Question

If y=2sinx+3cosx show that d²y/dx²+y=0.

Answer 1

Answer:

$\displaystyle \sf \longrightarrow \frac{d^2y}{dx^2}+y =0 \quad [ \ Shown \ ]$

Step-by-step explanation:

Given :

$\displaystyle \sf y=2\sin x+3 \cos x$

We have to show :

$\displaystyle \sf \frac{d^2y}{dx^2}+y=0$

We know :

$\displaystyle \sf \frac{d}{dx}\left(\sin x \right)=\cos x$

$\displaystyle \sf \frac{d}{dx}\left(\cos x \right)=-\sin x$

Now :

$\displaystyle \sf y=2\sin x+3 \cos x \quad ...(i)$

Diff. w.r.t. x :

$\displaystyle \sf \frac{dy}{dx} =(2\sin x+3 \cos x)'$

$\displaystyle \sf \frac{dy}{dx} =2\cos x-3 \sin x$

Finding second order derivative :

Diff. w.r.t x :

$\displaystyle \sf \frac{d^2y}{dx^2} =(2\cos x-3 \sin x)'$

$\displaystyle \sf \frac{d^2y}{dx^2} =-2\sin x-3 \cos x \quad ...(ii)$

Now adding ( i )  and  ( ii ) we get :

$\displaystyle \sf \frac{d^2y}{dx^2}+y =-2\sin x-3 \cos x+2\sin x+3 \cos x$

$\displaystyle \sf \longrightarrow \frac{d^2y}{dx^2}+y =0$

Hence shown!