# Differential Equation, finding values

• KCEWMG
In summary: After correcting that, divide both sides of your equation by x^2+k.After doing that, you get that y=-(x^2+k)+14/(2-x).Now what?In summary, the values of K for which y=x^2+k is a solution to the differential equation 2y-xy'=14 are 0, 14, and 28.

#### KCEWMG

Question:
Find the values of ω for which y=cosωt satisfies d^2y/dt^2 +81y = 0
where ω1<ω2

I guess I'm just having a lot of problem even figuring out how to do these at all. Since it's a second degree problem, I found the derivative of y=cosωt, which is -sin(ωt)*ω, and I found the derivative of that to be cos(ωt)*ω. I then set ωcos(ωt)=81cos(ωt).

Welcome to PF, KCEWMG!

You're doing fine, except that you dropped an omega factor from your 1st to your 2nd derivative.
After correcting that, divide both sides of your equation by cos wt.

Alright!

Then I get the -tan(ωt)-tω=81
This might be basic algebra, but where do I go from here? I don't think that I can take an ω out because it's the tangent of ω, right?

Errr... you have mangled your 2nd derivative now, I'm afraid.
You almost had it right before.

What did you do to get the derivative of -sin(ωt)*ω?

Well, I did it again and got a new answer...
After using the product rule, I got ((cos(ωt)ω)ω-sin(ωt)=81cos(ωt)
Which gets me ω^2=81+sin(ωt)
Right?

Let's focus on the derivative first.
Since you're differentiating with respect to t, the derivative of ω is not 1 but 0.

OH! So the final answer is positive and negative 9!
I understand. Thank you!

True!

Alright, the next problem is to find the values of K for which y=x^2+k is a solution to the differential equation 2y-xy'=14.
After that, I figured that 2(x^2+k)-x(2x+k)=14.
I then distributed and solved that k = 14/(2-x).
I put K into the equation y=x^2+(14/2-x).
After doing all of this, I got that y=0.
Now what?

Let's focus on the derivative of y first.
You seem to have made the same mistake as in the previous problem...