What is the solution for y^'=(x-5)e^{-2y},y(5)=ln(5) ?

The solution for y^'=(x-5)e^{-2y},y(5)=ln(5) is y=(ln(x^2-10x+50))/2

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y^'=(x-5)e^{-2y},y(5)=ln(5)

y^{^{'}} = (x - 5) e^{- 2 y}, y (5) = ln (5)

y = \frac{ln ( x ^{2} - 10 x + 50 )}{2}

Solution steps

y^{'} (x) = (x - 5) e^{- 2 y}

Solve separable ODE:

y = \frac{ln ( x ^{2} - 10 x + 50 )}{2}

y = \frac{ln ( x ^{2} - 10 x + 50 )}{2}

y^{^{''}} - 12 y^{^{'}} + 36 y = 2 t e^{6 t}

t ln (t) \frac{d r}{d t} + r = 9 t e^{t}

\frac{d}{d x} ((\frac{x - 9}{x + 2})^{\frac{3}{2}})

f (x) = - 9 x^{\frac{1}{3}} + 5, at x = 27

x \to \infty lim (\frac{7}{x} - 9)

What is the solution for y^'=(x-5)e^{-2y},y(5)=ln(5) ?
The solution for y^'=(x-5)e^{-2y},y(5)=ln(5) is y=(ln(x^2-10x+50))/2