Question

sin-1 dy/dx = x+y solve the following differential equation?

Answer 1

Here's the solution to your question.

Answer 2

Answer:

$-\frac{2}{tan\frac{x+y}{2}+1 } =x+c$

Step-by-step explanation:

We have, Sin¬1($\frac{dy}{dx}$) = x+y

⇒ $\frac{dy}{dx}=Sin(x+y)$ ...... (1)

To solve this differential equation, assume

x+y =z

Now, differentiating both sides with respect to x, we get,

$1+\frac{dy}{dx}=\frac{dz}{dx}$

⇒ $\frac{dy}{dx}=\frac{dz}{dx}-1$

Hence, the equation (1) becomes

$\frac{dz}{dx}-1=Sinz$

⇒$\frac{dz}{dx}=1+ Sinz$

⇒$\frac{dz}{1+Sinz}=dx$

Integrating both sides we get

⇒ ∫$\frac{dz}{1+Sinz}$= x+c {Where c is an integration constant}

⇒∫$\frac{Sec^{2}\frac{z}{2} dz }{Sec^{2}\frac{z}{2}+2 tan\frac{z}{2}   }$=x+c

{Since, Sin2x= 2Sinx Cosx}

⇒∫$\frac{2d(tan\frac{z}{2} )}{tan^{2}\frac{z}{2}+2 tan\frac{z}{2} +1  }$= x+c

{ Since, $Sec^{2}x- tan^{2}x=1$}

⇒2∫$\frac{d(tan\frac{z}{2}+1 )}{(tan\frac{z}{2} +1)^{2} } =x+c$

⇒$-\frac{2}{tan\frac{z}{2}+1 } =x+c$

Again putting the value of z, we get

⇒ $-\frac{2}{tan\frac{x+y}{2}+1 } =x+c$ (Answer)