Initial value problems!

[ectab]

We are taking courses online via classroom about differential equations but the pdf file the professor posted has only one example for initial value problems and that example is not clear and belongs to the second-order differential equations. So I have been assigned to solve these (attached) "initial value problem" of first-order.



If you can help with at least one (for example the third), it would be great.

 

[tkhunny]

How do you do it WITHOUT the initial value? Is it somehow different?

 

[ectab]

How do you do it WITHOUT the initial value? Is it somehow different?

Sorry, I don't understand what you mean.

 

[tkhunny]

You entitled the question "Initial value problems!" and then gave no other indication of anything.

I was wondering if you were able to solve for a general solution, which is a good start, but somehow became troubled by the application of the Initial Value.

We can help A LOT better if we see some of your own efforts.

 

[Steven G]

This is a free math help forum and not a free homework service. On this forum we help students solve their problems. If you had read the posting guidelines you would have received help by now and would have known that you have to share your work with us--even if you know it is wrong. If we see your work then we know the method you chose to solve your problem and we can see where you made any errors.

Please read the posting guideline and post back following those guidelines.

 

[ectab]

You entitled the question "Initial value problems!" and then gave no other indication of anything.

I was wondering if you were able to solve for a general solution, which is a good start, but somehow became troubled by the application of the Initial Value.

We can help A LOT better if we see some of your own efforts.

What I was trying to say is that I don't understand what the professor means by "initial value problem" at all. As I said there was only one example of the so called initial value in the file the professor had posted, and that example looks different from the homework problems above.

So what I hoped is that someone would help explaining it to me by solving one of the above, so that I would be able to do the rest.

I am sorry but I am a new member here so it seems that wasn't the right thing to do.

 

[ectab]

This is a free math help forum and not a free homework service. On this forum we help students solve their problems. If you had read the posting guidelines you would have received help by now and would have known that you have to share your work with us--even if you know it is wrong. If we see your work then we know the method you chose to solve your problem and we can see where you made any errors.

Please read the posting guideline and post back following those guidelines.

I am sorry for that. I actually had read the posting guidelines and it said it would help sharing my own work, but it didn't say that was necessary.
My problem is that I didn't understand the question at all. It is all new to me, so I thought maybe someone will explain it by solving one of those three problems.

Any way, I have tried to solve it with a help from a classmate, but I think it might be wrong. So I will repost the thread attaching my own work as well.

 

[tkhunny]

What I was trying to say is that I don't understand what the professor means by "initial value problem" at all. As I said there was only one example of the so called initial value in the file the professor had posted, and that example looks different from the homework problems above.

So what I hoped is that someone would help explaining it to me by solving one of the above, so that I would be able to do the rest.

Given a general solution to the differential equation, y(t), one would end up with an arbitrary constant of some sort. In many text books, this is portrayed as "+C", but it can take other forms. Your "Initial Value" is one tiny piece of information to determine what that "C" is for the specific problem.

Quick Example:
General Solution for Differential Equation: y(t) = 3t^2 + C
Given y(0) = 5
We have y(0) = 3(0)^2 + C = 5 ==> C = 5
The solution to the Initial Value Problem: y(t) = 3t^2 + 5

As far as showing YOUR work, pick more, not less. Pick better help, not making us guess what is needed.

 

[ectab]

Given a general solution to the differential equation, y(t), one would end up with an arbitrary constant of some sort. In many text books, this is portrayed as "+C", but it can take other forms. Your "Initial Value" is one tiny piece of information to determine what that "C" is for the specific problem.

Quick Example:
General Solution for Differential Equation: y(t) = 3t^2 + C
Given y(0) = 5
We have y(0) = 3(0)^2 + C = 5 ==> C = 5
The solution to the Initial Value Problem: y(t) = 3t^2 + 5

As far as showing YOUR work, pick more, not less. Pick better help, not making us guess what is needed.

Thank you, tkhunny

I have reposted the problem with my solution here, and, looking to your explanation, I think my solution is right. I was confused because of using t instead of the usual x.

 

[HallsofIvy]

There are two distinctly different kinds of differential equations.

A differential equation where all additional conditions at the same value of the independent variable, for example, y(a), y'(a),etc. are given, then the "fundamental existence and uniqueness theorem" says that the existence of a solution depends only on the differential equation. Those are "initial value problems" (so called because many applications have time, t, as independent variable and initial values at t= 0.

The initial value problem, \(\displaystyle \frac{d^2y}{dx^2}+ y= 0\), y(a)= P, y(a)= Q has a unique solution no matter what a, P, and Q are.

A "boundary value problem", on the other hand, has additional data, function values, derivative values, etc. at two or more different points. And whether of not a solution exists can depend on that additional data as well as the equation.

For example, \(\displaystyle \frac{d^2 y}{dx^2}+ y= 0\), y(0)= 0, \(\displaystyle y(\pi/2)= 1\) has a unique solutio, \(\displaystyle y= cos(x)\) but \(\displaystyle \frac{d^2 y}{dx^2}+ y= 0\), y(0)= 0, \(\displaystyle y(\pi)= 1\) has no solution while \(\displaystyle \frac{d^2 y}{dx^2}+ y= 0\), y(0)= 0, \(\displaystyle y(\pi)= 0\) has infinitely many solutions.