Question

For the following question verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.

$y=xsin3x:\frac{d^{2}y}{d{x}^2}+9y-6cos3x=0.$

Answer 1

Given, y=xsin3x ....(i)
On differentiating bothe sides w.r.t. x, we get
$\frac{dy}{dx}=x\frac{d}{dx}\left(sin3x\right)+sin3x\frac{d}{dx}\left(x\right)$ (using product rule of differentiation)
$\frac{dy}{dx}=xcos3x\times 3+sin3x$
Again, differentiating bothe sides w.r.t. x, we get
$\frac{d^{2}y}{d{x}^2}=3[x\frac{d}{dx}cos3x+cos3x\frac{d}{dx}x]+\frac{d}{dx}\left(sin3x\right)\Rightarrow \frac{d^{2}y}{d{x}^2}=3\left[x\right(-sin3x\times 3)+cos3x]+cos3x\times 3\Rightarrow \frac{d^{2}y}{d{x}^2}==9xsin3x+3cos3x+3cos3x\Rightarrow \frac{d^{2}y}{d{x}^2}=-9xsin3x+6cos3x\Rightarrow \frac{d^{2}y}{d{x}^2}=-9y+6cos3x(UsingEq.(i),y=xsin3x)\Rightarrow \frac{d^{2}y}{d{x}^2}+9y-6cos3x=0$
Hence, the given function is a solution of the corresponding differential equation.