Question

Solve:
$(x^{2}y-2x{y}^2)dx=(x^3-3x^{2}y)dy.$

Answer 1

$(x^{2}y-2x{y}^2)dx=(x^3-3x^{2}y)dy$

$(x^{2}y-2x{y}^2)dx-(x^3-3x^{2}y)dy=0$

Here,

$M=x^{2}y-2x{y}^2\Rightarrow \frac{\partial M}{\partial y}=x^2-4xy$

$N=-(x^3-3x^{2}y)\Rightarrow \frac{\partial N}{\partial x}=6xy-3x^2$

$\because \frac{\partial M}{\partial y}\ne \frac{\partial N}{\partial x}$

Therefore,

Integrating factor $=\frac{1}{Mx+Ny}=\frac{1}{x(x^{2}y-2x{y}^2)+y(3x^{2}y-x^3)}=\frac{1}{x^2{y}^2}$

Multiplying the given equation by the integrating factor, we get

$\left(\frac{1}{x^2{y}^2}\right)(x^{2}y-2x{y}^2)dx-\left(\frac{1}{x^2{y}^2}\right)(x^3-3x^{2}y)dy=0$

$\Rightarrow (\frac{1}{y}-\frac{2}{x})dx-(\frac{x}{y^2}-\frac{3}{y})dy=0$

Now again,

$M=(\frac{1}{y}-\frac{2}{x})\Rightarrow \frac{\partial M}{\partial y}=-\frac{1}{y^2}$

$N=-(\frac{x}{y^2}-\frac{3}{y})\Rightarrow \frac{\partial N}{\partial x}=-\frac{1}{y^2}$

Now the above equation is an exact differential equation.

Therefore,

Solution of the equation is-

$\int Mdx+\int \left(\text{terms in N not containing x}\right)dy=C$

$\Rightarrow \int (\frac{1}{y}-\frac{2}{x})dx+\int \left(\frac{3}{y}\right)dy=C$

$\Rightarrow \frac{x}{y}-2\log x+3\log y=C$

$\Rightarrow \frac{x}{y}-\log x^2+\log y^3=C$

$\Rightarrow \frac{x}{y}+\log \left(\frac{y^3}{x^2}\right)=C$