Therefore, y(t) = k1y1(t) + k2y2(t) is a solution to the given DE.

[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 don't know how to verify that its an equality. I am guessing it has something to do with the Linearity Principle, but I don't understand how to actually verify it.

 

[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?

 

[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.

 

[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