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

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

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(dy)/(dx)=((x-y))/((x+2y)),y(0)=1

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

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

Solution steps

\frac{d y}{d 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}

\int (\frac{1}{1 + x + y}) d x

y^{^{''}} + 2 y^{^{'}} + 18 y = 0, y (0) = 1, y^{^{'}} (0) = 0

\frac{\partial}{\partial x} (2 e^{x^{3} y})

\frac{\partial}{\partial x} (\frac{ln ( x )}{x})

\frac{d}{d x} (5 \cdot 2^{x})

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