# Second Order ODE - Initial Value Problem

Solve the initial value problem y''+3y'+2y = 3e$$^{2t}$$+1 with initial values y(0) = 1, y'(0) = 1.

I am unsure if I am going about the solution correctly.

1.) Find the characteristic equation.
r$$^{2}$$+3r+2=0 $$\Rightarrow$$ (r + 1)(r + 2) = 0
Therefore, y = c1•e$$^{-t}$$+c2•e$$^{-2t}$$

2.) Use method of undetermined coefficients for RHS. Attempt Y = Ae$$^{2t}$$+B. This means Y' = 2Ae$$^{2t}$$ and Y'' = 4Ae$$^{2t}$$

Substituting it all in and simplifying, I get...

12Ae$$^{2t}$$+2B = 3e$$^{2t}$$+1

Comparing coefficients, I get the following...

12A = 3 => A = 1/4
2B = 1 => B = 1/2

So this would yield a particular solution of:
y = c1•e$$^{-2t}$$+c2•e$$^{-t}$$+(1/4)e$$^{2t}$$+(1/2)

4.) Next, I take the derivative of the above equation, and then substitute the initial conditions. When I do this, I get a system of equations that yields the following constants. C1 = 3/4, C2 = -1/2.

5.) Then I just replace C1 and C2 in the equation for the particular solution.

Could somebody please tell me where I am going astray? Thank you VERY much!!!

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rock.freak667
Homework Helper
Well that is basically the correct method...so why do you think you are wrong?

HallsofIvy
Homework Helper
Unfortunately, although you are referring to an "initial value problem", you never bothered to tell us what those initial values are!

Ooops...added them now. Sorry! But thanks for looking!

HallsofIvy
Recheck you calculations for $C_1$ and $C_2$. I get $C_1= -\frac{3}{4}$ which also changes $C_2$.