Question

Solve the equation
x^2y^2(2ydx+xdy)-(5ydx+7xdy)=0

Answer 1

The above equation can be written as

$(2y^3x^2dx+x^3y^2dy)-(5ydx+7xdy)=0$

Reforming the equation as

$(y^3x^2dx+x^3y^2dy+y^3x^2dx)-(5ydx+7xdy)=0$

Multiplying and dividing by 3

$\frac{1}{3}(3y^3x^2dx+3x^3y^2dy)+y^3x^2dx-(5ydx+7xdy)=0$

Let total differential be $x^3y^3$

the $d(x^3y^3)=3x^2y^3dx+3y^2x^3dy$

$\frac{1}{3}d(x^3y^3)+x^2y^3dx-(5ydx+7xdy)=0$

$\frac{1}{3}d(x^3y^3)+x^2y^3dx-(5ydx+5xdy+2xdy)=0$

$\frac{1}{3}d(x^3y^3)+x^2y^3dx-5(ydx+xdy)-2xdy=0$

Lets select the total differential be xy then,

$d(xy)=xdy+ydx$

$\frac{1}{3}d(x^3y^3)+x^2y^3dx-5d(xy)-2xdy=0$

Consider the equation

$x^2y^3dx-2xdy=0$

Above equation can be written as,

$\int xdx=\int\frac{2}{y^3}dy$

Integrate both th sides

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

Where c i a constant

Hence the overall solution of given equation is,

$\frac{x^3y^3}{3}-5xy+\frac{x^2}{2}+\frac{1}{y^2}=c$