Answer to Question #206799 in Differential Equations for Leela krishna

Question

Reduce the second order linear differential equation d²y/dt²-7dy/dt+10y=0 to linear system of first order differential equation and hence solve the system of ODE's

Answer 1

$$\dfrac{d^2y}{dt^2}-7\dfrac{dy}{dt}+10y=0$$

$$y''-7y'+10y=0$$

Let $u=y, v=y'(t).$

Then $u'=y'=v, v'=y''=-10y+7y'=-10u+7v$

$$u'=0u+1v$$ $$v'=-10u+7v$$

$$\begin{bmatrix} u' \\ v' \end{bmatrix}= \begin{bmatrix} 0 & 1 \\ -10 & 7 \end{bmatrix} \begin{bmatrix} u \\ v \end{bmatrix}$$

$$\vec {x'}=\begin{bmatrix} u' \\ v' \end{bmatrix}, A=\begin{bmatrix} 0 & 1 \\ -10 & 7 \end{bmatrix}, \vec x=\begin{bmatrix} u \\ v \end{bmatrix}$$

The linear system of first order differential equation 

$$\vec {x'}= A\vec x$$

Find the eigenvalues and eigen vectors of $A=\begin{bmatrix} 0 & 1 \\ -10 & 7 \end{bmatrix}$

$$\det (A-\lambda I)=A=\begin{vmatrix} 0-\lambda & 1 \\ -10 & 7-\lambda \end{vmatrix}$$

$$=-\lambda(7-\lambda)+10=\lambda^2-7\lambda+10$$

$$\det (A-\lambda I)=0=>\lambda^2-7\lambda+10=0$$

$\lambda_1=5, \lambda_2=2$

These are the eigenvalues.

$\lambda=5$

$$\begin{bmatrix} 0-\lambda & 1 \\ -10 & 7-\lambda \end{bmatrix}=\begin{bmatrix} -5 & 1 \\ -10 & 2 \end{bmatrix}$$

$R_2=R_2-2R_1$

$$\begin{bmatrix} -5 & 1 \\ 0 & 0 \end{bmatrix}$$

$R_1=-R_1/5$

$$\begin{bmatrix} 1 & -1/5 \\ 0 & 0 \end{bmatrix}$$

If we take $q_2=1,$ then $q_1=1/5.$

$$\vec q=\begin{bmatrix} 1/5 \\ 1 \end{bmatrix}$$

$\lambda=2$

$$\begin{bmatrix} 0-\lambda & 1 \\ -10 & 7-\lambda \end{bmatrix}=\begin{bmatrix} -2 & 1 \\ -10 & 5 \end{bmatrix}$$

$R_2=R_2-5R_1$

$$\begin{bmatrix} -2 & 1 \\ 0 & 0 \end{bmatrix}$$

$R_1=-R_1/2$

$$\begin{bmatrix} 1 & -1/2 \\ 0 & 0 \end{bmatrix}$$

If we take $s_2=1,$ then $s_1=1/2.$

$$\vec s=\begin{bmatrix} 1/2 \\ 1 \end{bmatrix}$$

General solution is

$$\begin{bmatrix} u \\ v \end{bmatrix}=C_1e^{5t}\begin{bmatrix} 1 \\ 5 \end{bmatrix}+C_2e^{2t}\begin{bmatrix} 1 \\ 2 \end{bmatrix}$$

$$y=C_1e^{5t}+C_2e^{2t}$$