Trying To Learn Differential Equations

In summary: Thanks!Dao TuatIn summary, the person is trying to do a class that they were taking but were deployed before finishing and is now trying to do it on their own. They just received a book on the topic from back home and is doing some research when they have a chance. They have a few questions that they think someone could help them with.
  • #1
Dao Tuat
16
0
I was taking a class, but got deployed before finishing and now I am 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
 
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  • #2
You have to show some working to convince us that you're not just looking for homework answers...
 
  • #3
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.

Thanks for any help you can give,
Dao Tuat
 
  • #4
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
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, here's 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
 

What are differential equations?

Differential equations are mathematical equations that describe the relationship between a function and its derivatives. They are used to model many real-world situations, such as the growth of a population or the flow of electricity.

Why is it important to learn differential equations?

Differential equations are used in many scientific fields, including physics, engineering, and biology. They provide a powerful tool for understanding and predicting the behavior of complex systems.

What are some common techniques for solving differential equations?

Some common techniques for solving differential equations include separation of variables, substitution, and the use of integrating factors. Each technique is best suited for different types of differential equations.

What are the applications of differential equations?

Differential equations are used to model a wide range of phenomena, including motion, heat transfer, and population dynamics. They are also essential in fields such as control theory, where they are used to design systems that can respond to changing conditions.

How can I improve my understanding of differential equations?

To improve your understanding of differential equations, it is important to practice solving a variety of problems. You can also read textbooks or watch online tutorials to learn different techniques and applications. Collaborating with others and seeking help from a tutor or teacher can also be beneficial.

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