MATH 315 FALL 2006 TEST 3 SOLUTION KEY

Note: the following solutions are for one version of the test; some final solutions for the alternate test version problems are also given.

1. (20 pts.) Use Power series to solve the differential equation $y' + xy = 1 = x,$ with $y(0) = 1$.
   1. 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:
      $\begin{array}{rcrcrcrcrcrl} y'&=&a_1&+&2a_2x&+&3a_3x^2&+&4a_4x^3&+\cdots+&(n+1)... ...ts\ \hline = 1-x& =& 1&-& x&+& 0x^2&+& 0x^3&+\cdots+&0x^n&+\cdots \end{array}$,
      so $a_1 = 1$, $2a_2+a_0 = -1$ (alternate test version $a_1 = 2$, $2a_2+a_0 = 1$),
      and $(n+1)a_{n+1}+a_{n-1}=0$, or $a_{n+1}=-a_{n-1}/(n+1)$ for $n>1$.
   2. Determine the first six terms in the series for $y(x)$.
      Answer: using intial value $a_0=1$, then $a_1=1, a_2=(-1-a_0)/2=-1$,
      $a_3=-a_1/3=-1/3, a_4=-a_2/4=1/4, a_5=-a_3/5=1/15$;
      final solution: $y(x) \simeq 1 + x - x^2 - x^3/3 + x^4/4+x^5/15$
      (alternate test version $y(x) \simeq 2 + 2x - x^2/2 - 2x^3/3 + x^4/8+2x^5/15$).

2. (22 pts.) Use Laplace transforms to solve $y''+ 3y' + 2y = 2,$ with $y(0) = 1$, $y'(0) = -1$.
   Answer: the transformed equation becomes
   $s^2Y(s) - sy(0) - y'(0) + 3(sY(s) - y(0)) + 2Y(s) = 2/s$ or $(s^2+3s+2)Y(s)= 2/s + s+2$, and therefore $Y(s) = \frac{2}{s(s+1)(s+2)} + \frac{1}{s+1}$. Using partial fractions $\frac{2}{s(s+1)(s+2)}= \frac{A}{s}+ \frac{B}{(s+1)}+ \frac{C}{(s+2)}$, so $A(s^2+3s+2)+B(s^2+2s)+C(s^2+s)=2$, or $s^2(A+B+C)+s(3A+2B+C)+2A=2$,
   and therefore $A=1$, $2B+C=-3$, $B+C=-1$, so $B=-2$ and $C=1$.
   Then $Y(s) =\frac{1}{s}-\frac{2}{s+1}+\frac{1}{s+2}+\frac{1}{s+1}$; final solution: $y(t) = 1-e^{-t}+e^{-2t}$
   (alternate test version $y(t) = 1 + 2e^{-2t}-2e^{-t}$).

3. (13 pts.) Find the Laplace transform $Y(s)$ for the solution to the equation $y'''+ 2y'' + 2y' = e^{2t}$, with $y(0) = 0$, $y'(0) = 0$, $y''(0) = 2$.
   Answer: the transformed equation becomes
   $s^3Y(s)-s^2y(0)-sy'(0)-y''(0)+2(s^2Y(s)-sy(0)-y'(0))+2(sY(s) - y(0))=1/(s-2)$,
   or $(s^3+2s^2+2s)Y(s)= 2 + 1/(s-2)$; final solution: $Y(s)=\frac{2}{s(s^2+2s+2)}+\frac{1}{s(s-2)(s^2+2s+2)}$
   (alternate test version $Y(s)=\frac{3}{s(s^2+2s+5)}+\frac{1}{s(s+1)(s^2+2s+5)}$).

4. (15 pts.) Consider the equation $y'' + 2xy'+ y = 0 ,$ with $y(0) = 0$, $y'(0)=1$. 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:
   $\begin{array}{rcrcrcrcrcrl} y'' &=&2a_2&+&6a_3x&+&12a_4x^2&+&20a_5x^3&+\cdots+&... ...ts\ \hline = 0& =& 0&+& 0x&+& 0x^2&+& 0x^3&+\cdots+& 0x^n&+\cdots \end{array}$,
   so $(n+2)(n+1)a_{n+2}=-(2n+1)a_n$, or $a_{n+2}=-(2n+1)a_n/((n+1)(n+2)$
   (alternate test version $a_{n+2}=-a_n/(n+2)$), for $n \geq 0$.

5. (8 pts.) Convert the 4$^{th}$ order differential equation $y'''' + \sin(t)y'''- t^3y'+ \cos(y) = \log(1+2t)$, with $y(0) = 3$, $y'(0)=1$, $y''(0) = -2$, $y'''(0) = 1$, to a 1$^{st}$ order system of equations.
   Answer: let $x_1=y$, $x_2=y'$, $x_3=y''$, and $x_4=y'''$;
   final solution: $\left[\begin{array}{c}x_1\ x_2\ x_3\ x_4\end{array}\right]'= \left[\begin{ar... ...x_2\ x_3\ x_4\\ \log(1+2t) - \sin(t)x_4+ t^3x_2- \cos(x_1)\end{array}\right]$; ${\bf x}(0) = \left[\begin{array}{r}3\ 1\ -2\ 1\end{array}\right]$.

6. (22 pts.) Find the solution for the system of differential equations
   
   $${\bf x}' =\left[\begin{array}{cc}2&2\ 2&-1\end{array}\right... ...} {\bf x}(0) = \left[\begin{array}{c}1\ 3\end{array}\right] .$$
   
   
   Answer: eigenvalues determined from $(2-r)(-1-r)-4=0$, or $r^2-r-6=0$, so $r=-2,3$.
   Eigenvectors:
   $r = -2$: $A+2I= \left[\begin{array}{cc}4&2\ 2&1\end{array}\right]\sim \left[\begin{array}{cc}2&1\ 0&0\end{array}\right]$, so ${\bf v}_1= \left[\begin{array}{c}1\ -2\end{array}\right]$;
   $r = 3$: $A-3I= \left[\begin{array}{cc}-1&2\ 2&-4\end{array}\right]\sim \left[\begin{array}{cc}-1&2\ 0&0\end{array}\right]$, so ${\bf v}_2= \left[\begin{array}{c}2\ 1\end{array}\right]$, and ${\bf x}(t)=c_1{\bf v}_1e^{-2t}+c_2{\bf v}_2e^{3t}$.
   Constants $c_1$ and $c_2$ are determined from initial values using ${\bf x}(0)=c_1{\bf v}_1+c_2{\bf v}_2$, or
   $\left[\begin{array}{ccc}1&2&1\ -2&1&3\end{array}\right]\sim \left[\begin{array}{ccc}1&2&1\ 0&5&5\end{array}\right]$, so $c_2=1$, $c_1=-1$; final solution: ${\bf x}(t)=-{\bf v}_1e^{-2t}+{\bf v}_2e^{3t}$
   (alternate test version ${\bf x}(t)=2{\bf v}_1e^{t}-{\bf v}_2e^{-t}$, with ${\bf v}_1= [1\ 1]'$, ${\bf v}_1= [1\ 3]'$ ).