Help with 2nd order differential equation with 2 constants

[mj.c]

Consider the differential equation: 2y'' - 13y' - 7y = 0

a. Show that, for any constants A and B, the following is a solution to the differential equation: y = Ae^(-9x) + Be^(x/3)

b. Find the values A and B that make the above general solution into a solution for the following initial value problem:
2y'' - 13y' - 7y = 0; y(0) = 3, y'(0) = -5

my thoughts / attempts so far:
My process started with finding the first and second derivatives of y and inputting those into the original equation. I then got 272Ae^(-9x) - (100Be^(x/3))/9
I am unsure if this is correct, and even less sure where to go from there to begin solving part b. thank you for your help!

 

[LCKurtz]

There appears to be something wrong with that problem. If that were a solution you would get identically zero. What is given is not a solution. Probably a mixup somewhere in posing or typing the problem.

 

[mj.c]

There appears to be something wrong with that problem. If that were a solution you would get identically zero. What is given is not a solution. Probably a mixup somewhere in posing or typing the problem.

Oh - You're saying the equation for y given in part a is not a solution? I believe I typed it in right but in case I didn't, I've attached a screenshot of the original problem.
And I've double checked the derivatives I got for it, and the expression I got by substituting them in the original equation. I got the same things both times, and am fairly certain I've got the calculations right. (I've double checked with calculators as well)

I'm not sure why there's something wrong there if the given expression itself seems to be where the problem is?
either way, could you briefly describe how I would even start to go about solving part b (assuming for a moment that the expression in part a was correct), because I'm very lost there, and don't know where I would even begin. If the first expression for y were proven to have been true, then where do I go from there? I'm unsure how the given values for y(0) and y'(0) come into play.

 

[LCKurtz]

The problem with this homework is that the proposed solution is not a solution. So nothing is going to work right. There are many sites on the internet that address constant coefficient differential equations. One such link is:

Constant Coefficients

The general second‐order homogeneous linear differential equation has the form

www.cliffsnotes.com

If you follow the method given in that link you will find that the characteristic equation for [imath]2y''-13y'-7y =0[/imath] is [imath]2r^2-13r-7=0[/imath] having roots of [imath]r=-\frac 1 2,~r=7[/imath]. So the solution would be in the form [math]y=Ae^{-\frac 1 2 x}+Be^{-7x}[/math]If you use that solution you might try the problem again.

 

[HallsofIvy]

Consider the differential equation: 2y'' - 13y' - 7y = 0

a. Show that, for any constants A and B, the following is a solution to the differential equation: y = Ae^(-9x) + Be^(x/3)

b. Find the values A and B that make the above general solution into a solution for the following initial value problem:
2y'' - 13y' - 7y = 0; y(0) = 3, y'(0) = -5

my thoughts / attempts so far:
My process started with finding the first and second derivatives of y and inputting those into the original equation. I then got 272Ae^(-9x) - (100Be^(x/3))/9
I am unsure if this is correct, and even less sure where to go from there to begin solving part b. thank you for your help!

Well that's not 0 so that is NOT a solution to the equation.