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Popular >

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

Solution

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

Solution

y = \frac{ln ( x ^{2} - 8 x + c _{1} )}{2}

Solution steps

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

Solve separable ODE:

y = \frac{ln ( x ^{2} - 8 x + c _{1} )}{2}

y = \frac{ln ( x ^{2} - 8 x + c _{1} )}{2}

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x \to - 1 lim (\frac{1}{- x - 1})

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\int (2 x^{2} - 4 x)^{9} (x - 1) d x

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\int 8 x^{3} + 30 x - 1 d x

Frequently Asked Questions (FAQ)

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