Solve this initial value problem...
y' = x^3 (1 - y) where y(0) = 3

Question

Solve this initial value problem...
y' = x^3 (1 - y) where y(0) = 3

anonymous · Answer

$$\int\limits\limits_{}^{}dy/(1-y) = \int\limits\limits_{}^{}x^{3}dx$$

anonymous · Answer

$$\ln 1 -y = x ^{4}/4 + C$$

anonymous · Answer

$$1-y = \exp(x ^{4}/4) + expC$$

anonymous · Answer

$$1-y = C \exp(x ^{4}/4)$$

anonymous · Answer

y(0) = 3, C=-2
$$1 + 2\exp(x ^{4}/4)=y$$
But the back of the book shows -(x^4/4).
Where did I go wrong??

turingtest · Answer

this is linear and I think an integrating factor will work

anonymous · Answer

At the second line  the result of integral of dy ( 1-y) = - ln ( 1-y)

anonymous · Answer

yes... the dreaded minus sign

anonymous · Answer

You're right. Thank you

turingtest · Answer

$$y'=x^3(1-y)=x^3-yx^3$$$$y'-x^3y=x^3$$$$\mu=e^{\int x^3dx}=e^{x^4/4}$$haha I went the long way but it will get the same result :P

anonymous · Answer

Yeah, separate and integrate should work.

turingtest · Answer

yeah. much easier than an IF

anonymous · Answer

But, you're a champ for finding alternatives.

turingtest · Answer

thank ya kindly

anonymous · Answer

IF confuses me

turingtest · Answer

you'll get there

anonymous · Answer

I haven't seen the proof so i feel like the theorem comes out of thin air.

turingtest · Answer

anonymous · Answer

Thanks!

turingtest · Answer

welcome!