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Popular >

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

Solution

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

Solution

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

Solution steps

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

Solve separable ODE:

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

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

Graph

Popular Examples

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\int_{- π}^{π} ∣ x ∣ cos (n x) d x

integral from 0 to 1 of (x^4+4)/x

\int_{0}^{1} \frac{x ^{4} + 4}{x} d x

limit as x approaches 4 of 1/((4-x)^2)

x \to 4 lim (\frac{1}{( 4 - x ) ^{2}})

Frequently Asked Questions (FAQ)

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