Question

[ Cos x Log ( 2y-8) + 1/x ] dx + sin x /(y-4) dy = 0

given y(1)=9/2


please solve this differential equation ​

Answer 1

Step-by-step =sinxlog(2y−8)+logx=C

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,

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

• (cos x−sin x)+(sin x+cos x )

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

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

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

$\sf 2 \dfrac{dy}$

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

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 ,

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

Now our required ,

LHS

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

$\begin{gathered}\\ :\implies \sf -2e^x.sin\ x+2e^x.cos\ x-(4e^x.cos\ x)+2e^x.sin\ x+2e^x.cos\ x\\\end{gathered}$

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

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

$\orange{\bigstar}\sf Hence\; proved \green{\bigstar}$