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Differential equation help.?
y'=y+e^x
I have no idea how to solve this and I'm stuck here.
dy/dx = y + e^x
dy/dx = ce^x +e^x
dy = (ce^x + e^x)dx
⌠dy = ⌠(ce^x + e^x) dx
y= ?
The answer should be y= ce^x +(e^x)x
I don't know where the last x comes from.
Thank you for your help.
1 Answer
- RapidfireLv 710 years agoFavorite Answer
I do not understand the method you are using to solve this.
Solve this differential equation by using an integrating factor:
y' = y + ℮˟
dy / dx = y + ℮˟
dy / dx - y = ℮˟
dy / dx + P(x)y = f(x)
P(x) = -1
f(x) = ℮˟
I(x) = ℮^[∫ P(x) dx]
I(x) = ℮^(∫ -1 dx)
I(x) = ℮ˉ˟
I(x) = 1 / ℮˟
I(x)y = ∫ I(x)f(x) dx
y / ℮˟ = ∫ 1 dx
y / ℮˟ = x + C
y = (x + C)℮˟