Thanks,

Could you give me some guidance on another question?

Question:

t^2y''+2ty'-2y=0

y1(t)=t

it asks to find the general solution

cheers

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- Aug 21st 2011, 07:01 AMDownesSecond-Order DE with Variable Coefficients
Thanks,

Could you give me some guidance on another question?

Question:

t^2y''+2ty'-2y=0

y1(t)=t

it asks to find the general solution

cheers - Aug 21st 2011, 07:07 AMProve ItRe: Coupled Differential Equations
- Aug 21st 2011, 10:20 AMHallsofIvyRe: Coupled Differential Equations
- Aug 21st 2011, 11:01 AMDownesRe: Coupled Differential Equations
Thanks for the advice, i got y=t, is this correct?

- Aug 22nd 2011, 02:19 AMHallsofIvyRe: Coupled Differential Equations
The condition "y1(t)= t" makes no sense. First there is no "y1" in your problem, second, it states an entire function, not an iniitial condition. Please tell us what it really was.

**Unless**the problem is NOT "to solve the equation" but "given that y1(t)= t is a solution to this equation, find a second, independent solution". If that is the problem, then, of course, "y= t" is NOT a correct answer since that is just the function you were already given. - Aug 22nd 2011, 02:29 AMHallsofIvyRe: Coupled Differential Equations
By the way, going back to your first problem, y'+ z= x, z- 4y= 0, y(0)= 1, z(0)= 1, can be rewritten as a matrix problem. If we let $\displaystyle Y= \begin{bmatrix}y \\ z\end{bmatrix} $, then $\displaystyle Y'= \begin{bmatrix}y' \\ z'\end{bmatrix}$ and the system becomes

$\displaystyle \begin{bmatrix}y' \\ z'\end{bmatrix}= \begin{bmatrix} 0 & -1 \\ 4 & 0\end{bmatrix}\begin{bmatrix}y \\ z\end{bmatrix}+ \begin{bmatrix} x \\ 0 \end{bmatrix}$

Which we can think of as just "Y'= AY+ F" where A and F are the given 2 by 2 and 1 by 2 matrices. That's relatively easy to solve by finding the eigenvalues of A.