# Solve the following IVP for the second order linear equations y''-4y'+9y=0, y(0)=0, y'(0)=-8

Second order linear equations
Solve the following IVP for the second order linear equations
$$y''-4y'+9y=0,\ \ y(0)=0,\ \ y'(0)=-8$$

2021-01-11

Given:
$$y''-4y'+9y=0$$
$$y(0)=0$$
$$y'(0)=-8$$
The auxiliary equation of the givendifferential equation is
$$m^2-4m+9=0$$
$$m=\frac{-(-4)+-\sqrt{(-4)^2-(4)(1)(9)}}{2}$$
$$=\frac{4\pm\sqrt{16-36}}{2}$$
$$=\frac{(4\pm i2\sqrt5)}{2}$$
$$m=2\pm i\sqrt5$$
Hence the general solution is
$$y=e^{2x}(A\cos\sqrt5x+B\sin\sqrt5x)$$
$$y(0)=0$$
$$\Rightarrow0=1(A(1)+B(0))$$
$$\Rightarrow A=0$$
$$y=e^{2x}(B\sin\sqrt5x)$$
$$y'(0)=-8$$
$$y'=2e^x(B\sin\sqrt5x)+\sqrt5e^x(B\cos\sqrt5x)$$
$$-8=0+\sqrt5(B)$$
$$B=\frac{-8}{\sqrt5}$$
Hence, the solution is
$$y=-e^{2x}(\frac{8}{\sqrt5}\sin\sqrt5x)$$