What is the solution for y^'=((x-y))/((x+2y)),y(0)=1 ?

The solution for y^'=((x-y))/((x+2y)),y(0)=1 is y=(-x+sqrt(3x^2+4))/2

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y^'=((x-y))/((x+2y)),y(0)=1

y^{^{'}} = \frac{( x - y )}{( x + 2 y )}, y (0) = 1

y = \frac{- x + 3 x ^{2} + 4}{2}

Solution steps

y^{'} (x) = \frac{( x - y )}{( x + 2 y )}

Solve exact ODE:

y = \frac{- x + 3 x ^{2} + 4}{2}

y = \frac{- x + 3 x ^{2} + 4}{2}

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What is the solution for y^'=((x-y))/((x+2y)),y(0)=1 ?
The solution for y^'=((x-y))/((x+2y)),y(0)=1 is y=(-x+sqrt(3x^2+4))/2