# [Homework Help] Solving Seperable Differential Equations

1. Dec 13, 2009

### IareBaboon

Hi, this is my first time posting a homework help problem, so please excuse any mistakes I happen to make :) I actually have two problems I'm stuck on. I have been trying to figure it out, and I'm sure it's a very simple problem, but I guess I'm just not that bright. Anyways, here they are.

1. The problem statement, all variables and given/known data

1) Find the particular solution of the differential equation for:

$$11x-6y\sqrt{x^2+1}\frac{dy}{dx}=0$$

2) Find all values of k for which the function $$y=sin(kt)$$ satisfies the differential equation $$y''+16y=0$$

2. Relevant equations

For the first question, you're told to use the following initial condition: $$y(0)=4$$

3. The attempt at a solution

So for the first question, what I did was bring the x's to the right side and then got stuck at the integral part, which is this:

$$\int-6ydy=\int\frac{-11x}{\sqrt{x^2+1}}dx$$

The left hand side is easy to integrate, however how would I go about integrating the right hand side? That's where I'm getting stuck at. A quick rundown would be very appreciated.

As for the second question, what I think I'm supposed to do is take the derivative of $$y=sin(kt)$$ twice, so that I have the second derivative of y, then substitute it into the original equation. Would that be right?

Thank you very much for any help. Anything would be appreciated :)

Last edited: Dec 14, 2009
2. Dec 13, 2009

### mg0stisha

I'd use a u-substitution to integrate the right side.

3. Dec 14, 2009

### n1person

as for the second one, if I am understanding you right, that is the right approach

4. Dec 14, 2009

### IareBaboon

Alright. Thanks guys :) I think I can do the first question now. But the second question is a tad bit confusing. So let's say I take the first derivative of $$y=sin(kt)$$ which becomes $$y'=kcos(kt)$$ right? And then I take the second derivative of that, which is $$y''=-k^2sin(kt)$$ right?

After finding the second derivative, I plug in $$y=sin(kt)$$ and $$y''=-k^2sin(kt)$$ into $$y''+16y=0$$ which then makes it into $$-k^2sin(kt)+16sin(kt)=0$$. What do I do at this point to find all values of k that would satisfy the differential equation $$y''+16y=0$$?

Sorry if it's confusing to read. But thanks a lot for the help :)

5. Dec 14, 2009

### snipez90

Factor out the sin(kt) term on the left hand side and it comes down to finding the zeros of both sin(kt) and 16-k^2, which is easy.

6. Dec 14, 2009

### IareBaboon

Oh wow I can't believe I missed something so obvious. Lol thanks a lot for pointing that out to me. I got the right answer now :)

7. Dec 14, 2009

### IareBaboon

Really sorry for double posting, but I'm still having troubles with the first question. This is what I've done so far after moving the x's to the right side:

$$\int-6ydy=\int\frac{-11x}{\sqrt{x^2+1}}dx$$

$$-3y^2=\int\frac{-11x}{\sqrt{x^2+1}}dx$$

U-substitution for right side
$$u=x^2+1$$
$$du=2xdx$$
$$-5.5du=-11xdx$$

$$-3y^2=-5.5\int(u^\frac{-1}{2})du$$

$$-3y^2=-5.5\frac{u^\frac{1}{2}}{\frac{1}{2}}$$

$$-3y^2=-11(u^\frac{1}{2})$$

$$y=\sqrt{\frac{-11(x^2+1)^\frac{1}{2}}{-3}$$

That's what I'm getting. I know it's wrong, because the intitial condition given was $$y(0)=4$$ and when I plug in 0 into the x's, I'm not getting 4. What am I doing wrong? (Sorry if I'm bothering you guys by now. I just really need to understand this stuff as I have an exam soon...)

8. Dec 14, 2009

### oinkbanana

all your integration steps are missing the "+C"

use y(0)=4 to solve for "C", and add it to your equation!

9. Dec 14, 2009

### IareBaboon

Ah, right, forgot about the C. But I dont think that's what the question is asking for. Here's the question (My homework is posted on the web,

EDIT: Woops, just saw the last line of your last post just now. Let me do that and I'll get back to you.

Last edited: Dec 14, 2009
10. Dec 14, 2009

### IareBaboon

Okay, so I got the C which is 2.085145784. So what I get now is $$y=\sqrt{\frac{-11(x^2+1)^\frac{1}{2}}{-3}}+2.085145784$$ yet when I submit it online, the webworks says I am wrong. I really don't get what I'm doing wrong.

11. Dec 14, 2009

### oinkbanana

edit:///

i dunno.

Last edited: Dec 14, 2009
12. Dec 14, 2009

### IareBaboon

lol its fine. thank you for the help though, much appreciated.

13. Dec 14, 2009

### phsopher

The C should be inside the square root because it comes in integration when y is still squared:

$$\int-6ydy=\int\frac{-11x}{\sqrt{x^2+1}}dx \Leftrightarrow 3\int d(y^2)= 11\int{d(\sqrt{x^2+1)} \Leftrightarrow y^2 =\frac{11}{3} \sqrt{x^2+1} + C$$

Now solve for C using the initial condition.

14. Dec 14, 2009

### IareBaboon

Ah, thanks to everyone! I got the answer now :)

15. Jan 25, 2010

### dovec

An airplane is flying with a velocity of 82.0 at an angle of 24.0 above the horizontal. When the plane is a distance 108 directly above a dog that is standing on level ground, a suitcase drops out of the luggage compartment. Help please!!!!!!!!!!!