# Thread: Second-Order DE with Variable Coefficients

1. ## 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

2. ## Re: Coupled Differential Equations

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?

3. ## Re: Coupled Differential Equations

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.

4. ## Re: Coupled Differential Equations

Thanks for the advice, i got y=t, is this correct?

5. ## 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.

6. ## 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.