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Popular >

y^'=e^{x-y},y(0)=2

Solution

y^{^{'}} = e^{x - y}, y (0) = 2

Solution

y = ln (e^{x} + e^{2} - 1)

Solution steps

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

Solve separable ODE:

y = ln (e^{x} + e^{2} - 1)

y = ln (e^{x} + e^{2} - 1)

Graph

Popular Examples

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\int \frac{x ^{4} + 1}{x ( x ^{2} + 2 ) ^{2}} d x

limit as x approaches pi of ln|cos(x)-9|

x \to π lim (ln ∣ cos (x) - 9 ∣)

Frequently Asked Questions (FAQ)

What is the solution for y^'=e^{x-y},y(0)=2 ?
The solution for y^'=e^{x-y},y(0)=2 is y=ln(e^x+e^2-1)