = 1 \neq 0$. ✓

10.

Solve the initial value problem

\[ y'' - xy' - y = 0, \quad y(0) = 3, \quad y'(0) = 0 \]

using the power series method and identify the solution in closed form.

Verify that $x = 0$ is an ordinary point. Substitute $y = \displaystyle\sum_{n=0}^\infty a_n x^n$ and show that the recurrence relation is $a_{n+2} = \dfrac{a_n}{n+2}$ for $n \geq 0$.
Apply the initial conditions $y(0) = 3$, $y'(0) = 0$ to set $a_0 = 3$ and $a_1 = 0$. Use the recurrence to find $a_2$, $a_4$, $a_6$, and $a_8$.
Show that the general term for the nonzero coefficients is $a_{2k} = \dfrac{3}{2^k\,k!}$, recognize the series as $y(x) = 3e^{x^2/2}$.
Verify directly that $y = 3e^{x^2/2}$ satisfies both the ODE and the initial conditions.

SolutionSolution

(a) Standard form: $y'' + (-x)y' + (-1)y = 0$. Both $P(x) = -x$ and $Q(x) = -1$ are polynomials, so $x = 0$ is an ordinary point. Substituting $y = \displaystyle\sum_{n=0}^\infty a_n x^n$:

\[y'' = \sum_{n=0}^{\infty}(n+2)(n+1)\,a_{n+2}\,x^n, \quad xy' = \sum_{n=0}^{\infty} n\,a_n\,x^n, \quad y = \sum_{n=0}^{\infty} a_n\,x^n\]

So $y'' - xy' - y = 0$ becomes:

\[\sum_{n=0}^{\infty}\left[(n+2)(n+1)\,a_{n+2} - n\,a_n - a_n\right]x^n = \sum_{n=0}^{\infty}\left[(n+2)(n+1)\,a_{n+2} - (n+1)\,a_n\right]x^n = 0\]

By the identity theorem each coefficient must vanish: $(n+2)(n+1)\,a_{n+2} = (n+1)\,a_n$. Dividing by $(n+1)$ (which is nonzero):

\[\boxed{a_{n+2} = \dfrac{a_n}{n+2}, \quad n \geq 0}\]

(b) $y(0) = a_0 = 3$ and $y'(0) = a_1 = 0$. Since $a_1 = 0$, all odd-indexed coefficients vanish. Even terms:

\[a_2 = \dfrac{a_0}{2} = \dfrac{3}{2}, \quad a_4 = \dfrac{a_2}{4} = \dfrac{3}{8}, \quad a_6 = \dfrac{a_4}{6} = \dfrac{3}{48} = \dfrac{1}{16}, \quad a_8 = \dfrac{a_6}{8} = \dfrac{3}{384} = \dfrac{1}{128}\]

(c) The recurrence $a_{n+2} = \dfrac{a_n}{n+2}$ for even indices means:

\[a_{2k} = \dfrac{a_{2(k-1)}}{2k} = \dfrac{a_{2(k-2)}}{(2k)(2k-2)} = \cdots = \dfrac{a_0}{2\cdot 4\cdot 6\cdots(2k)} = \dfrac{3}{2^k\,k!}\]

Therefore:

\[y(x) = \sum_{k=0}^{\infty} a_{2k}\,x^{2k} = \sum_{k=0}^{\infty}\dfrac{3}{2^k\,k!}\,x^{2k} = 3\sum_{k=0}^{\infty}\dfrac{(x^2/2)^k}{k!}\]

\[\boxed{y(x) = 3\,e^{x^2/2}}\]

since $e^u = \displaystyle\sum_{k=0}^\infty \dfrac{u^k}{k!}$ with $u = x^2/2$.

(d) Direct verification. Compute the derivatives of $y = 3e^{x^2/2}$:

\[y' = 3x\,e^{x^2/2}, \qquad y'' = 3e^{x^2/2} + 3x^2 e^{x^2/2} = 3(1+x^2)e^{x^2/2}\]

Substitute into $y'' - xy' - y$:

\[3(1+x^2)e^{x^2/2} - x\cdot 3x\,e^{x^2/2} - 3e^{x^2/2} = e^{x^2/2}\bigl[3 + 3x^2 - 3x^2 - 3\bigr] = 0 \;\checkmark\]

Initial conditions: $y(0) = 3e^0 = 3$ ✓ and $y'(0) = 3(0)e^0 = 0$ ✓.

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