Originally Posted by

**entrepreneurforum.co.uk** Differential equation problem:

Question;

show any function $\displaystyle x=x(t)$ which satisfies $\displaystyle xe^{tx} = C $ is a solution of the differential equation $\displaystyle (1+tx)x' = x^2$

This is what I have so far;

$\displaystyle C' = x^2e^{tx}$ W.R.T. (t)

$\displaystyle x' = t$

so I have;

$\displaystyle C' = (xt * xt)e^{tx}$

$\displaystyle -x^2 = c'e^{tx}$

$\displaystyle -x^2=x^2 * e^{tx}$

therefore;

$\displaystyle -x^2 = [[xt*xt]e^{tx}]e^{tx}$

then I end up with

$\displaystyle -x^2 = xt(t)e^{2tx}$

$\displaystyle -x^2 = xt(x')$

What happens to the $\displaystyle e^{2tx}$ ???