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

The solution for y^'=e^{3x+2y},y(2)= 1/3 is y=-(ln(-2/3 e^{3x}-(-2e^{20/3}-3)/(3e^{2/3)}))/2

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y^'=e^{3x+2y},y(2)= 1/3

y^{^{'}} = e^{3 x + 2 y}, y (2) = \frac{1}{3}

y = - \frac{ln ( - \frac{2}{3} e ^{3 x} - \frac{- 2 e ^{\frac{20}{3}} - 3}{3 e ^{\frac{2}{3}}} )}{2}

Solution steps

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

Solve separable ODE:

y = - \frac{ln ( - \frac{2}{3} e ^{3 x} - \frac{- 2 e ^{\frac{20}{3}} - 3}{3 e ^{\frac{2}{3}}} )}{2}

y = - \frac{ln ( - \frac{2}{3} e ^{3 x} - \frac{- 2 e ^{\frac{20}{3}} - 3}{3 e ^{\frac{2}{3}}} )}{2}

What is the solution for y^'=e^{3x+2y},y(2)= 1/3 ?
The solution for y^'=e^{3x+2y},y(2)= 1/3 is y=-(ln(-2/3 e^{3x}-(-2e^{20/3}-3)/(3e^{2/3)}))/2