Solve $y'+ty=t^3y^3$ ?

Question

1826 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-13T17:06:05

See below.

Making the substitution

$y = t/z rArr 1/2t d/(dt)(z^2)-(t^2+1)z^2+t^6=0$ now considering

$xi = z^2$ we have the linear differential equation

$1/2 t xi'-(t^2+1) xi+t^6 = 0$

now solving for $xi$

$xi = t^2(C_0 e^(t^2)+t^2+1)$ and then

$z = abst sqrt(C_0 e^(t^2)+t^2+1)$ and finally

$y = t/(abs t sqrt(C_0 e^(t^2)+t^2+1))$

Solve y'+ty=t^3y^3y'+ty=t^3y^3?