Question

〖(e〗^2y-y cos⁡〖 xy) dx+(2xe^2y 〗- x cos⁡〖 xy+2y) dy=0〗.

Answer 1

here is your answer and please follow me

Answer 2

Concept:

To solve this question , We need to recall the concept of Exact differential equation.

A differential equation of the form P(x,y)dx+Q(x,y)dy = 0             ........(1)

is said to be exact if and only if    $\frac{dP}{dy} = \frac{dQ}{dx}$ .

The required solution of exact differential equation is given by:

$\int_{regarding \;y\;as\;constant} P dx + \int_{terms \;independent\;of\;x} Qdy$ = C                .........(2)

where C is Constant.

Given:

The differential equation :

$[e^{2y} - ycosxy] dx +[2xe^{2y} -xcosxy+2y]dy = 0$ .

To find:

The solution of the given differential equation.

Solution:

On comparing the given differential equation with equation (1).

We get P = $[e^{2y} - ycosxy]$

and      Q = $[2xe^{2y} -xcosxy+2y]$

Now,  $\frac{dP}{dy}= 2e^{2y}-cosxy+xysinxy\\\frac{dQ}{dx}= 2e^{2y}-cosxy+xysinxy$

So,      $\frac{dP}{dy} = \frac{dQ}{dx}$

The given differential equation is exact.

The solution of differential equation is given by

$\int[e^{2y} - ycosxy] dx +\int[2y]dy = C$                               (using (2))

         $[xe^{2y} - \frac{ysinxy}{y}] +[\frac{2y^{2}}{2}] = C$

            $[xe^{2y} - sinxy] +[y^{2}]= C$

Hence , the solution is :  $[xe^{2y} - sinxy] +[y^{2}]= C$ .