Consider $y^{''}+y^{'}-2y=0$. Find the general solution to this differential equation. Convert the second order equation to a system of linear equations corresponds to the general solution of the second order equation.

Solution so far

Substitution $y_1=y$, $y_2 =\dot{y}$ yields an equivalent 1st order system

\begin{cases} \dot{y_1}=y_2, \\ \dot{y_2}=2y_1 -y_2 \end{cases} i.e $\dot{y}=Ay$, where $A=\begin{bmatrix} 0& 1 \\ 2 & -1 \end{bmatrix}$

Eigenvalues $\lambda_{1}=1$ , $\lambda_{2}=-2$ eigenvectors $v_{1}=[1,1]^{T}$, $v_{2}=[1,-2]^{T}$ . Hence $P=\begin{bmatrix} 1& 1 \\ 1 & -2 \end{bmatrix}$ with $P^{-1}=\begin{bmatrix} \frac{2}{3}& \frac{1}{3} \\ \frac{1}{3} & \frac{-1}{3} \end{bmatrix}$

Thus the solution of the IVP is given by $y(t)= P \begin{bmatrix} e^t& 0 \\ 0 & e^{-2t} \end{bmatrix} P^{-1} y(0)$

Is what I have correct?

  • 2
    $\begingroup$ The conversion you did is correct. You can write $Y(t) = c_1 e^{1 t} v_1 + c_2 e^{-2 t} v_2$ as the solution. Not sure why you diagonalized for this problem. $\endgroup$ – Moo Feb 22 '16 at 16:00

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