Question

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

Answer 1

`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)`.