Question

Solve. dx/dt=-2x+7y,dy/dt=3x+2ysubject to the condition x(0)=9 and y(0)=-1

Answer 1

Let us solve the sysytem:

$\begin{cases} \frac{dx}{dt}=-2x+7y \\ \frac{dy}{dt}=3x+2y \end{cases}$ $x(0)=9$ and $y(0)=-1$

The second eqution is equivalent to $x=\frac{1}{3}( \frac{dy}{dt}-2y)$, and therefore, $\frac{dx}{dt}=\frac{1}{3}( \frac{d^2y}{dt^2}-2\frac{dy}{dt})$ . Let us put these in the first equation:

$\frac{1}{3}( \frac{d^2y}{dt^2}-2 \frac{dy}{dt})=-2\cdot\frac{1}{3}( \frac{dy}{dt}-2y)+7y$

$\frac{1}{3} \frac{d^2y}{dt^2}-\frac{2}{3} \frac{dy}{dt}=-\frac{2}{3}\frac{dy}{dt}+\frac{4}{3}y+7y$

$\frac{1}{3} \frac{d^2y}{dt^2}=\frac{25}{3}y$

$\frac{d^2y}{dt^2}-25y=0$

The characteristic equation $k^2-25=0$ of the last differential eqution has the solutions $k=5$ and $k=-5.$ Consequently, its general solution is of the form

$y(t)=C_1e^{5t}+C_2e^{-5t}$

$\frac{dy}{dt}=5C_1e^{5t}-5C_2e^{-5t}$

Thus, $x=\frac{1}{3}( \frac{dy}{dt}-2y)=\frac{1}{3}( 5C_1e^{5t}-5C_2e^{-5t}-2(C_1e^{5t}+C_2e^{-5t}))=C_1e^{5t}-\frac{7}{3}C_2e^{-5t}.$

If $x(0)=9$ and $y(0)=-1$, then $9=x(0)=C_1-\frac{7}{3}C_2$ and $-1= y(0)=C_1+C_2$ .

It follows that $C_1=2,\ C_2=-3.$

Therefore, the solution is of the form:

$\begin{cases} x(t)=2e^{5t}+7e^{-5t}\\ y(t)=2e^{5t}-3e^{-5t} \end{cases}$