# DE#6

#### idontknow

(1) $$\displaystyle \: y''(1+yy')=y'(1+y'^2 )$$.

(2) $$\displaystyle \: y''-y'+e^{4x}y=0$$.

#### skipjack

Forum Staff
(2) $y'' - y' + e^{4x}y = 0$
Let $y = e^{x/2}u$, then the equation becomes $u'' + (e^{4x} - 1/4)u = 0$.
Let $(1/2)e^{2x} = t$, then the equation becomes $t^2d^2u/dt^2 + tdu/dt + (t^2 - 1/4^2)u = 0$,
whch is a standard equation whose general solution is a linear combination of Bessel functions.

idontknow

#### idontknow

(1) $$\displaystyle y''+yy'y''=y'+y'^{3} \:$$ or $$\displaystyle \: dy'+yy'dy'=dy+y'^{2} dy$$.

$$\displaystyle y'-y=c+ 2\int y'^{2} dy -\int y'd(yy')$$.
how to continue ?

Last edited: