Question

e^2x (2cosydx - sinydy)=0​

Answer 1

Step-by-step explanation:

this is the required solution of this question.

om sai ram

Answer 2

Step-by-step explanation:

Given:

$e^{2x} (2cosydx - sinydy)=0$

To find:

The corresponding particular solution of $e^{2x} (2cosydx - sinydy)=0$.

Solution:      

Given that,

$e^{2x} (2cosydx - sinydy)=0$

It can be simplified as follows,

⇒ $e^{2x} (2cos(y)dx - sin(y)dy)=0$

⇒ $e^{2x} 2cos(y)dx - e^{2x} sin(y)dy=0$  

Its general form is,  

→ $M(x,y)dx+M(x,y)dy=0$    

∴ $2e^{2x}cos(y)dx - (e^{2x} sin(y)dy)=0$  

Where,

$M=2e^{2x}cos(y)dx$,$N=e^{2x} sin(y)dy$  

Check: $\frac{dM}{dy}= \frac{dN}{dx}$

⇒ $\frac{dM}{dy}=-2e^{2x} sin(y),\frac{dN}{dx} =-2e^{2x}sin(y)$ ⇒ Exact  

and,

$f(y,x)=$ ∫ $M dx+g(y)$  ⇒ Where,  $f(x,y)=C, g(y)=Integer constant$

$f(x,y)=$ ∫ $2e^{2x} cos(y)dx+g(y)=e^{2x} cos(y)+g(y)$

 $\frac{df}{dy}=N$ ⇒ To get $g(y)$  

∴  $\frac{df}{dy} =-e^{2x} siny+g'(y), N=-e^{2x} siny$              

∴ $g'(y)=0$  ⇒ $g(y)=C$

$f(x,y)=e^{2x} cos(y)=C$  

∴ The solution is $y=cos^{-1} (\frac{C}{e^{2x} } )$

Answer:

The corresponding particular solution of $e^{2x} (2cosydx - sinydy)=0$ is  $y=cos^{-1} (\frac{C}{e^{2x} } )$.