What is the solution for y^'=9*t*e^{-y},y(0)=3 ?

The solution for y^'=9*t*e^{-y},y(0)=3 is y=ln((9t^2)/2+e^3)

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y^'=9te^{-y},y(0)=3

y^{^{'}} = 9 \cdot t \cdot e^{- y}, y (0) = 3

y = ln (\frac{9 t ^{2}}{2} + e^{3})

Solution steps

y^{'} (t) = 9 \cdot t \cdot e^{- y}

Solve separable ODE:

y = ln (\frac{9 t ^{2}}{2} + e^{3})

y = ln (\frac{9 t ^{2}}{2} + e^{3})

y^{^{'}} = csc (y) cos (\frac{x}{2}), y (3 π) = \frac{3 π}{2}

\frac{d y}{d x} tan (y) = x

\frac{d x}{d y} = y x, y (1) = 7

3 y \frac{d y}{d x} + \frac{1}{y} = 0

\frac{d y}{d x} = \frac{x y}{x ^{2} - 1}

What is the solution for y^'=9*t*e^{-y},y(0)=3 ?
The solution for y^'=9*t*e^{-y},y(0)=3 is y=ln((9t^2)/2+e^3)