Second Order ODE, With Initial Conditions

[checkmatechamp]

Homework Statement

y'' + 4y = t² + 6e^t; y(0) = 0; y'(0) = 5

Homework Equations

The Attempt at a Solution

So, getting the general solution, we have r² + 4 = 0, so r = +/- 2i

So the general solution is y_c = sin(2t) + cos(2t)

I then used the method of undetermined coefficients to figure that the particular solution had a form At² + Bt + C + De^t + E = 0, and found that A = 0.25, B = 0 , C = 0, D = 1.2, E = 2

So the overall solution is y = c₁sin(2t) + c₂cos(2t) + 0.25t² + 1.2e^t + 2

The derivative is y' = 2c₁*cos(2t) - 2c₂*sin(2t) + 0.5t + 1.2e^t

So I got that 0 = 0 + c₂ + 0 + 1.2 + 2, which means that c₂ + 3.2 = 0, which means that c₂ = -3.2

5 = 2c₁ - 0 + 0 + 1.2, which means that 2c₁ = 3.8, and c₁ = 1.9

So the overall equation should be y = 1.9sin(2t) - 3.2*cos(2t) + 0.25t² + 1.2e^t + 2.

But when I enter it as an answer, it tells me it's wrong. Where did I make a mistake?

 

[SteamKing]

Homework Statement

y'' + 4y = t² + 6e^t; y(0) = 0; y'(0) = 5

Homework Equations

The Attempt at a Solution

So, getting the general solution, we have r² + 4 = 0, so r = +/- 2i

So the general solution is y_c = sin(2t) + cos(2t)

No, this is only the complementary solution, i.e. the solution to the homogeneous equation.

I then used the method of undetermined coefficients to figure that the particular solution had a form At² + Bt + C + De^t + E = 0, and found that A = 0.25, B = 0 , C = 0, D = 1.2, E = 2

Why do you have two different constant terms?

So the overall solution is y = c₁sin(2t) + c₂cos(2t) + 0.25t² + 1.2e^t + 2

The derivative is y' = 2c₁*cos(2t) - 2c₂*sin(2t) + 0.5t + 1.2e^t

So I got that 0 = 0 + c₂ + 0 + 1.2 + 2, which means that c₂ + 3.2 = 0, which means that c₂ = -3.2

5 = 2c₁ - 0 + 0 + 1.2, which means that 2c₁ = 3.8, and c₁ = 1.9

So the overall equation should be y = 1.9sin(2t) - 3.2*cos(2t) + 0.25t² + 1.2e^t + 2.

But when I enter it as an answer, it tells me it's wrong. Where did I make a mistake?

The general solution of the ODE is the sum of the particular solution and the complementary solution.

Also, you can't apply the initial conditions solely to the complementary solution. These must be applied to the general solution of the ODE to work out the values of the undetermined coefficients.

http://tutorial.math.lamar.edu/Classes/DE/UndeterminedCoefficients.aspx

 

[checkmatechamp]

Alright, so for the particular solution:
yp = At^2 + Bt + C + De^t
yp' = 2At + B + De^t
yp'' = 2A + De^t

2A + De^t + 4(At^2 + Bt + C + De^t) = t^2 + 6e^t

2A + De^t + 4At^2 + 4Bt + 4C + De^t = t^2 + 6e^t

So grouping the terms together, you get 4A = 1, so A = 0.25

De^t = 6e^t, so D = 6

2A + 4C = 0
2(1) + 4C = 0
4C = -2
C = -0.5

So yp = 0.25t^2 - 0.5 + 6e^t

y = c1sin(2t) + c2cos(2t) + 6e^t + 0.25t^2 - 0.5

y' = 2c1*cos(2t) - 2c2*sin(2t) + 0.5t + 6e^t

0 = 0 + c2 + 6 + 0 - 0.5
-5.5 = c2

5 = 2c1 + 6

2c1 = -1

c1 = -0.5

So the final solution is y = -0.5sin(2t) -5.5 cos(2t) + 6e^t + 0.25t^2 - 0.5. Does this look correct?

 

[SteamKing]

You can check it yourself. It's good practice.

Does it satisfy the original ODE?

Does it satisfy the initial values prescribed for y and y'?

 

[checkmatechamp]

After some checking, I finally entered the answer y = 1.9*sin(2t) - 1.075*cos(2t) + 0.25t^2 - 0.125 + 1.2e^t. Thank you so much for your help! :)