Question

Separable ODE:
xy'=y^2+y

Answer 1

It says in the problem to set y/x=u

I tried that and got:
y'=y^2/x+y/x
=uy+u

since y/x=u, y=ux and y'=u'x+u

equating y', I get:
uy+u=u'x+u
uy=u'x

I'm stuck here, can't integrate since there are 3 variables.

Answer 2

\[x \frac{ dy }{dx }=y ^{2}+y\]
in the statement variable separable
\[\frac{ dy }{y \left( y+1 \right) }=\frac{ dx }{x }\]
make partial fractions of L.H.S and integrate

Answer 3

That would be one way of solving it, but the book wants me to use substitution, setting u=y/x

Answer 4

\[\frac{ dy }{ dx }=xy ^{2}+xy\]
\[\frac{ dy }{ dx }-xy=xy ^{2}\]
\[divide by y ^{2}\]
\[y ^{-2}\frac{ dy }{ dx }+y ^{-1}x=x\]
\[substitute y ^{-1}=u,-y ^{-2}\frac{ dy }{dx}=\frac{ du }{dx }\]
it serves your purpose?

Answer 5

\[u=\frac{ y }{x }ory=ux,\frac{ dy }{dx }=u+x \frac{ du }{dx }\]
\[x \left( u+x \frac{ du }{dx} \right)=u ^{2}x ^{2}+ux\]
\[u+x \frac{ du }{dx }=u ^{2}x+u\]
i think now you can solve, i have told you many ways.