Initial condition DE

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March 12th 2009, 09:01 PM
mollymcf2009

Initial condition DE

This is probably really easy, but I am not getting the right answer. Can someone check this please?
Thanks!!

Solve.

$\frac{dy}{dx} = \frac{x}{y}$ , $y(0)=-8$

$\int ydy = \int xdx$

$y = \frac{x^2}{2} + C$

$C = -8$

So I get:

$y = \frac{x^2}{2} - 8$

Where am I screwing this up?
March 12th 2009, 09:17 PM
Chris L T521

Quote:

Originally Posted by mollymcf2009

[snip]
Solve.

$\frac{dy}{dx} = \frac{x}{y}$ , $y(0)=-8$

$\int ydy = \int xdx$

${\color{red}\frac{1}{2}}y^{{\color{red}2}} = \frac{x^2}{2} + C$

[/snip]

See my correction in red.

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