# How to solvethis problem? (3y^2+2xy)dx - (2xy+x^2)dy=0classify the equation: linear, nonlinear, separable,exact, homogeneous, or one that requires an integration factor.

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### 1 Answer

You need to write the new form of the separable initial value differential equation such that:

`dy/dx = (3y^2+2xy)/(2xy+x^2)`

`` `dy/dx = (3y^2)/(x(2y + x)) + (2xy)/(x(2y + x))`

`dy/dx = (3y^2)/(x(2y + x)) + (2y)/(2y + x)`

You should come up with the substitution`y/(2y + x) = t ` `=gt y = 2yt + xt =gt x = (y - 2yt)/t =gt x = y(1-2t)/t =gt tx = y(1-2t) =gt t/(1-2t) = y/x`

You need to write the right side in terms of t only such that:

`dy/dx = (3t^2)/(1-2t) + 2t`

You need to bring the terms to a common denominator such that:

`dy/dx = (3t^2 + 2t - 4t^2)/(1-2t)=gtdy/dx = (2t - t^2)/(1-2t)`

You need to integrate both sides such that:

`y = int ((2t - t^2)/(1-2t))dt`

You may write the fraction such that:

`(2t - t^2)/(1-2t) = t/2 - 3/4 + 3/(4(1-2t))`

Hence `int ((2t - t^2)/(1-2t))dt = int (tdt)/2 - (3/4)int dt + (3/4)int (dt)/(1-2t)`

`` `int ((2t - t^2)/(1-2t))dt = (t^2)/4 - 3t/4 - (3/4)ln|1-2t| + c`

Pluggin `t =y/(2y + x)` in the expression yields:

`y = (y^2)/(4(2y+x)) - 3/(4(2y+x)) - (3/4)ln|1-(2y)/(2y+x)| + c`

**Hence, the solution to differential equation is `y = (y^2-3)/(4(2y+x)) - (3/4)ln|1-(2y)/(2y+x)| + c` .**