# Second-Order DE with Variable Coefficients

• August 21st 2011, 07:01 AM
Downes
Second-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, 07:07 AM
Prove It
Re: Coupled Differential Equations
Quote:

Originally Posted by Downes
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

Is y a function of t?
• August 21st 2011, 10:20 AM
HallsofIvy
Re: Coupled Differential Equations
Quote:

Originally Posted by Downes
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

That's a "Cauchy-Euler" type equation. The substitution y= $t^r$ reduces it to an algebraic equation for r. Also, the change of variable x= ln(t) changes this to an equation with constant coefficients.
• August 21st 2011, 11:01 AM
Downes
Re: Coupled Differential Equations
Thanks for the advice, i got y=t, is this correct?
• August 22nd 2011, 02:19 AM
HallsofIvy
Re: 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, 02:29 AM
HallsofIvy
Re: 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 $Y= \begin{bmatrix}y \\ z\end{bmatrix}$, then $Y'= \begin{bmatrix}y' \\ z'\end{bmatrix}$ and the system becomes
$\begin{bmatrix}y' \\ z'\end{bmatrix}= \begin{bmatrix} 0 & -1 \\ 4 & 0\end{bmatrix}\begin{bmatrix}y \\ z\end{bmatrix}+ \begin{bmatrix} x \\ 0 \end{bmatrix}$

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.