What is the solution for (e^{-3y})y^'=e^{8x},y(0)=12 ?

The solution for (e^{-3y})y^'=e^{8x},y(0)=12 is y=-(ln(1/(e^{36)}+3/8-3/8 e^{8x}))/3

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(e^{-3y})y^'=e^{8x},y(0)=12

(e^{- 3 y}) y^{^{'}} = e^{8 x}, y (0) = 12

y = - \frac{ln ( \frac{1}{e ^{36}} + \frac{3}{8} - \frac{3}{8} e ^{8 x} )}{3}

Solution steps

(e^{- 3 y}) y^{'} (x) = e^{8 x}

Solve exact ODE:

y = - \frac{ln ( \frac{1}{e ^{36}} + \frac{3}{8} - \frac{3}{8} e ^{8 x} )}{3}

y = - \frac{ln ( \frac{1}{e ^{36}} + \frac{3}{8} - \frac{3}{8} e ^{8 x} )}{3}

\frac{\partial}{\partial x} (yz e^{x z} - 2)

\int_{0}^{\infty} x e^{- \frac{x}{y}} d x

y^{^{''}} - 6 y^{^{'}} + 7 y = 0

\frac{\partial}{\partial x} (arctan (x) y)

(x - 1)^{3}, x - 1

What is the solution for (e^{-3y})y^'=e^{8x},y(0)=12 ?
The solution for (e^{-3y})y^'=e^{8x},y(0)=12 is y=-(ln(1/(e^{36)}+3/8-3/8 e^{8x}))/3