Question

Find all solutions of y'=1+x-(1+2x)y+xy^2.

Answer 1

@SithsAndGiggles

Answer 2

Looks like a Bernoulli eq to me.

Answer 3

So how do I work it out?

Answer 4

\[\begin{align*}y'&=1+x-(1+2x)y+xy^2\\\\
y^{-2}y'&=y^{-2}+xy^{-2}-(1+2x)y^{-1}+x\\\\
-v'&=v^2+xv^2-(1+2x)v+x&\text{setting }v=y^{-1}\\&&\text{then }v'=-y^{-2}y'\\
-v'&=v^2-v+x(v^2-2v+1)\\\\
-v'&=v(v-1)+x(v-1)^2\\\\
-\frac{1}{(v-1)^2}v'&=\frac{v}{v-1}+x
\end{align*}\]
Try another substitution, like \(t=\dfrac{1}{v-1}\), then \(t'=-\dfrac{1}{(v-1)^2}v'\).
\[\begin{align*}
t'&=\left(\frac{1}{t}+1\right)t+x\\\\
t'&=1+t+x\\\\
t'-t&=1+x
\end{align*}\]
which is linear.

Answer 5

Thank you!