MATH 315 FALL 2006 TEST 2 KEY

1. (20 pts.) Consider the differential equation $y''' - y'' - 2y' = 0$.
   1. Find the general solution for the equation.
      Answer: characteristic polynomial is $r(r^2-r-2)=r(r+1)(r-2)$, with roots 0, -1, 2.
      The general solution is therefore $y(t)=c_1+c_2e^{-t}+c_3e^{2t}.$

   2. Determine the Wronskian for your fundamental solution set.
      
      $$W = \left\vert \begin{array}{ccc} 1&e^{-t}&e^{2t}\\ 0&-e^{-t}... ...^{-t}&4e^{2t} \end{array}\right\vert = -4e^{t}-2e^{t}=-6e^{t}.$$
      
      

   3. Determine the solution that satisfies initial conditions $y(0) = 1$, $y'(0) = 1$, $y''(0)=5$.
      $$\begin{eqnarray*} \mbox{\bf Answer:}\hspace{4mm} y(t)=c_1+c_2e^{-t}+c_3e^{2t},& ... ...\ y''(t)=c_2e^{-t}+4c_3e^{2t}, & \mbox{ so } & y(0)=c_2+4c_3=5; \end{eqnarray*}$$
      
      
      the solution to the linear system is $c_1=-1$, $c_2=1$, $c_3=1$.
      
      Final solution: $y(t)=-1+e^{-t}+e^{2t}.$

2. (16 pts.) Find the general solution for the differential equation $y'''' - y = 4e^{t}$.
   Answer: characteristic polynomial is $r^4-1 = (r^2-1)(r^2+1)$, with roots $\pm 1, \pm i$, so
   
   $y_c(t) = c_1e^{t}+c_2e^{-t}+c_3\cos(t)+c_4\sin(t).$
   
   A particular solution should have the form
   $$\begin{eqnarray*} Y(t) & = &\ Ate^{t};\\ Y'(t) & = &Ate^{t}+ \ Ae^{t},\\ Y'... ... &(At+4A-At)e^{t}\\ & = &4Ae^{t}=4e^{2t}, \mbox{\ so\ } A = 1. \end{eqnarray*}$$
   
   
   Final solution: $y(t) = y_c(t) + te^{t}$.

3. (15 pts.) Suppose the general solution to a $7^{th}$ order homogeneous equation $L[y]=0$ is
   
   $$y_c(t) = c_1e^{t}+c_2t\sin(t)+c_3t\cos(t)+c_4\cos(t)+c_5\sin(t) +c_6e^{2t}+c_7te^{2t}.$$
   
   
   Determine the general form that would be used with the method of undetermined coefficients for the solution to the nonhomogeneous equation $L[y] = 5te^{2t} - 2e^{-t}\sin(t)+\cos(t).$
   $$\begin{eqnarray*} \mbox{\bf Answer:}\hspace{4cm} g_1(t) = 5te^{2t}, &\mbox{\ so\... ..._3(t) = \cos(t), &\mbox{\ so\ }& Y_3(t) = t^2(E\cos(t)+F\sin(t)) \end{eqnarray*}$$
   
   
   Final solution: $y(t) = y_c(t) + Y_1(t)+Y_2(t)+Y_3(t).$

4. (23 pts.) Find the general solution for the differential equation $y'' - 2y' + 5y = 5t^2 - 4e^{3t}$.
   Answer: characteristic polynomial is $r^2-2r+5$, with roots $1\pm 2i$, so
   
   $y_c(t)=e^{t}(c_1\cos(2t)+c_2\sin(2t))$.
   
   For $g_1=5t^2$, a particular solution should have the form
   $$\begin{eqnarray*} Y_1(t) & = &A+Bt+Ct^2; \mbox{\ then\ }\\ Y_1'(t) & = &B+2Ct... ...2(B+2Ct)+5(A+Bt+Ct^2) \\ & = &2C-2B+5A+(5B-4C)t+5Ct^2 = 5t^2 . \end{eqnarray*}$$
   
   
   Comparing terms, $C=1$, $B=4/5$, $A=-2/25$.
   For $g_2=-4e^{3t}$, a particular solution should have the form
   $$\begin{eqnarray*} Y_2(t) & = &\ De^{3t}; \mbox{\ then\ }\\ Y_2'(t) & = &3De^{... ...Y_2 & = &De^{3t}(9 - 6 + 5) = -4e^{3t}, \mbox{\ and\ } D = -1/2. \end{eqnarray*}$$
   
   
   Final solution: $y(t) = y_c(t) - \frac{1}{2}e^{3t}+ t^2 + \frac{4}{5}t - \frac{2}{25}.$

5. (6 pts.) Determine intervals where the solution for the following problem is sure to exist:
   
   $$(t^2-4)y'' - t(t+3)y'+(t+4)y = e^t.$$
   
   
   Answer: the equation in standard form is $y'' - t(t+3)y'/(t^2-4) +(t+4)y/(t^2-4) = e^t/(t^2-4)$, so the coefficient functions for $y'$ and $y$, and the $g$ function do not exist when $t = \pm 2$.
   There should be a unique solution when $t \in (-\infty,-2)$ or $t \in (-2,2)$ or $t \in (2,\infty)$.

6. (20 pts.) Use the variation of parameters method to solve the differential equation
   
   $y''+2y'+y = t^{-1}e^{-t},$
   
   with initial values $y(1) = 0$, and $y'(1) = 0$.
   Answer: characteristic polynomial is $r^2+2r+1= (r+1)^2$, with roots $-1, -1$.
   $y_c(t) = c_1e^{-t}+c_2te^{-t}$, so $Y(t)= u_1e^{-t}+u_2te^{-t}.$
   The linear system is
   $u_1'e^{-t}+u_2'te^{-t}=0$, $-u_1'e^{-t}+u_2'(e^{-t}-te^{-t}=t^{-1}e^{-t}$.
   Adding first equation to second equation , the result is
   $u_2'e^{-t}=e^{-t}/t$, so $u'_2=1/t$ and $u_2=\ln(t)$; using first equation,
   $u_1' = -tu_2' = - 1$, so $u_1 = -t$. Therefore
   $y(t) = c_1e^{-t}+c_2te^{-t}-te^{-t}+\ln(t)te^{-t}$; now
   $y(1) = c_1e^{-1}+c_2e^{-1}-e^{-1}+0=0$, and
   $y'(1) = -c_1e^{-1}+c_2e^{-1}-c_2e^{-1}+e^{-1}=0$, so $c_1=1$ and $c_2=0$.
   Final solution: $y(t) = e^{-t}-te^{-t}+\ln(t)te^{-t}$.