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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- August 21st 2011, 08:01 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 - August 21st 2011, 08:07 AMProve ItRe: Coupled Differential Equations
- August 21st 2011, 11:20 AMHallsofIvyRe: Coupled Differential Equations
- August 21st 2011, 12:01 PMDownesRe: Coupled Differential Equations
Thanks for the advice, i got y=t, is this correct?

- August 22nd 2011, 03:19 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. - August 22nd 2011, 03:29 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 , then and the system becomes

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.