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Solve y'+y=e^2x, y(0)=2

User Zaz Gmy
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Answer:


y = (1)/(3)e^(2x) + (5)/(3)e^(-x).

Explanation:

Let's find a particular solution
y_p = ae^(bx):


y' = abe^(bx), so


abe^(bx) + ae^(bx) = e^(2x)


ae^(bx)(b + 1) = e^(2x), then, b=2 and 3a = 1, so a= 1/3.

Our particular solution is
y_p = (1)/(3)e^(2x). Now, we are going to find the solution of the homogeneus equation with constants coefficients.

Let y =
e^(\lambda x)


y' = \lambda e^(\lambda x), so


\lambda e^(\lambda x)+ e^(\lambda x) = 0


e^(\lambda x)(\lambda + 1) = 0


\lambda= -1. Then
y_h = Ce^(-x) and the solution is


y = y_p + y_h =  (1)/(3)e^(2x) + Ce^(-x). Now, we use the initial condition to find C:


y(0) = (1)/(3)e^(0) + Ce^(0) = (1)/(3) + C = 2


C = 2- (1)/(3) = (5)/(3). The final result is


y = (1)/(3)e^(2x) + (5)/(3)e^(-x).

User Seunghee
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