# Solve for a variable, linear algebra

• Jun 5th 2008, 12:31 PM
Kim Nu
Solve for a variable, linear algebra
Heres the system:

(D + 2)x + (D^2 + 2D)y = 5e^(-t)
(D + 1)x - (D + 2)y = 0

I want to solve for y with no fractional answer on the "x and D" side. The only solution I can seem to come up with is y = x times D something divided by some second degree polynomial of D. This doesn't do me any good. Any suggestions?
• Jun 5th 2008, 01:26 PM
TheEmptySet
Quote:

Originally Posted by Kim Nu
Heres the system:

(D + 2)x + (D^2 + 2D)y = 5e^(-t)
(D + 1)x - (D + 2)y = 0

I want to solve for y with no fractional answer on the "x and D" side. The only solution I can seem to come up with is y = x times D something divided by some second degree polynomial of D. This doesn't do me any good. Any suggestions?

try $\displaystyle -(D+1)\cdot E_1+(D+1)\cdot E_2$

That will eliminate the x with no fractions.
• Jun 5th 2008, 03:27 PM
Kim Nu
Hey thanks for thought EmptySet, but I don't want to eliminate the x, but rather I want to write it in the form of y=Dx something. For example:

1. (2D - 3)x - 2Dy = t
(2D + 3)x + (2D + 8)y = 2

2. Add these two to get:
(4D)x + 8y = t + 2

3. Simplify to get:
y = (1/8)t + (1/4) - (1/2)(D)x

Can I do this with?:
(D + 2)x + (D^2 + 2D)y = 5e^(-t)
(D + 1)x - (D + 2)y = 0
• Jun 6th 2008, 07:41 AM
TheEmptySet
Quote:

Originally Posted by Kim Nu
Hey thanks for thought EmptySet, but I don't want to eliminate the x, but rather I want to write it in the form of y=Dx something. For example:

1. (2D - 3)x - 2Dy = t
(2D + 3)x + (2D + 8)y = 2

2. Add these two to get:
(4D)x + 8y = t + 2

3. Simplify to get:
y = (1/8)t + (1/4) - (1/2)(D)x

Can I do this with?:
(D + 2)x + (D^2 + 2D)y = 5e^(-t)
(D + 1)x - (D + 2)y = 0

Hello Kim,

It cannot be done on every equation. It worked in the above because you were albe to reduce out ALL the derivative of y. What you have found is a solution to the equation for y. In the proposed you have a mixture of y and its first and second derivative .

Conclusion: What you want is only possible if the ratio's of the coeffients of the 1,2,3...nth derivatives in both equations is constant and the function is not.

for example consider

$\displaystyle E_1: (D^4+4)x+(D^3+2D+1)y=t^5$
$\displaystyle E_2: (D^2-1)x+(5D^3+10D-27)y=\frac{t}{6}e^{-\frac{t}{5}}\sin(3t)$

Since the ratio's of the coeffeint's of 3rd (2nd not present) 1st derviative is constant at 5 and the ratio of 1 and 27 is not. You would be able to do what you wanted. I hope this clears it up. Good luck.