What is the solution for 4x^2y^'=y^'+9xe^{-y} ?

The solution for 4x^2y^'=y^'+9xe^{-y} is y=ln(9/8 ln(4x^2-1)+c_{1})

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4x^2y^'=y^'+9xe^{-y}

4 x^{2} y^{^{'}} = y^{^{'}} + 9 x e^{- y}

y = ln (\frac{9}{8} ln (4 x^{2} - 1) + c_{1})

Solution steps

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

Solve separable ODE:

y = ln (\frac{9}{8} ln (4 x^{2} - 1) + c_{1})

y = ln (\frac{9}{8} ln (4 x^{2} - 1) + c_{1})

\int 1 x d x

y = 19 sin (x) cos (x)

tan (ln (a x + b))

y = 6 - 3 x^{2}, y = 3 x^{2}, x = - 2, x = 2

f (x) = - x^{3} + 8, - 8, 1

What is the solution for 4x^2y^'=y^'+9xe^{-y} ?
The solution for 4x^2y^'=y^'+9xe^{-y} is y=ln(9/8 ln(4x^2-1)+c_{1})