Question

I need to solve and verify this y"-9y = 0

Answer 1

The characteristic polynomial is \(r^2-9=0\) and has roots \(r=\pm3\). What would the solution look like?

Answer 2

this is what you need to solve the DE?

Answer 3

Yeah, for a homogeneous equation, i.e. of the form
\[a_ny^{(n)}+a_{n-1}y^{(n-1)}+\cdots+a_1y'+a_0y=0\]
you get a characteristic polynomial
\[a_nr^n+a_{n-1}r^{n-1}+\cdots+a_1r+a_0=0\]
If the roots are \(r_1,r_2,\cdots,r_n\), the solution to the DE would be
\[y=C_1e^{r_1 t}+C_2e^{r_2 t}+\cdots+C_ne^{r_n t}\]

Answer 4

so the answer would be y=ce^3t ?

Answer 5

There were two roots to the equation...

Answer 6

so it is y'= c1e^3t + c2e^-3t ?

Answer 7

yup

Answer 8

Thank you how to I verify that?

Answer 9

*y = , not y' =

Answer 10

To check if a certain solution is in fact a solution to a DE, just plug in the solution.

You find your solution to be \(y=C_1e^{3t}+C_2e^{-3t}\), so you have
\[y''=9C_1e^{3t}+9C_1e^{-3t}\]
Subbing into the equation, you have
\[y''-9y=0~~\iff~~9C_1e^{3t}+9C_1e^{-3t}-9\left(C_1e^{3t}+C_2e^{-3t}\right)=0\]

Answer 11

or you can do separately by : y_1 = C1e^3t
y_1'
y_1"
then test whether y_1" - 9y_1 =0 or not. If it is =0, then y_1 is one of the solution
do the same with y_2 = C2 e^(-3t)
:)

Answer 12

Right, and that's due to the superposition principle. \(y=C_1e^{3t}\) and \(y=C_2e^{-3t}\) can work individually as solutions.

Answer 13

so how do you solve something like this y"-2y'+y=0