What is the solution for y^'=e^{3y+3t},y(0)=-12 ?

The solution for y^'=e^{3y+3t},y(0)=-12 is y=-(ln(-e^{3t}+e^{36}+1))/3

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y^'=e^{3y+3t},y(0)=-12

y^{^{'}} = e^{3 y + 3 t}, y (0) = - 12

y = - \frac{ln ( - e ^{3 t} + e ^{36} + 1 )}{3}

Solution steps

y^{'} (t) = e^{3 y + 3 t}

Solve separable ODE:

y = - \frac{ln ( - e ^{3 t} + e ^{36} + 1 )}{3}

y = - \frac{ln ( - e ^{3 t} + e ^{36} + 1 )}{3}

y^{^{'}} + \frac{1}{1 + x} y = \frac{2}{1 + x}

11 x - 8 y x^{2} + 1 \frac{d y}{d x} = 0, y (0) = 3

y^{^{'}} = 1 - x + y - x y, y (0) = 0

5 y^{^{'}} = y cos (x), y (0) = 2

\frac{d v}{d t} = 9.81 - 3 v

What is the solution for y^'=e^{3y+3t},y(0)=-12 ?
The solution for y^'=e^{3y+3t},y(0)=-12 is y=-(ln(-e^{3t}+e^{36}+1))/3