# [Homework Help] Solving Seperable Differential Equations

• IareBaboon
In summary, the plane is flying at a velocity of 82.0 and 24.0 above the horizontal, and a suitcase drops out of the luggage compartment when the plane is 108 directly above the dog.
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.

## Homework Statement

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 $$y''+16y=0$$

## Homework Equations

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

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

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

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 :)

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.

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 :)

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...)

all your integration steps are missing the "+C"

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

Ah, right, forgot about the C. But I don't 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:
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.

edit:///

i dunno.

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

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.

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

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!

## What is a seperable differential equation?

A seperable differential equation is a type of ordinary differential equation in which the variables can be separated on either side of the equation. This allows for the equation to be solved by integrating both sides separately.

## How do I know if an equation is seperable?

An equation is seperable if it can be written in the form dy/dx = f(x)g(y), where f(x) and g(y) are functions of x and y respectively. This indicates that the variables can be separated on either side of the equation.

## What is the general process for solving a seperable differential equation?

The general process for solving a seperable differential equation is as follows:

1. Identify the variables that can be separated on either side of the equation.
2. Multiply both sides of the equation by dx to move all terms containing y to one side and all terms containing x to the other side.
3. Integrate both sides separately.
4. Add a constant of integration on one side.
5. Solve for y.

## What if the equation is not in the form dy/dx = f(x)g(y)?

If the equation is not in the form dy/dx = f(x)g(y), you may need to use algebraic techniques to rearrange the equation into this form. This may involve factoring, substitution, or other methods.

## Are there any common mistakes to avoid when solving seperable differential equations?

One common mistake to avoid is forgetting to add a constant of integration when integrating both sides of the equation. Another mistake to watch out for is incorrectly separating the variables on either side of the equation. It is important to carefully check each step of the process to ensure accuracy.

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