# Second Order Linear Equations

1. Nov 19, 2009

### KevinL

I understand how to solve a normal second order linear equation, but this question in the homework is a bit more theoretical and I'm a bit confused.

"Suppose y1(t) and y2(t) are solutions of y'' + py' + qy = 0

Verify that y(t) = k1y1(t) + k2y2(t) is also a solution for any choice of constants k1 and k2."

If we were given actual functions (i.e. e^-t) this would be very simple. Just plug it into the DE and see that you have an equality. But when given arbitrary functions I dont know how to verify that its an equality. Im guessing it has something to do with the Linearity Principle, but I dont understand how to actually verify it.

2. Nov 19, 2009

### pbandjay

Since it's homework, I probably shouldn't give the answer. But a hint...

Since y_1 and y_2 are solutions, then

y1'' + py1' + qy1 = 0
y2'' + py2' + qy2 = 0

How about plugging (k1y1 + k2y2) into the original equation and rearranging a bit?

3. Nov 19, 2009

### KevinL

That got me started on the right track, I think. With the two equations you have, solve each one for y''. Then plug that into the other equation and everything nicely cancels so that 0=0.

4. Nov 19, 2009

### pbandjay

Sounds like you have it. You just need to show that inserting y into the equation will equal zero as well.

(k1y1 + k2y2)'' + p(k1y1 + k2y2)' + q(k1y1 + k2y2) = 0

k1y1'' + k2y2'' + pk1y1' + pk2y2' + qk1y1 + qk2y2 = 0

k1(y1'' + py1' + qy1) + k2(y2'' + py2' + qy2) = 0

k1*0 + k2*0 = 0 ==> 0 = 0