Question

y = e^x(sin x + cos x), prove that d^2y/dx^2 -2 dy/dx + 2y = 0

Answer 1

Step-by-step explanation:

2xe - xy ety – e - ty ety + e - ry 2x € 2xy +1 - exy + e - xy у dy dx fx ( x , y ) fy ( x , y ) х dạy X dy dx y 2y Also , dx2 X x2 § 11.14 . IMPLICIT FUNCTION BY A SINGLE FUNCT

Answer 2

Given :

$\bullet\ \; \sf y=e^x(sin\ x+cos\ x)$

To Prove :

$\bullet\ \; \sf \dfrac{d^2y}{dx^2}-2 \dfrac{dy}{dx}+2y=0$

Proof :

$\sf y=e^x(sin\ x+cos\ x)$

On differentiating wrt 'x' on b.s , we get ,

$\\ \to \sf \dfrac{dy}{dx}=e^x(cos\ x-sin\ x)+(sin\ x+cos\ x) e^x$

$\\ \to \sf \dfrac{dy}{dx}=e^x.cos\ x-e^x.sin\ x+e^x .sin\ x+e^x . cos\ x$

$\to \sf \dfrac{dy}{dx}=2 e^x.cos\ x$

Again differentiating b.s wrt 'x' , we get ,

$\to \sf \dfrac{d^2y}{dx^2}=-2e^x.sin\ x+2e^x.cos\ x...(1)$

Now $\sf 2\ \dfrac{dy}{dx}$ ,

$\to \sf 2( 2e^x.cos\ x )$

So ,

$\sf \to 2\ \dfrac{dy}{dx}=4e^x.cos\ x...(2)$

Now 2y ,

$\to \sf 2(e^x(sin\ x+cos\ x))$

$\to \sf 2(e^x.sin\ x+e^x.cos\ x)$

So ,

$\to \sf 2y=2e^x.sin\ x+2e^x .cos\ x ...(3)$

Now our required ,

LHS

$:\implies \sf \dfrac{d^2y}{dx^2}-2\ \dfrac{dy}{dx}+2y$

$\\ :\implies \sf -2e^x.sin\ x+2e^x.cos\ x-(4e^x.cos\ x)+2e^x.sin\ x+2e^x.cos\ x$

$:\implies \sf 4e^x.cos\ x-4e^x.cos\ x$

$:\implies \sf 0\ \; \purple{\bigstar}$

RHS

$\orange{\bigstar}$ ... Hence proved  ... $\green{\bigstar}$