Question

Solve y'=(y^2+2xy)/x^2 explicitly.

Answer 1

y'=y^2/x^2+(2/x)y
y'-(2/x)y=y^2/x^2

Answer 2

WAFFLES
lol im lying

Answer 3

@SithsAndGiggles

Answer 4

Do I multiply x^2 by the whole equation?

Answer 5

No, I think a substitution should work for this one.
\[\begin{align*}
y'&=\frac{y^2+2xy}{x^2}\\
y'&=\frac{y^2}{x^2}+\frac{2y}{x}
\end{align*}\]
Let \(u=\dfrac{y}{x}\), or \(y=ux\), then \(y'=u'x+u\):
\[\begin{align*}
u'x+u&=u^2+2u\\
u'x&=u^2+u\\
\frac{du}{u^2+u}&=\frac{dx}{x}
\end{align*}\]
Separable... Partial fractions when integrating, or a trig sub.

Answer 6

I have never used this kind of substitution. Thanks.

Answer 7

Here's a whole link on it: http://tutorial.math.lamar.edu/Classes/DE/Substitutions.aspx

You're welcome!