Question

Solve the following differential equation.
(3x +2y +y2)dx+(x+4xy+5y2)dy = 0.
Help please

Answer 1

Solution

Check the exactness of the equation:

$M=3x+2y+y^2$

$N=x+4xy+5y^2$

$\frac{dM}{dy}=2+2y$

$\frac{dN}{dx}=1+4y$

Clearly,

$\frac{dM}{dy}\neq\frac{dN}{dx}$

The equation is not exact.

Take $4x+4y^2$ as the integrating factor;

$M'=(3x+2y+y^2)(4x+4y^2)$

$M'=12x^2+8xy+16xy^2+8y^3+4y^4$

Hence

$\frac{dM'}{dy}=8x+32xy+24y^2+16y^3$

$N'=(x+4xy+5y^2)(4x+4y^2)$

$N'=4x^2+16x^2y+24xy^2+16xy^3+20y^4$

$\frac{dN'}{dx}=8x+32xy+24y^2+16y^3$

in which

$\frac{dM'}{dy}=\frac{dN'}{dx}$

The equation is now exact.

The general solution is

$\int M'dx+\int($ Terms in N independent of x$)dy=C$

$\int(12x^2+8xy+16xy^2+8y^3+4y^4)dx+\int20y^4dy=C$

$=4x^3+4x^2y+8x^2y^2+8xy^3+4xy^4+4y^5=C$