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Popular Calculus >

-y^'-y=e^x

Solution

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

Solution

y = - \frac{e ^{x}}{2} + \frac{c _{1}}{e ^{x}}

Solution steps

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

Solve linear ODE:

y = - \frac{e ^{x}}{2} + \frac{c _{1}}{e ^{x}}

y = - \frac{e ^{x}}{2} + \frac{c _{1}}{e ^{x}}

Graph

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Frequently Asked Questions (FAQ)

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