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

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

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

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

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

Solution steps

\frac{d y}{d 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}

\int_{0}^{1} (x + 1)^{2} d x

\int_{0}^{π} tan^{2} (x) d x

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

y = 2 x^{2}, y = 16 x - 32, y = 0

\frac{d y}{d t} = \frac{y + 1}{t + 1}

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