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

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

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

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

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

Solution steps

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

Solve separable ODE:

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

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

\frac{d y}{d x} = 4 e^{x - y}, y (0) = ln (7)

\frac{d y}{d x} = - \frac{6}{4} x^{5} y^{6}

a x^{^{'}} + b x = 0

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

(x^{2} + x - 6) y + (x + 3) y^{^{'}} + (x - 2) y = 0

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