What is the solution for y^'=e^{4y+4x},y(0)=9 ?

The solution for y^'=e^{4y+4x},y(0)=9 is y=-(ln(-e^{4x}+1/(e^{36)}+1))/4

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y^'=e^{4y+4x},y(0)=9

y^{^{'}} = e^{4 y + 4 x}, y (0) = 9

y = - \frac{ln ( - e ^{4 x} + \frac{1}{e ^{36}} + 1 )}{4}

Solution steps

y^{'} (x) = e^{4 y + 4 x}

Solve separable ODE:

y = - \frac{ln ( - e ^{4 x} + \frac{1}{e ^{36}} + 1 )}{4}

y = - \frac{ln ( - e ^{4 x} + \frac{1}{e ^{36}} + 1 )}{4}

y^{^{'}} = \frac{x ^{2} + 3 x}{2 x y}

(x y + 2 y + x + 2) d y - (x^{2} - 4) d x = 0

y^{^{'}} = \frac{3 x ^{2} + 4 x + 2}{2 y + 1}

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

y^{^{'}} = e^{t + y}, y (ln (2)) = 1

What is the solution for y^'=e^{4y+4x},y(0)=9 ?
The solution for y^'=e^{4y+4x},y(0)=9 is y=-(ln(-e^{4x}+1/(e^{36)}+1))/4