Differential Equations
Find the equation of the solution to dy/dx = x^3 y through the point (x,y) = (1,2)

Question

Differential Equations
Find the equation of the solution to dy/dx = x^3 y through the point (x,y) = (1,2)

anonymous · Answer

What I did: it's separable - 1/y dy = x^3 dx. 
Integrate -> ln|y| = 1/4 x^4 +c
take e^both -> y = e^(1/4 x^4 +c) = ce^(1/4x^4)

anonymous · Answer

is (x,y) = (1,2) the same as: y(1) = 2 = ce^(1/4(1)^2) ?

empty · Answer

Yep!

anonymous · Answer

So c = 2e^(-1/4)

empty · Answer

Yes, exactly.

anonymous · Answer

then y = 2e^(-1/4) * e^(1/4 x^4)

anonymous · Answer

I'm getting a response that this is wrong

empty · Answer

Provided of course you're talking about the second c and not the first c you wrote, since you're reusing the letter to represent two different things, you probably should have renamed it K or something else when you took the logarithm

empty · Answer

Oh is it expecting you to simplify it more perhaps? You can combine them all into one exponent with exponent rules.

anonymous · Answer

How so?

empty · Answer

$$y=2e^{(x^4-1)/4}$$

It also might be that you're not using parenthesis I'm not sure, but we can check that this does indeed satisfy the differential equation by differentiating it and plugging it in and we can also plug in x=1 and see that we get $e^0=1$ which means we'll have y=2 like we expect.

empty · Answer

What I mean by the parenthesis bit is if you're typing in:

$$y=2e^{-1/4}e^{1/4x^4}$$

The computer grading mabob might be interpretting this as:


$$y=2e^{-1/4}e^{1/(4x^4)}$$

So you might not have to simplify it at all, but this is just myguess

anonymous · Answer

it wanted it simplified. Thanks for your help!

empty · Answer

Cool yeah ask anything anytime I'll do my best to help out, it's rare to see people who actually have tried to solve their problem and are using their mind ya know? haha xD

anonymous · Answer

Haha, thanks.