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

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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))#