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

The solution for y^'=(x-2)e^{-2y},y(2)=ln(2) is y=(ln(x^2-4x+8))/2

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

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

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

Solution steps

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

Solve separable ODE:

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

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

\frac{\partial}{\partial y} (x e^{\frac{x}{y}})

\int_{25}^{50} 20 d x

f (x) = \frac{x ^{2} + 1}{x + 4}

f (x) = x^{\frac{5}{2}} (2 - 5 x^{3})

e^{(- 2 x)} (3 x)^{2}

What is the solution for y^'=(x-2)e^{-2y},y(2)=ln(2) ?
The solution for y^'=(x-2)e^{-2y},y(2)=ln(2) is y=(ln(x^2-4x+8))/2