Solving 2nd order DE with initial condition

[DaxInvader]

Hello Guys, We haven't yet covered on how to solve 2nd order equation in class however we have this assignment given to us. Any tips would be appreciated for these 2 little problems.

1. Homework Statement
We have this initial Equation: d²y/dt²−7dy/dt+ky=0, and we need to find the values of k in which the solution y=e^3t applies and the general solution.

The Attempt at a Solution

[/B]
In this case, I simply started to find k but substituting y into the equations.

y(t) = e^3t, y'(t) = 3*e^3t, y''(t) = 9*e^3t

We get: 9e^3t - 7*(3*e^3t) + k*e^3t = 0

=> 9*e^3t-21*e^3t + k*e^3t = 0
=>e^3t * (k-12) = 0.

I find the value of K in which y = e^3t is a solution to be 12.

Where I am lost is to find the general solution? Do I already have the necessary information?

Homework Statement

[/B]
Find all values of k for which the function y=sin(kt) satisfies the differential equation y′′+11y=0. Hint: There are more than 2 values of k

The Attempt at a Solution

We know that y is a solution of the DE.

y'(t) = sin(kt), y'(t) = kcos(kt), y''(t)= -k²sin(kt)

By substitution:

-k²sin(kt) + 11sin(kt) = 0
=> sin(kt)(11-k²)=0

I find that √11 for k is a solution, 0 is a solution, and any multiple of π is a solution as well. I tried entering the following:
√11, 0, nπ

And It is not correct. Any tips to see if I did something wrong?Thanks for your help!

 

[kuruman]

Where I am lost is to find the general solution? Do I already have the necessary information?

You are given a hint that the solution involves an exponential. What if you tried a solution that looks like y = e^βt and were asked to find what β is? The solutions (there are two of them because this is a second order diff. eq.) should be a function of k.
For the other problem, why have you considered only positive values for k?

 

[DaxInvader]

For the other problem, why have you considered only positive values for k?

doesn't n*pi/t cover that? we need sin(kt) = 0 so we need kt = 0 or any multiple of pi right? so let's assume kt= n*pi for n part of ℤ. therefore k = n*pi/t for any n in ℤ

EDIT: Oooohhh... do you mean + or - sqrt(11)?

 

[DaxInvader]

For the First problem, Let's see.. I got y=e^βt, y'=βe^βt, y''=β²e^βt

By substitution we get e^βt(β²-7β+k) = 0

The only way this is true is that (β²-7β+k) = 0 and if we Isolate k we get: k = β(7-β) But how does that bring me closer to getting a general solution y = something ?

Thank for your time. Appreciated.

 

[kuruman]

(β²-7β+k) = 0 and if we Isolate k we get: k = β(7-β)

You are not looking for k, it is assumed given and part of the general solution. The other part are the values of β that satisfy the diff. eq. For what values of β is the (quadratic) equation β²-7β+k = 0 satisfied?

 

[kuruman]

DIT: Oooohhh... do you mean + or - sqrt(11)?

Yes.

 

[DaxInvader]

You are not looking for k, it is assumed given and part of the general solution. The other part are the values of β that satisfy the diff. eq. For what values of β is the (quadratic) equation β²-7β+k = 0 satisfied?

Mmh I get β = (7 ± √(49-4k))/2 I am not seeing how you could simplify it more. Are you suggesting that I replace the β in the solution y = e^βt?

 

[kuruman]

Yes, that's what I am suggesting. Note that there two solutions, one with a plus sign in front of the radical and one with a minus sign. Also note that when k = 12, the radical is 1 in which case β₁ = 3 (you knew that already) and β₂ = 4 (new solution).

 

[DaxInvader]

Thanks, I understand now!

 

[kuruman]

Good. If you haven't formally solved differential equations in class, this problem is a good introduction that will help you make sense of what follows.