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

The solution for y^'=e^{(3x+2y)} is y=-(ln(-2/3 e^{3x}-c_{1}))/2

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

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

y = - \frac{ln ( - \frac{2}{3} e ^{3 x} - c _{1} )}{2}

Solution steps

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

Solve separable ODE:

y = - \frac{ln ( - \frac{2}{3} e ^{3 x} - c _{1} )}{2}

y = - \frac{ln ( - \frac{2}{3} e ^{3 x} - c _{1} )}{2}

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What is the solution for y^'=e^{(3x+2y)} ?
The solution for y^'=e^{(3x+2y)} is y=-(ln(-2/3 e^{3x}-c_{1}))/2