# Trying To Learn Differential Equations

1. Sep 18, 2006

### Dao Tuat

I was taking a class, but got deployed before finishing and now im trying to do it on my own so that I can take it online and finish it fairly fast once I get back home. I just received my book in the mail from back home not too long ago and I've been doing a little whenever I have a chance. There are a few questions, however, that have completely stumped me and I picked out 3 that I think if someone could work out for me, then I would be able understand how to approach them and do all right on my own. So, if anyone could work these out for me I would really appreciate it:

dy/dx = (y + sqrt [(x^2) - (y^2)])/x with Y(1) = 0,

Find the general solution of (t^3)(dx/dt) + 3(t^2)x = t

and, Determine whether siny + xy - (x^2) = 2 is an implicit solution to y'' = (6xy' + [(y')^3] siny - 2[(y')^2]) / [(3x^2) - y]

Thanks for the help,
Dao Tuat

2. Sep 19, 2006

### J77

You have to show some working to convince us that you're not just looking for homework answers...

3. Sep 19, 2006

### Dao Tuat

Show work?

Perhaps you misunderstood, I am asking how to do these problems. I would like someone who has some free time and wouldn't mind helping me out, to work them out for me - or at least get me started. It's been a few years since I've had a class like this so I'm a bit rusty. As I said before, I'm hoping to have a good understanding of this topic as well as a few others by the time I'm able to go back to school and finish my degree. If you or anyone else don't want to help me, or just plain out won't then just tell me so and I'll stop wasting my time here and look for some people who don't mind helping someone learn. However, I'm sure this is not the case here and you simply misunderstood my intentions, which is my fault to not stating my purpose as clearly as I should have.

Dao Tuat

4. Sep 19, 2006

### HallsofIvy

Or perhaps you misunderstood. You said you wanted someone to do the problems for you. One of the rules of the forum is that we don't do people's work for them- even as examples. It is far better that you try yourself first. Then, if you have problems seeing your work will help us see what hints might work.

I will try to get you started.

In the first, problem, if you go ahead and divide that 'x' in the denominator into the numerator you will see that the right hand side can be written as a function of y/x. Try the substitution v= y/x (and look in the index of your textbook for "homogeneous first order equations". Be careful, there are two different meanings for "homogeneous" in d.e. You want the one associated with first order equations.)

For the second think "exact equation".

The third is the simplest. You are not asked to solve the equation, just decide if the given function is a solution. Find the derivative of the function, plug into the equation, and see if it satisfies the equation.

5. Sep 19, 2006

### Dao Tuat

Thanks for your help. I would also like to say sorry if I seemed a bit rude. The only reason I didn't put out any work was because its so hard for my to type it all out. Anyways, heres what I have:

For the first problem, I got it to dy/dx = y/x + (sqrt((x^3)-(y^2)))/x and then set v = y/x. I then have (dv/dx)x + v = dy/dx. I still don't understand what to do on this problem.

As for the second problem, I tried it two different ways:
1. I got it in the form of (t^3)dx + [(3t^2)x]dt - t(dt) = 0, but I'm not sure if this is what you implying when you said exact. If it is, then I have no idea what to do next.

2. I did the problem again doing it this way:
(dx/dt) + [(3t^2)/(t^3)]x = t/t^3
which, of course changes to:
dx/dt + (3/t)x = 1/(t^2)
I then found μ(x) to be 3lnx and set the equation up as:
3lnx(dx/dt) + 3lnx[(3t^2)/(t^3)]x = 3lnx[1/(t^2)]
From here, again, I'm lost.

The last problem I made a lot of progress on, but then got stuck towards the end. Heres what I did:

y' = [(3x^2)-y]/(cosy+x) y'' = (6x-y')/(1-siny)

And then pluged it all in:

[(3x^2)-y(1-siny(dy/dx))]/[(cosy+x)(cosy+x)]

Which can be changed to:

([(3x^2)-y]/(cosy+x)) + [(1-siny(dy/dx))/(cosy+x)]

Then:

(dy/dx)[(1-siny(dy/dx))/(cosy+x)]

Which led me to:

[y'(6x-y')(y')^2]/[y'([3x^2]-y)]

And then:

(6xy'-(y')^2)y'/[(3x^2)-y]

I feel like a few of these steps I messed up somewhere, but It also seems as if I'm getting close.

Once again I would appreciate any help.

Thanks