y′=12x^3 √(1+x^4)(1+y^2)
Man, after the last one, this should not have caused you any headaches. It's basically the same thing
dy = 12x^3 √(1+x^4)(1+y^2) dx
dy/(1+y^2) = 12x^3 √(1+x^4)
arctan(y) = 2(1+x^2)^(3/2) + c
y = tan(2(1+x^2)^(3/2) + c)
y(0)=1 ⇒ tan(2+c) = 1 ⇒ 2+c = π/4
y = tan(2(1+x^2)3/2 + π/4-2)
Better check my math
y′=12x^(3)sqrt(1+x^4)(1+y^2), y(0)=1
⇒ y=
Thanks!
2 answers
oops typo that should be √(1+x^4) everywhere. I seem to have slipped in some 1+x^2...