- #1
jmg498
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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!
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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