# Second Order ODE - Initial Value Problem

1. Dec 16, 2007

### jmg498

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!!!

Last edited: Dec 16, 2007
2. Dec 16, 2007

### rock.freak667

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

3. Dec 16, 2007

### HallsofIvy

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

4. Dec 16, 2007

### jmg498

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

5. Dec 17, 2007

### HallsofIvy

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