Answer to Question #233790 in Differential Equations for Phyroehan

Question

Find the general/particular solution of the following Differential Equations.

(Exact D.E)

[x Cos ( x + y ) + Sin ( x + y )] dy + x Cos (x+y) dy=0

Answer 1

$(x \cos ( x + y ) + \sin ( x + y )) dx+ x\cos (x+y) dy=0$

$\dfrac{\partial P}{\partial y}=-x\sin(x+y)+\cos(x+y)$

$\dfrac{\partial Q}{\partial x}=\cos(x+y)-x\sin(x+y)$

$\dfrac{\partial P}{\partial y}=-x\sin(x+y)+\cos(x+y)=\dfrac{\partial Q}{\partial x}$

The differential equation 

$(x \cos ( x + y ) + \sin ( x + y )) dx+ x\cos (x+y) dy=0$

is an exact equation.

Then we write the system of two differential equations that define the function $u(x,y)$

$\begin{cases} \dfrac{\partial u}{\partial x}=x \cos ( x + y ) + \sin ( x + y ) \\ \\ \dfrac{\partial u}{\partial y}=x\cos(x+y) \end{cases}$

Integrate the first equation over the variable $x$

$u(x, y)=\int(x \cos ( x + y ) + \sin ( x + y ))dx+\varphi(y)$

$\int x\cos(x+y)dx=x\sin(x+y)-\int\sin(x+y)dx$

$=x\sin(x+y)+\cos(x+y)+C_1$

$\int \sin(x+y)dx=-\cos(x+y)+C_2$

Then

$u(x, y)=\int(x \cos ( x + y ) + \sin ( x + y ))dx+\varphi(y)$

$=x\sin(x+y)+\cos(x+y)-\cos(x+y)+\varphi(y)$

$=x\sin(x+y)+\varphi(y)$

$u(x,y)=x\sin(x+y)+\varphi(y)$

Differentiate with respect to $y$

$\dfrac{\partial u}{\partial y}=x\cos(x+y)+\varphi'(y)=x\cos(x+y)$

$\varphi'(y)=0$

$\varphi(y)=-C$

The general solution of the exact differential equation is given by

$x\sin(x+y)=C$