Question

Solve the initial-value problem y'=2xy(1+y^2), y(0)=1.

Answer 1

dy/(y(1+y^2))=2x dx

Answer 2

Partial fractions?

Answer 3

Yes, but notice that 1+y^2=0,

Answer 4

Have you dealt with partial fractions that involve nonlinear terms like this?

Answer 5

No.

Answer 6

It seems like you're trying to use the method that involves plugging in a value of \(y\) that'll help with solving for the coefficients. That won't work for nonlinear factors.

Try this way:
\[\frac{1}{y(1+y^2)}=\frac{A}{y}+\frac{By+C}{1+y^2}\]

Answer 7

*irreducible nonlinear factors

Answer 8

Actually I think this one works out really well if you make the substitution y=tan(theta).

Answer 9

Yeah a trig sub is another good way to approach the left side integration.

Answer 10

You mean u=1+y^2?

Answer 11

Nope, I mean \[\LARGE y= \tan \theta\]\[\LARGE dy = \sec^2 \theta d \theta\]\[\LARGE \int\limits \frac{dy}{y(1+y^2)}=\int\limits \frac{\sec^2 \theta d \theta}{\tan \theta(1+\tan^2\theta)}\]

See I just took its derivative, and then plugged in y and dy into the integral everywhere. Notice that the denominator simplifies and you get a really easy integral.

Answer 12

Thanks, everyone.

Answer 13

\[\frac{ 1 }{ 2} \int\limits \frac{ 2 y dy }{ y^2\left( 1+y^2 \right) }=x^2+c\]
\[put~y^2=t,2y~dy=dt\]
\[L.H.S.=\frac{ 1 }{ 2 }\int\limits \frac{ dt }{ t \left( 1+t \right) }\]
make partial fractions and solve.