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

The solution for y^'=(2-e^x)/(3+2y),y(0)=0 is y=(-3+sqrt(8x-4e^x+13))/2

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y^'=(2-e^x)/(3+2y),y(0)=0

y^{^{'}} = \frac{2 - e ^{x}}{3 + 2 y}, y (0) = 0

y = \frac{- 3 + 8 x - 4 e ^{x} + 13}{2}

Solution steps

y^{'} (x) = \frac{2 - e ^{x}}{3 + 2 y}

Solve separable ODE:

y = \frac{- 3 + 8 x - 4 e ^{x} + 13}{2}

y = \frac{- 3 + 8 x - 4 e ^{x} + 13}{2}

y = \frac{1}{x + k e ^{x}}

8

x \to + 3 lim (2 x - 5)

tan^{2} (x)

\frac{d}{d x} (x^{2} - ln (2))

What is the solution for y^'=(2-e^x)/(3+2y),y(0)=0 ?
The solution for y^'=(2-e^x)/(3+2y),y(0)=0 is y=(-3+sqrt(8x-4e^x+13))/2