- #1

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## Homework Statement

y' = (2x) / y+(x^2)y y(0) = -2

i realize this is sort of algebra prob, but i cant seem to separate x's from y's..

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- Thread starter darryw
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- #1

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y' = (2x) / y+(x^2)y y(0) = -2

i realize this is sort of algebra prob, but i cant seem to separate x's from y's..

- #2

Mark44

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## Homework Statement

y' = (2x) / y+(x^2)y y(0) = -2

i realize this is sort of algebra prob, but i cant seem to separate x's from y's..

## Homework Equations

## The Attempt at a Solution

[tex]y' = \frac{2x}{y(1 + x^2)}[/tex]

Now try separating it.

- #3

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wow. that shouldve been obvious.

thanks.

thanks.

- #4

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find solution in explicit form:

y' = (2x) / y+(x^2)y initial conditions: y(0) = -2

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

y dy = 2x + (2/x) dx

integrate both sides...

(1/2)y^2 = x^2 + 2ln|x| + c

y^2 = 2x^2 + 4ln|x| + 2c

y = root [2x^2 + 4ln|x| + c]

apply IC:

-2 = root [2(0)^2 + 4ln|0| + c]

-2 = (root 4) + c

c = -4 (i am reasong that root c is same as c because whether its root c or just c, both are still constants.. right?)

so my solution is:

y = root [2x^2 + 4ln|x| -4]

- #5

Mark44

Mentor

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Mistake in line above. It looks like you are trying to say that 2x/(1 + x^2) = 2x/1 + 2x/x^2.Please check my work thanks alot..

find solution in explicit form:

y' = (2x) / y+(x^2)y initial conditions: y(0) = -2

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

y dy = 2x + (2/x) dx

That's just like saying that 2/(4 + 8) = 2/4 + 2/8 = 1/2 + 1/4 = 3/4, which I hope you can see is not true.

If you are working with differential equations, you shouldn't be having problems with basic algebra manipulation.

integrate both sides...

(1/2)y^2 = x^2 + 2ln|x| + c

y^2 = 2x^2 + 4ln|x| + 2c

y = root [2x^2 + 4ln|x| + c]

apply IC:

-2 = root [2(0)^2 + 4ln|0| + c]

-2 = (root 4) + c

c = -4 (i am reasong that root c is same as c because whether its root c or just c, both are still constants.. right?)

so my solution is:

y = root [2x^2 + 4ln|x| -4]

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