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

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