Math 315 Fall 1998 Test 3 Solution Key

MATH 315 FALL 1998 TEST 3 SOLUTION KEY

1.

(10 pts.) Use Power series to solve the differential equation y' + y = 3x , with y(0) = 1.

(a)

Find the recursion formula for the coefficients a_n in the power series representation of the solution $y(x) = \sum_{n=0}^\infty a_nx^n$.

Answer:
Substituting y(x) into the equation $\sum_{n=0}^\infty na_nx^{n-1}+ \sum_{n=0}^\infty a_nx^n = 3x$, so
$\sum_{n=0}^\infty (n+1)a_{n+1}x^n+ \sum_{n=0}^\infty a_nx^n = 3x$, or $\sum_{n=0}^\infty ((n+1)a_{n+1}+a_n)x^n = 3x$.
Therefore a₁+a₀ = 0, 2a₂+a₁=3, and (n+1)a_n+1+a_n = 0 for n >1.
The recursion formula is a_n+1 = -a_n/(n+1), for n>1.

(b)

Use this recursion formula to determine a general formula for the coefficient a_n.

Answer:
Using y(0)=1, a₀ = 1, so a₁ = -1, a₂ = (3+1)/2=2, a₃=-2/3,
$a_4=2/(3\cdot 4), a_5 = -2/(3\cdot 4 \cdot 5)$, and therefore a_n = (-1)ⁿ4/n!, for n>1.

2.

(10 pts.) Determine the Laplace transform for the function

$$f(t)=\left\{ \begin{array}{cc}-2&\mbox{if }0\leq t<1\\ 3&\mbox{if }t>1\end{array}\right. .$$

Answer:

$$\begin{eqnarray*}{\cal L}\{f(t)\} & = &\int_0^{\infty}f(t)e^{-st}dt = \int_0^1... ...^{-s}-1) + \frac{3}{s}e^{-s} = \frac{5}{s}e^{-s} - \frac{2}{s}. \end{eqnarray*}$$

3.

(20 pts.) Use Power series to solve the differential equation

y'' + 4y = 0 ,

with y(0) = 0, y'(0)=1.

(a)

Find the recursion formula for the coefficients a_n in the power series representation of the solution $y(x) = \sum_{n=0}^\infty a_nx^n$.

Answer:
Substituting y(x) into the equation $\sum_{n=0}^\infty (n-1)na_nx^{n-2}+ 4\sum_{n=0}^\infty a_nx^n = 0$, so
$\sum_{n=0}^\infty (n+1)(n+2)a_{n+2}x^n+ 4\sum_{n=0}^\infty a_nx^n = 0$, or $\sum_{n=0}^\infty ((n+1)(n+2)a_{n+2}+4a_n)x^n = 0$.
Therefore (n+1)(n+2)a_n+2+4a_n=0 for $n \geq 0$, and the
recursion formula is a_n+2 = -4a_n/((n+1)(n+2)), for $n \geq 0$.

(b)

Use this recursion formula to determine a general formula for the coefficient a_n.
Answer:
Using y(0)=0, a₀ = 0, so a_2n = 0, for all $n \geq 0$. Using y'(0)=1, a₁ = 1, so $a_3 = -4/(2\cdot 3)$. $a_5=4^2/(2\cdot 3\cdot 4\cdot 5)$, and therefore a_2n+1 = (-4)ⁿ/(2n+1)!, for all $n \geq 0$.

4.

(20 pts.) Use Laplace transforms to solve the differential equation

y'' + y = e^t ,

with y(0) = 1, y'(0) = 0.

Answer:
Taking the Laplace transform, $s^2Y(s)-s+Y(s)=\frac{1}{s-1}$, so $Y(s) = \frac{s}{s^2+1} + \frac{1}{(s-1)(s^2+1)}$.
Using partial fractions $\frac{1}{(s-1)(s^2+1)}=\frac{A}{s-1}+\frac{Bs+C}{s^2+1}$, so As²+A+Bs²-Bs+Cs-C=1 and
therefore A+B=0, C-B=0 and A-C=1, so $A=\frac{1}{2}$, $B=C=-\frac{1}{2}$.
Therefore $y(t)=\cos(t)+\frac{1}{2}e^t-\frac{1}{2}\cos(t)-\frac{1}{2}\sin(t), so y(t) = \frac{1}{2}( e^t +\cos(t)-\sin(t))$.

Math 315 Fall 1998 Test 3 Solution Key

5.

(20 pts.) Find the solution for the system of differential equations

$${\bf x}' = \left[\begin{array}{cc}1&2\\ 2&1\end{array}\right]... ...} {\bf x}(0) = \left[\begin{array}{c}2\\ 1\end{array}\right] .$$

Answer:
First find the eigenvalues: $\left\vert\begin{array}{cc}1-r&2\\ 2&1-r\end{array}\right\vert = (1-r)^2-4=r^2-2r-3= (r-3)(r+1)=0$, so r₁=3 and r₂=-1. Next find the eigenvectors.
r₁=3: $\left[\begin{array}{cc}1-3&2\\ 2&1-3\end{array}\right] = \left[\begin{array}{cc... ...-2\end{array}\right] \sim \left[\begin{array}{cc}1&-1\\ 0&0\end{array}\right]$, so ${\bf v}_1 = \left[\begin{array}{c}1\\ 1\end{array}\right]$;
r₂=-1: $\left[\begin{array}{cc}1+1&2\\ 2&1+1\end{array}\right]= \left[\begin{array}{cc}2&2\\ 2&2\end{array}\right]\sim \left[\begin{array}{cc}1&1\\ 0&0\end{array}\right]$, so ${\bf v}_2 = \left[\begin{array}{c}1\\ -1\end{array}\right]$.
The general solution is therefore ${\bf x}(t)=c_1{\bf v}_1e^{3t}+c_2{\bf v}_2e^{-t}$.
Using the initial value, $\left[\begin{array}{c}2\\ 1\end{array}\right] = c_1\left[\begin{array}{c}1\\ 1\end{array}\right] + c_2\left[\begin{array}{c}1\\ -1\end{array}\right]$, so
reduce $\left[\begin{array}{ccc}1&1&2\\ 1&-1&1\end{array}\right]\sim \left[\begin{arra... ...{array}\right]\sim \left[\begin{array}{ccc}1&0&3/2\\ 0&1&1/2\end{array}\right]$.
The final solution is therefore ${\bf x}(t)=(3{\bf v}_1e^{3t}+{\bf v}_2e^{-t})/2$.

6.

(20 pts.) Find the general solution for the system of differential equations

$${\bf x}' = \left[\begin{array}{ccc}2&0&0\\ 1&3&0\\ 0&1&-1\end{array}\right]{\bf x}.$$

Answer:
First find the eigenvalues: $\left\vert\begin{array}{ccc}2-r&0&0\\ 1&3-r&0\\ 0&1&-1-r\end{array}\right\vert = -(2-r)(3-r)(r+1)=0$,
so r₁=2, r₂=3 and r₃=-1. Next find the eigenvectors.
r₁=2: $\left[\begin{array}{ccc}2-2&0&0\\ 1&3-2&0\\ 0&1&-1-2\end{array}\right] = \left[... ...y}\right] \sim \left[\begin{array}{ccc}1&0&3\\ 0&1&-3\\ 0&0&0\end{array}\right]$so ${\bf v}_1 = \left[\begin{array}{c}3\\ -3\\ -1\end{array}\right]$;
r₂=3: $\left[\begin{array}{ccc}2-3&0&0\\ 1&3-3&0\\ 0&1&-1-3\end{array}\right] = \left[... ...y}\right] \sim \left[\begin{array}{ccc}1&0&0\\ 0&1&-4\\ 0&0&0\end{array}\right]$so ${\bf v}_2 = \left[\begin{array}{c}0\\ 4\\ 1\end{array}\right]$;
r₃=-1: $\left[\begin{array}{ccc}2+1&0&0\\ 1&3+1&0\\ 0&1&1-1\end{array}\right] = \left[\... ...ay}\right] \sim \left[\begin{array}{ccc}1&0&0\\ 0&1&0\\ 0&0&0\end{array}\right]$so ${\bf v}_3 = \left[\begin{array}{c}0\\ 0\\ 1\end{array}\right]$.
The general solution is therefore ${\bf x}(t)=c_1{\bf v}_1e^{2t} +c_2{\bf v}_2e^{3t}+c_3{\bf v}_3e^{-t}$.