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

The solution for y^'=e^{2x+3y},y(0)=0 is y=-(ln(-3/2 e^{2x}+5/2))/3

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

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

y = - \frac{ln ( - \frac{3}{2} e ^{2 x} + \frac{5}{2} )}{3}

Solution steps

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

Solve separable ODE:

y = - \frac{ln ( - \frac{3}{2} e ^{2 x} + \frac{5}{2} )}{3}

y = - \frac{ln ( - \frac{3}{2} e ^{2 x} + \frac{5}{2} )}{3}

\frac{d P}{d t} = 0.1 P (1 - \frac{P}{40})

y^{^{'}} = - 4 y + 2

\frac{d y}{d x} + 12 x y = - x

x^{2} \frac{d y}{d x} + 8 x \frac{d y}{d x} + 6 y = 0

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

What is the solution for y^'=e^{2x+3y},y(0)=0 ?
The solution for y^'=e^{2x+3y},y(0)=0 is y=-(ln(-3/2 e^{2x}+5/2))/3