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Popular Calculus >

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}

Graph

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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