Hello, just another equation sorry about the hassle.. Find the solution (x+y)y'=10y-x which satisfies y(3)=2 not sure how to go about doing this..
Last edited by biffer; Mar 4th 2008 at 06:08 AM.
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$\displaystyle y' = \frac{{3y - x}} {{x - y}} = \frac{{\dfrac{{3y}} {x} - 1}} {{1 - \dfrac{y} {x}}}.$ Now consider $\displaystyle y=xz.$
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