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(x^2)dy + 2xy dx = (x^2) dx

The solution given is: (3x^2)y = x^3 + c

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- Thread starter bryanosaurus
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- #1

- 15

- 0

(x^2)dy + 2xy dx = (x^2) dx

The solution given is: (3x^2)y = x^3 + c

- #2

rock.freak667

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[tex]x^2dy + 2xy dx = x^2 dx[/tex]

[tex]\equiv x^2 \frac{dy}{dx}+2xy=x^2[/tex]

[tex]\equiv \frac{dy}{dx}+\frac{2}{x}y=1[/tex]

this is in a form that you should know how to solve and it isn't a separation of variables type.

If you don't know how to solve ODE's in this form check this link http://en.wikipedia.org/wiki/Integrating_factor" [Broken]

[tex]\equiv x^2 \frac{dy}{dx}+2xy=x^2[/tex]

[tex]\equiv \frac{dy}{dx}+\frac{2}{x}y=1[/tex]

this is in a form that you should know how to solve and it isn't a separation of variables type.

If you don't know how to solve ODE's in this form check this link http://en.wikipedia.org/wiki/Integrating_factor" [Broken]

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arildno

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[tex]u=\frac{y}{x}[/tex]

Then, we have:

[tex]\frac{du}{dx}=\frac{1}{x}\frac{dy}{dx}-\frac{u}{x}\to\frac{dy}{dx}=x\frac{du}dx}+u[/tex]

We therefore get the diff.eq:

[tex]x\frac{du}{dx}+u+2u=1\to\frac{1}{1-3u}\frac{du}{dx}=\frac{1}{x}[/tex], which is separable.

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