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Popular >

y^'+y+e^{-x}=0

Solution

y^{^{'}} + y + e^{- x} = 0

Solution

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

Solution steps

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

Solve linear ODE:

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

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

Popular Examples

y^{^{'}} = 2 x y + 1

\frac{y ^{^{'}}}{t + 3 y} = 3

1 + x y = x y^{^{'}}

Find the Antiderivative f(x)=5x-6 integral

f (x) = 5 x - 6

Using L'Hospital's Rule limit as x approaches 0 of (e^(5x)-1-5x)/(x^2)lhopital

x \to 0 lim (\frac{( e ^{5 x} - 1 - 5 x )}{( x ^{2} )})

Frequently Asked Questions (FAQ)

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