Question

Solve the differential equation dy/dx = 2x^3/y with initial condition y(1) = 2.

Answer 1

I believe the answer is 1/2 y^2 = 1/2x^4 + 4

Answer 2

@Loser66

Answer 3

Or it might be (1/2) y^2 = (1/2)x^4 + 3

Answer 4

i think it's the latter

Answer 5

How do you know if it is +3 or +4?

Answer 6

checked on wolfram alpha

Answer 7

i used to do calculations like this but haven't done them in months so i'm a little rusty.

Answer 8

Can you send me the link?

Answer 9

http://www.wolframalpha.com/input/?i=dy%2Fdx+%3D+2x%5E3%2Fy+with+initial+condition+y%281%29+%3D+2

Answer 10

Well sqrt(x^4+3) s not the same as x^4+3

Answer 11

Actually wait. I understand you square both sides.

Answer 12

it is if you take the square root on both sides, you you y^2 on one side. that cancels each other out doesn't it?

Answer 13

I saw that mistake soon after I made it.

Answer 14

Thank you very much!

Answer 15

it's ok, mistakes are easily made in math. I do it all the time

Answer 16

\[\frac{dy}{dx} = \frac{2x^3}{y}\]
\[\int\limits y dy = \int\limits2x^3 dx\]
\[\frac {y^2}{2} = \frac{x^4}{2} + C\]
y(1) = 2
\[\frac{2^2}{2} = \frac{1^4}{2} + C; C = \frac{3}{2}\]
\[y^2 = x^2 + 3\]