What is the solution for y^'=e^{-y}-xe^{-y}cos(x^2) ?

The solution for y^'=e^{-y}-xe^{-y}cos(x^2) is y=ln(x-1/2 sin(x^2)+c_{1})

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

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

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

Solution steps

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

Solve separable ODE:

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

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

4 y^{^{'}} - 4 e^{2 x} y = 0, y (0) = 3

4 (1 + y^{2}) d x - x y d y = 0

\frac{d y}{d t} + y t sin (t) = 0

\frac{d y}{d x} = \frac{3 x ^{2} + 4 x + 2}{2 y - 2}

\frac{d y}{d x} = - \frac{7}{4} x^{8} y^{6}

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