dy/dx=x^2+y^(1/2), y(0)=1

**Wolvenmoon** · March 29th 2011, 09:04 PM

http://www.wolframalpha.com/input/?i=dy%2Fdx%3Dx^2*%28y^%281%2F2%29%29%2Cy%280%29%3D 1&asynchronous=false&equal=Submit

Alright, so I have wolfram alpha out and am trying to figure this problem out. It's one I missed on my review.

The problem: dy/dx=x^2*y^(1/2)

My answer y=x^6/36+c^2, C=1

My review sheet's answer: 2*y*(1/2) = 1/3 * x^3 + 2.

I'm in the center of it quite confused and would like to see the problem worked step by step to the review sheet answer, and beyond if any of you are feeling really nice. Thanks!

**pickslides** · March 29th 2011, 09:15 PM

The two answers look very similar. Are they the same thing?

This equation looks like it separates, show us your working.

**TheEmptySet** · March 29th 2011, 09:18 PM

Originally Posted by Wolvenmoon

http://www.wolframalpha.com/input/?i=dy%2Fdx%3Dx^2*%28y^%281%2F2%29%29%2Cy%280%29%3D 1&asynchronous=false&equal=Submit

Alright, so I have wolfram alpha out and am trying to figure this problem out. It's one I missed on my review.

The problem: dy/dx=x^2*y^(1/2)

My answer y=x^6/36+c^2, C=1

My review sheet's answer: 2*y*(1/2) = 1/3 * x^3 + 2.

I'm in the center of it quite confused and would like to see the problem worked step by step to the review sheet answer, and beyond if any of you are feeling really nice. Thanks!

First note that the problem in the title is different than the one in your post.

The one in the post is

$\displaystyle \frac{dy}{dx}=x^2\sqrt{y}$

This equation is separable

$\displaystyle \frac{dy}{\sqrt{y}}=x^2dx \iff 2\sqrt{y}=\frac{1}{3}x^3+c$

Now using the initial condition

$\displaystyle 2\sqrt{1}=\frac{1}{3}(0)^3+c \implies c=2$

So the solution is

$\displaystyle 2\sqrt{y}=\frac{1}{3}x^3+2$

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