Answer to Question #194962 in Differential Equations for shiva

Question

d2x/dt2=x-2y, d2y/dt2=4x+5y find general solution using matrix method.

Answer 1

$z''=\begin{pmatrix} 1 & -2 \\ 4 & 5 \end{pmatrix}z$

$\begin{vmatrix} 1-\lambda & -2 \\ 4 & 5-\lambda \end{vmatrix}=0$

$(1-\lambda)(5-\lambda)+8=0$

$\lambda^2+6\lambda -13=0$

$\lambda=\frac{-6\pm \sqrt{36+52}}{2}=-3\pm \sqrt{22}$

for $\lambda_1=-3-\sqrt{22}$ :

$(4+\sqrt{22})x-2y=0$

$4x+(8+\sqrt{22})y=0$

$x\sqrt{22}-(10+\sqrt{22})y=0$

eigenvector:

$\begin{pmatrix} 1 \\ 1+10/\sqrt{22} \end{pmatrix}$

for $\lambda_2=-3+\sqrt{22}$ :

$(4-\sqrt{22})x-2y=0$

$4x+(8-\sqrt{22})y=0$

$-x\sqrt{22}-(10-\sqrt{22})y=0$

eigenvector:

$\begin{pmatrix} 1 \\ 1-10/\sqrt{22} \end{pmatrix}$

$z'(t)=c_1e^{-(3+\sqrt{22})t}\begin{pmatrix} 1 \\ 1+10/\sqrt{22} \end{pmatrix}+c_2e^{-(3-\sqrt{22})t}\begin{pmatrix} 1 \\ 1-10/\sqrt{22} \end{pmatrix}$

$z(t)=C_1e^{-(3+\sqrt{22})t}\begin{pmatrix} 1 \\ 1+10/\sqrt{22} \end{pmatrix}+C_2e^{-(3-\sqrt{22})t}\begin{pmatrix} 1 \\ 1-10/\sqrt{22} \end{pmatrix}$