# General solution for a differential equation $x = t x^\prime + (x^\prime)^2$

I am studying for an exam and I don't know how to determine a general solution for a differential equation. For example, $$x = t x^\prime + (x^\prime)^2$$ Can anyone help?

One approach is to assume a Taylor series expansion for the solution, substitute and equate coefficients. I haven't worked it out to the end and proved that these are all the solutions, but I can find 2 solutions like this. Another, simpler way, is to differentiate the ODE once, and see what you can deduce from the resulting 2nd-order ODE - the solutions drop out directly this way.

A more straightforward approach is the following:

$$(x')^2 +t(x') - x = 0$$

Thus by taking the roots of the quadratic in $\displaystyle x'$

$$x' = \frac{-t \pm \sqrt{t^2 + 4x}}{2}$$

Thus

$$2x' + t = \pm \sqrt{t^2 + 4x}$$

Setting $\displaystyle y = 4x+t^2$ gives us

$$\frac{y'}{2} = \pm \sqrt{y}$$

This should give some of the solutions and in order to get all, the substitution $y = 4x + t^2$ suggests itself, which I believe gives us the differential equation

$$(y')^2 - 4y = 0$$

• Very nice! $\quad$ – Matt E Jan 17 '11 at 20:11
• @Matt: Thank you. For some reason I missed your comment! – Aryabhata Jul 11 '11 at 15:29