# DE#7

#### idontknow

(1) If $$\displaystyle |f(t)|\leq \frac{c}{t^{1+a}}$$ , show that the DE $$\displaystyle u''=u-uf(u)$$ has solutions :

$$\displaystyle u_1 (t) =e^t [1+\mathcal{O}(t^{-a})]$$ and $$\displaystyle u_2 (t) =e^{-t}[1+\mathcal{O}(t^{-a})]$$ as $$\displaystyle t\rightarrow \infty$$.

(2) $$\displaystyle 2xy''+y'-2y=0$$.

(3) $$\displaystyle yy'' +y'^2 =\phi(x)$$.

(4) $$\displaystyle xy''+xy'^2 =yy'$$.

#### idontknow

My approach : about (4) I tried to set $$\displaystyle yy'=z$$ then $$\displaystyle xz'=z$$ or $$\displaystyle z=yy'=cx$$.
$$\displaystyle ydy=cxdx$$ ; $$\displaystyle \; \; dy^2 =dcx^2$$ ; $$\displaystyle \; \; y^2=c_0 +cx^2$$.

about (3) : $$\displaystyle yy''+y'^2 = (yy')'=\phi(x)$$ ; $$\displaystyle \; \; yy'=c_1 +\int \phi(x) dx$$ ; $$\displaystyle \; \; y^2 = c_2 +c_1 x +\int \int \phi(x) dx dx$$.

How to continue with (2) and (1) ?

#### skipjack

Forum Staff
(2) Have you tried substituting $x = t^2$ or $x = -t^2\hspace{2px}$?

#### SDK

You seem to have a big interest in solving obscure ODEs and this doesn't appear to be homework. Many of the DE's you have posted have been solved by some clever change of variables which seems to come from nowhere (usually provided by skipjack). This is what ODE theory looked like before ~1900s which explains why so many of these transforms are named after 19th century mathematicians. However, it was realized in the past ~120 years that there is a general theory which governs these transforms and a method for understanding exactly which clever change of variables to use, and similarly, when an equation won't admit a solution via any clever change of variables. This was the start of the theory of Lie groups and Lie algebras and I would highly recommend that you pick up an introductory book on these topics, especially if you have ever wondered where skipjack gets all of these transformations.

#### idontknow

(2) Have you tried substituting $x = t^2$ or $x = -t^2\hspace{2px}$?
So what is the y=?
for $$\displaystyle x=t^2$$ I got $$\displaystyle y''(t)+y'(t)[\frac{1}{t}-1]-4y(t)=0$$ which one I cannot solve for the moment.
By wolframalpha the solution is $$\displaystyle y=c_1 \cosh (2\sqrt{x})+ic_2 \sinh (2\sqrt{x})$$.

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#### skipjack

Forum Staff
You made a mistake while doing that substitution.

You should have got $$\displaystyle \frac{d^2y}{dt^2} - 4y = 0$$.

idontknow