Question

Show that $y=A\cos nx+B\sin nx$, is a solution of the differential equation: $\frac{d^{2}y}{d{x}^2}+n^{2}y=0$.

Answer 1

given differential equation is $\frac{d^{2}y}{d{x}^2}+n^{2}y=0$

$y=A\cos nx+B\sin nx...\left(2\right)$

Differentiating on bot sides

$\frac{dy}{dx}=-An\sin nx+Bn\cos nx$

Again differentiating on both sides

$\frac{d^{2}y}{d{x}^2}=-A{n}^2\sin nx-B{n}^2\sin nx$

Substituting in (1)

LHS $=-A{n}^2\sin nx-B{n}^2\sin nx+n^2(A\cos nx+B\sin nx)=0$

RHS$=0$

LHS=RHS

$\therefore$ $y=A\cos nx+B\sin nx$ is a solution of (1)