Use the substitution u=x^2+y^2 to solve 2yy’=x^2-2x



Answer to a math question Use the substitution u=x^2+y^2 to solve 2yy’=x^2-2x

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u = x^2 + y^2
\frac{du}{dx} = 2x + 2yy'
2yy' = x^2 - 2x
\frac{du}{dx} = 2x + (x^2 - 2x)
\frac{du}{dx} = x^2
\int \frac{du}{dx} dx = \int x^2 dx
u = \frac{x^3}{3} + C
x^2 + y^2 = \frac{x^3}{3} + C

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