MATH528Oct 17, 2018

Instructions: Exam is closed book. Present your answers legibly, show and briefly comment steps in your solution.

1. Find the general solution of the following ODEs:

   1. $y^{\text{'}}+2.5y=1.6x$

   2. $y^{\text{'}}-0.4y=29\sin x$

   3. $25y{y}^{\text{'}}-4x=0$

   Classify each equation. ( $3\times 2$ points = 6 points)

   Solution.

   1. $y^{\text{'}}+2.5y=1.6x$ is a linear, first-order, explicit, inhomogeneous ODE. Solution of homogeneous equation $y^{\text{'}}+2.5y=0$ is $y\left(x\right)=c{e}^{-2.5x}.$ Apply variation of parameters, $y\left(x\right)=c\left(x\right)e^{-2.5x}$ to find general solution of inhomogeneous equation

      ${\displaystyle y^{\text{'}}+2.5y=c^{\text{'}}{e}^{-2.5x}=1.6x\Rightarrow c\left(x\right)=1.6x{e}^{2.5x}\Rightarrow c\left(x\right)=1.6\int x{e}^{2.5x}dx+a,}$

      and integration by parts $\int udv=uv-\int vdu$, $u=x$, $dv=d\left(\frac{1}{2.5}{e}^{2.5x}\right)$ gives

      ${\displaystyle c\left(x\right)=\frac{1.6}{2.5}(x{e}^{2.5x}-\int e^{2.5x}dx)+a=\frac{1.6}{2.5}(x-\frac{1}{2.5})e^{2.5x}+a=1.6(0.4x-0.16)e^{2.5x}+a.}$

      General solution is

      ${\displaystyle y\left(x\right)=1.6(0.4x-0.16)+a{e}^{-2.5x}.}$

   2. $y^{\text{'}}-0.4y=29\sin x$ is a linear, first-order, explicit, inhomogeneous ODE. Solution of homogeneous equation $y^{\text{'}}-0.4y=0$ is $y=c{e}^{0.4x}$. Use method of undetermined coefficients to seek general solution as

      ${\displaystyle y\left(x\right)=c{e}^{0.4x}+a\sin x+b\cos x.}$

      Replacing,

      ${\displaystyle y^{\text{'}}-0.4y=(a-0.4b)\cos x-(b+0.4a)\sin x=29\sin x\Rightarrow a-0.4b=0,b+0.4a=-29}$ ${\displaystyle a=0.4b,1.16b=29\Rightarrow b=\frac{-29}{1.16}=-25,a=\frac{-29}{2.9}=-10}$

      and find the general solution

      ${\displaystyle y\left(x\right)=c{e}^{0.4x}-10\sin x-25\cos x.}$

   3. $25y{y}^{\text{'}}-4x=0,$ is a non-linear, first-order, implicit, separable, inhomogeneous ODE. Direct integration gives

      ${\displaystyle \frac{25}{2}{y}^2-2x^2=c\Rightarrow y=\pm \frac{1}{5}\sqrt{c+4x^2}.}$

2. Solve the IVP $y^{\text{'}\text{'}\text{'}}-y^{\text{'}\text{'}}-y^{\text{'}}+y=0$, $y\left(0\right)=0,y^{\text{'}}\left(0\right)=1,y^{\text{'}\text{'}}\left(0\right)=0.$ (3 points)

   Solution. Third-order, constant-coefficient, linear ODE. Trying solution of form $y\left(x\right)=e^{rx}$, the characteristic equation

   ${\displaystyle r^3-r^2-r+1=(r-1)(r^2-1)=0}$

   results with roots $r_{1,2}=1$ (double root), $r_3=-1$. The general solution is

   ${\displaystyle y\left(x\right)=c_1{e}^x+c_{2}x{e}^x+c_3{e}^{-x}.}$

   Applying initial conditions gives

   ${\displaystyle c_1+c_3=0,c_1+c_2-c_3=1,c_1+2c_2+c_3=0\Rightarrow c_1=\frac{1}{2},c_2=0,c_3=-\frac{1}{2}}$

   so the solution is

   ${\displaystyle y\left(x\right)=\frac{1}{2}(e^x-e^{-x})=\sinh x}$

3. Find the general solution and determine the type of critical points of the system

   ${\displaystyle \begin{array}{l}{y}_1^{\text{'}}=3y_1+4y_2\\ y_2^{\text{'}}=3y_1+2y_2\end{array}.}$

   (3 points)

   Solution. Solve the eigenproblem, $AX=X\Lambda$

   ${\displaystyle A=\left(\begin{array}{ll}3& 4\\ 3& 2\end{array}\right),\det (A-\lambda I)=(3-\lambda )(2-\lambda )-12={\lambda }^2-5\lambda -6=(\lambda -6)(\lambda +1)=0\Rightarrow {\lambda }_1=-1,{\lambda }_2=6.}$ ${\displaystyle A-{\lambda }_{1}I=\left(\begin{array}{ll}4& 4\\ 3& 3\end{array}\right)\Rightarrow x_1=\left(\begin{array}{l}1\\ -1\end{array}\right),A-{\lambda }_{2}I=\left(\begin{array}{ll}-3& 4\\ 3& -4\end{array}\right)\Rightarrow x_2=\left(\begin{array}{l}4\\ 3\end{array}\right),}$ ${\displaystyle \Lambda =\left(\begin{array}{ll}-1& 0\\ 0& 6\end{array}\right),X=\left(\begin{array}{ll}{x}_1& x_2\end{array}\right)=\left(\begin{array}{ll}1& 4\\ -1& 3\end{array}\right),X^{-1}=\frac{1}{7}\left(\begin{array}{ll}3& -4\\ 1& 1\end{array}\right).}$

   The system $y^{\text{'}}=Ay=X\Lambda X^{-1}y$, is equivalent to $z^{\text{'}}=\Lambda z$, with $z=X^{-1}y$ and solution

   ${\displaystyle z\left(x\right)=\left(\begin{array}{l}{c}_1{e}^{-x}\\ c_2{e}^{6x}\end{array}\right),y=Xz=\left(\begin{array}{ll}1& 4\\ -1& 3\end{array}\right)\left(\begin{array}{l}{c}_1{e}^{-x}\\ c_2{e}^{6x}\end{array}\right)=\left(\begin{array}{l}{c}_1{e}^{-x}+4c_2{e}^{6x}\\ -c_1{e}^{-x}+3c_2{e}^{6x}\end{array}\right)=\left(\begin{array}{ll}{e}^{-x}& 4e^{6x}\\ -e^{-x}& 3e^{6x}\end{array}\right)\left(\begin{array}{l}{c}_1\\ c_2\end{array}\right).}$

   Since $\det \left(A\right)\ne 0$, the only solution of $y^{\text{'}}=0$ is $y=0$, and the origin is a saddle point (one positive, one negative eigenvalue of $A$).