What is the solution for y^'=9x^2e^{-y},y(0)=1 ?

The solution for y^'=9x^2e^{-y},y(0)=1 is y=ln(3x^3+e)

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y^'=9x^2e^{-y},y(0)=1

y^{^{'}} = 9 x^{2} e^{- y}, y (0) = 1

y = ln (3 x^{3} + e)

Solution steps

y^{'} (x) = 9 x^{2} e^{- y}

Solve separable ODE:

y = ln (3 x^{3} + e)

y = ln (3 x^{3} + e)

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n = 1 \sum \infty \frac{n}{6 ^{n}}

x \to 0 lim (\frac{( 1 + x ) ^{9} - 1}{x})

\frac{\partial}{\partial x} (\frac{42}{( 5 + x ^{2} + y ^{2} )})

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What is the solution for y^'=9x^2e^{-y},y(0)=1 ?
The solution for y^'=9x^2e^{-y},y(0)=1 is y=ln(3x^3+e)