Answer to Question #96310 in Differential Equations for Olajide Olaitan

Question

Find the general solution frac{dy}{dx}= frac{2y^2+3xy}{x^2}

Answer 1

This is a Bernoulli differential equation.

$\frac{dy}{dx}= \frac{2y^2}{x^2}+\frac{3y}{x}$

Let's divide both sides by $\,-y^2$ $\,$

$-\frac{dy}{y^2dx}= -\frac{2}{x^2}-\frac{3}{xy}$

Using the substitution $v=y^{-1}$ , $v'=-\frac{y'}{y^2}$ will lead us to $\,$

$v'=-\frac{2}{x^2}-\frac{3v}{x}$

This is a first-order linear equation which can be solved using an integrating factor $\,$

$M=e^{\int\frac{3}{x}dx}=e^{3\ln x}=x^3$

Multiplying both sides by M we get $\,$

$x^3v'+3x^2v=-2x\\$

$(x^3v)'=-2x\\$

$x^3v=-x^2+C$

$v=\frac{C-x^2}{x^3}\\$

Hence $\,$

$y=\frac{1}{v}=\frac{x^3}{C-x^2}$