Solving Forced Equations: y'' + 2y' = 3t+2

[KevinL]

I need to compute the general solution to:

y'' + 2y' = 3t+2

The only method we have been taught is to first find the general solution of the unforced equation and then find the particular solution.

For general solution, the characteristic polynomial is L^2 + 2L = 0 (where L= lambda)

L^2 = -2L
L = -2

So general solution of unforced equation is y(t) = k1e^-2t +K2te^-2t

I am fairly certain about the above. The particular solution is a bit trickier for me.

guess: yp(t) = At + b

Plug in: 2A = 3t + 2
A = (3/2)t + 1

Is that correct? It seems like an odd answer, and B sort of just disappeared.

 

[LCKurtz]

Don't forget that the general solution to the NH equation consists of the general solution to the homogeneous equation plus a particular solution to the NH equation.

 

[HallsofIvy]

I need to compute the general solution to:

y'' + 2y' = 3t+2

The only method we have been taught is to first find the general solution of the unforced equation and then find the particular solution.

For general solution, the characteristic polynomial is L^2 + 2L = 0 (where L= lambda)

L^2 = -2L
L = -2

Well, no. You get L= -2 if you divide by L which you cannot do if L= 0. The two roots are L= -2 and L= 0.

So general solution of unforced equation is y(t) = k1e^-2t +K2te^-2t

No, the general solution to the unforced equation is $y(t)= k_1e^{-2t}+ K_2e^{0t}$$= k_1e^{-2t}+ K_2$.

I am fairly certain about the above. The particular solution is a bit trickier for me.

guess: yp(t) = At + b

Plug in: 2A = 3t + 2
A = (3/2)t + 1

Is that correct? It seems like an odd answer, and B sort of just disappeared.

No that is not correct. The whole point of writing "y= At+ B" is that A and B are to be constants. A cannot be "(3/2)t+ 1" so your particular solution is NOT "(3/2)t^2+ 2t".

If you set y= At+ B, then y'= A and y"= B. That means your equation becomes y"+ 2y'= 0+ B= 3t+ 2. That's impossible because we have no "t" on the left. The problem is that the "3t+ 2" correspond to multiples of $e^{0t}= 1$ which is already a solution to the associated homogeneous equation. Just like with a multiple root, you should try multiplying by "t". If $y(t)= At^2+ Bt$, then $y'= 2At+ B$ and $y"= 2A$. Now the equation becomes $y"+ 2y= 2A+ 4At+ 2B= 4At+ (2A+ 2B)= 2t+ 3$. For that to be true for all t, we must have 4A= 2 and 2A+ 2B= 3. From the first equation, A= 1/2. Then the second equation becomes 1+ 2B= 3 so B= 1.

Now, write down the specific solution and the general solution to the entire equation.