# Math Help - Linear system--I have this 1/2 solved...

1. ## Linear system--I have this 1/2 solved...

E1: (2D + 1)x1 + (D^2 - 4)x2 = -7e^(-t)
E2: Dx1 - (D + 2)x2 = -3e^(-t)

Here's how I solved for x1:

1. Multiply E2 by (D - 2) to get:

(Dx1)(D-2) - (D^2 - 4)x2 = -3e^(-t) * (D-2)

2. Simplify the above and then add to E1 to get:

(D^2 + 1)x1 = 2e^(-t)

3. x1c = c1cos (t) + c2sin (t)

4. Then using Ae^(-t), x1p = e^(-t)

5. x1 = c1cos (t) + c2sin (t) + e^(-t)

How do I solve for x2? I'm lost again on this one! Thanks,

Kim

2. Originally Posted by Kim Nu
E1: (2D + 1)x1 + (D^2 - 4)x2 = -7e^(-t)
E2: Dx1 - (D + 2)x2 = -3e^(-t)

Here's how I solved for x1:

1. Multiply E2 by (D - 2) to get:

(Dx1)(D-2) - (D^2 - 4)x2 = -3e^(-t) * (D-2)

2. Simplify the above and then add to E1 to get:

(D^2 + 1)x1 = 2e^(-t)

3. x1c = c1cos (t) + c2sin (t)

4. Then using Ae^(-t), x1p = e^(-t)

5. x1 = c1cos (t) + c2sin (t) + e^(-t)

How do I solve for x2? I'm lost again on this one! Thanks,

Kim
Go back to your original equations and try to eliminate $x_1$

Hint try $D \cdot E_1+[-(2D+1)]E_2$ and see what happens.

3. Hey thanks Empty Set,

Now my posts are definitely getting annoying but:

I took your suggestion, which is the correct course of action; I end up with

x2 = c3cos(t) + c4sin(t) + c3e^(-2t) + 2e^(-t)

Then I take the before mentioned solved x1 value and x2 value and plug them into E2: Dx1 - (D + 2)x2 = -3e^(-t)

I work this all out and end up with:
sin (t)[-c1 + c3 -2c4] + cos (t)[c2 - c4 - 2c3) = -8e^(-t)

The answer in the back of the book is:

x1 = 5c1cost(t) + 5c2sin(t) + e^(-t)
x2 = (c1 + 2c2)cost + (-2c1 + c2)sin(t) + c3e^(-2t) + 2e^(-t)

So my answers are close to being correct, I just don't understand how those coefficient were arrived at. Do you have any suggestions? Thanks,

Kim

4. Originally Posted by Kim Nu
Hey thanks Empty Set,

Now my posts are definitely getting annoying but:

I took your suggestion, which is the correct course of action; I end up with

x2 = c3cos(t) + c4sin(t) + c3e^(-2t) + 2e^(-t)

Then I take the before mentioned solved x1 value and x2 value and plug them into E2: Dx1 - (D + 2)x2 = -3e^(-t)

I work this all out and end up with:
sin (t)[-c1 + c3 -2c4] + cos (t)[c2 - c4 - 2c3) = -8e^(-t)

The answer in the back of the book is:

x1 = 5c1cost(t) + 5c2sin(t) + e^(-t)
x2 = (c1 + 2c2)cost + (-2c1 + c2)sin(t) + c3e^(-2t) + 2e^(-t)

So my answers are close to being correct, I just don't understand how those coefficient were arrived at. Do you have any suggestions? Thanks,

Kim
I think $x_2$ should be this

$x_2 = c_3\cos(t) + c_4\sin(t) + c_5e^{-2t} + 2e^{-t}$

And your $x_1$ from your 1st post is
$x_1 = c_1\cos (t) + c_2\sin (t) + e^{-t}$

So if we put these into $E_2$ we get...

$-c_1\sin(t)+c_2\cos(t)-e^{-t}-(-c_3\sin(t)+c_4\cos(t)-2c_5e^{-2t}-2e^{-t})-2(c_3\cos(t) +$
$c_4\sin(t) + c_5e^{-2t} + 2e^{-t})=-3e^{-t}$

$(-c_1+c_3-2c_4)\sin(t)+(c_2-c_4-2c_3)\cos(t)=0$

Now we have

$-c_1+c_3-2c_4=0 \\\ c_2-c_4-2c_3=0$

We need to solve these for $c_3,c_4$

To get $c_3$ so we want to eliminate $c_4$ so we multiply the 2nd by -2 and add it to the first to get

$-2(c_2-c_4-2c_3)-c_1+c_3-2c_4=0 \iff -c_1-2c_2+5c_3=0$

$5c_3=c_1+2c_2 \iff c_3=\frac{c_1+c_2}{5}$

Now we need to find $c_4$ so we want to dliminate $c_3$ so now we multiply the first by 2 and add it to the 2nd

$2(-c_1+c_3-2c_4)+c_2-c_4-2c_3=0 \iff -2c_1+c_2-5c_4=0$

$c_4=\frac{-2c_1+c_2}{5}$

Now we finally can get the final solution

$x_1 = c_1\cos (t) + c_2\sin (t) + e^{-t}$

$x_2 = \frac{c_1+c_2}{5}\cos(t) + \frac{-2c_1+c_2}{5}\sin(t) + c_5e^{-2t} + 2e^{-t}$

The answers are equivalent if you replace $c_1=5k_1 \\\ c_2=5k_2$ in both equations you will get the same answer.

Good luck.