Question

Then $y=e^{ax}sinbx$ solve the equation:
$\frac{d^{2}y}{d{x}^2}-2a$ $\frac{dy}{dx}+(a^2+b^2)y=0$

Answer 1

Given, $y=e^{ax}sinbx...\left(i\right)$

Diff.w.r.t.$x$ of both sides,

$\Rightarrow \frac{dy}{dx}=e^{ax}\times \frac{d}{dx}\left(\sin bx\right)+\sin bx\times \frac{d}{dx}\left(e^{ax}\right)$

$=e^{ax}\times \cos bx\times \frac{d}{dx}\left(bx\right)+\sin bx{e}^{ax}\times \frac{d}{dx}\left(ax\right)$

$=e^{ax}\times \cos bxb+e^{ax}\sin bx\times a$

$=b{e}^{ax}\cos bx+ay...\left(ii\right)$

Again, Diff.w.r.t.$x$ of both sides,

$\frac{d^{2}y}{d{x}^2}=b[e^{ax}\times \frac{d}{dx}(\cos bx)+\cos bx\times \frac{d}{dx}(e^{ax}\left)\right]+a\times \frac{dy}{dx}$

$b\left[e^{ax}\right(-\sin bx\left)\frac{d}{dx}\right(bx)+(\cos bx)+\cos bx\times e^{ax}\times \frac{d}{dx}(e^{ax}\left)\right]+a\frac{dy}{dx}$

$b\{-e^{ax}\sin bx.b+e^{ax}\cos bx\times a\}+a\frac{dy}{dx}$

$-b^2\left(e^{ax}\sin bx\right)+a\left(b{e}^{ax}\cos bx\right)+a\frac{dy}{dx}$

$=-b^{2}y+a[\frac{dy}{dx}-ay]++a\frac{dy}{dx}$ [from $\left(i\right)$ and $\left(ii\right)$]

$=-b^{2}y+a\frac{dy}{dx}-a^{2}y+a\frac{dy}{dx}$

$=2a\frac{dy}{dx}-(a^2+b^2)y$

Hence, $\frac{d^{2}y}{d{x}^2}-2a$ $\frac{dy}{dx}+(a^2+b^2)y=0$