$y'' + (y')^2 = y$
and
$y*y''= (y')^2 - (y')^3$
I know that both of these equations can be solved by using $p(y) = y'$, but Im getting stuck in the middle, and can't seem to solve the thing after I get an expression which looks like something: $p^2$
any tips please?
$\endgroup$3 Answers
$\begingroup$The solutions of the first ODE can be expressed on the form of $x(y)$ as a function defined by an integral. It is doubtfull that a closed form could be derived.
The solutions of the second ODE can be expressed on a closed form thanks to a special function :
For the first equation, $y'' + (y')^2 = y$, you might try the substitution $w(y)=(y')^2$ instead.
$$\frac{d}{dy}w(y)=\frac{d}{dy}(y')^2\\ =\frac{dy'}{dy}\frac{d}{dy'}(y')^2\\ =\frac{dx}{dy}\frac{dy'}{dx}\frac{d}{dy'}(y')^2\\ =\frac{1}{y'}(y'')(2y')\\ =2y''.$$
Hence the equation becomes the 1st order linear inhomogeneous equation with constant coefficients, $w'(y)+2w(y)=2y$. The general solution to this equation is found to be:
$$w(y)=c_1e^{-2y}+y-\frac12.$$
The equation $(y')^2=c_1e^{-2y}+y-\frac12$ can then be used to find an implicit formula for $y(x)$ in the form of an integral, but for most values of $c_1$ an explicit equation for $y(x)$ is likely unobtainable.
$\endgroup$ $\begingroup$Hello everyone I am currently training and I have to solve a nonlinear differential equation and I totally lost I have a BAC + 2 and I never saw her going after large search I are always happens not:
I would like to solve this equation
A *y''(t) - B *(y'(t))^2 + C * y(t)
AND I AM FRENCH SORRY FOR THE BAD ENGLISH
$\endgroup$
