Use the elimination method to find a general solution for...

Use the elimination method to find a general solution for the given linear system, where differentiation is with respect to t. $(D+1)[u]-(D+1)[v] = e^t$ $(D-1)[u]+(3D + 1)[v] = 8$ Eliminate v and solve the remaining differential equation for u. Choose the correct answer below. OA. The system is degenerate. O B. $u(t)=C_1+C_2e^t+\frac{1}{2}e^t+2t$ OC. $u(t)=C_1e^t+C_2e^{-t}+\frac{1}{2}e^t+2t$ D. $u(t) = C_1+C_2e^{-t}+\frac{1}{2}e^t+2t$ OE. $u(t)=C_1+C_2te^{-t}+\frac{1}{2}e^t+2t$ Now find v(t) so that v(t) and the solution for u(t) found in the previous step are a general solution to the system of differential equations. Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. v(t) = OB. The system is degenerate.

Transcript

00:01 Okay, we're supposed to solve this pair of equations using the elimination method, which is probably not how i would do it, but nonetheless.

00:12 So we wanna eliminate v, get a single equation for u, and then solve it, okay? so what i would do, i would take the first equation and multiply it by four, and then add things together.

00:35 And then the d term on v is gonna cancel out.

00:40 So i get four times d plus one plus d minus one on u minus four v plus v is equal to four e to the t plus four.

01:10 Okay, so this here is minus three v.

01:17 I'm going to bring that over to the right -hand side and bring the four e to the t plus four over to the left and combine some terms over here.

01:27 So we get five d plus four minus one, so it's plus three hitting on u minus four times e to the t plus one.

01:45 And that's equal to three v.

01:48 So then i divide by three, so i get v is equal to that.

02:12 Then what i wanna do, i'm gonna take its derivative and then i'm gonna plug it back into one of these equations and i'll get a single equation for u.

02:27 So dv is therefore equal to one third, and then we get five d squared plus three d on u minus four e to the t.

02:54 And then the one term is just zero when i take the derivative, okay? so then, doesn't matter which equation i take, but i think i'm gonna take the second one.

03:31 So i get d minus one on u plus four times dv, which is all that.

03:57 And then i get a plus v and that equals, so this is from the second equation.

04:30 I'm gonna multiply through by three because it'll make life a little easier.

04:36 I won't have to carry around the fractions.

04:44 So i get a three out here, then we lose this three and we lose this one.

04:54 And then on the right -hand side, we get four times three.

05:09 So now we'll collect terms on du and d squared u, and i'll bring the terms that don't involve u over to the right -hand side.

05:26 So i got 20 d squared, and then we got three d here and 12 d here, and then we got minus three.

05:46 Oh, we got another five, that's 20.

06:01 And then we got minus three u plus three.

06:07 Well, that's kind of nice...

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