I'm trying to figure out a way to solve the following second-order nonlinear ODE:


with the initial conditions

$$y(1)=0, y'(1)=1$$

I've considered multiplying both sides by $y'$, but that doesn't seem particularly illuminating (for me at least) due to the $y'$ and the $t$ already in the problem.

I've also thought about maybe some kind of integrating factor, or maybe making a substitution like $u=y^2$ to make this linear, but that only seems to complicate things. Currently, I'm thinking that I might need to convert this to a first-order vector equation, but even doing so has me a little stumped when dealing with the resulting first-order scalar equations.

Any thoughts? I also welcome any general recommendations for resources on solving second-order differential equations. I'm good with second-order linear homogeneous equations with constant coefficients, but once it gets more complicated than that, I panic. Thank you!

  • $\begingroup$ I am not sure if this has analytical solution. Most nonlinear ode's are solved by numerical methods! $\endgroup$ – chandu1729 Feb 16 '16 at 1:30
  • $\begingroup$ You can look for a series solution. From the initial values we get $y''(1)=2$. Then $y=(t-1)+(t-1)^2+\dots$ You can obtain a recurrence relation for the coefficients of the higher order terms. $\endgroup$ – Julián Aguirre Feb 16 '16 at 16:32

You can transform the homogeneous form of the second order ODE into a (nonlinear) first order ODE by introducing $w(y) = y'$ (see here), which yields \begin{equation} w \frac{\text{d} w}{\text{d} y} = w + y^2. \end{equation} This is a form of the Abel equation of the second kind. The solvable versions of this equation are listed here, Table 1: however, your version ($s=0$, $m=2$) does not appear on that list.

Based on the above, I don't think your second order ODE has an analytical solution.


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