- #1

IareBaboon

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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) Find the particular solution of the differential equation for:

[tex]11x-6y\sqrt{x^2+1}\frac{dy}{dx}=0[/tex]

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

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

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:

[tex]\int-6ydy=\int\frac{-11x}{\sqrt{x^2+1}}dx[/tex]

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 [tex]y=sin(kt)[/tex] 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 :)

## Homework Statement

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

[tex]11x-6y\sqrt{x^2+1}\frac{dy}{dx}=0[/tex]

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

## Homework Equations

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

## 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:

[tex]\int-6ydy=\int\frac{-11x}{\sqrt{x^2+1}}dx[/tex]

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 [tex]y=sin(kt)[/tex] 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 :)

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