But then again, I could be wrong.

[epenguin]

Sorry.

Have you checked it?

If it is right then the problem is solved for y and z too because

y(t) + z(t) = 1 + t - x(t)

and y = z at all times.

 

[masterchiefo]

Sorry.

Have you checked it?

If it is right then the problem is solved for y and z too because

y(t) + z(t) = 1 + t - x(t)

and y = z at all times.

yes it works.
=)

 

[ehild]

Sorry.

Have you checked it?

Of course, I always do.

If it is right then the problem is solved for y and z too because

y(t) + z(t) = 1 + t - x(t)

and y = z at all times.

It has been solved by the OP, see Post #27

 

[epenguin]

It has been solved by the OP, see Post #27

Sorry I got lost on the way there.

Bit of a pity in such a long thread that the working of the essential bit was not shown. However I was trying to work it out not using integrating factors but a couple of other methods. Late because it was one of those days you may have had where it came out something like the OP's (correct) solution but not exactly right, t n finding the error the next attack came nearly but not quite right ands so on. In the end it is quite simple.

The first was the way it's usually taught (mostly for second order) for linear non-homogeneous d.e.'s with constant coefficients.

Denoting differentiation with ' to the equation

x' + 3x =. t. +. 2

following teaching, we assume a particular solution of form x = at + b. Then putting this into the above equation we get

a + 3(at + b) = 2 + t

From which we find we must have

a = 1/3 , b = 5/9

This is to be added to the general solution to

x' + 3x = 0

Which is x = Ae^-3t

Into the solution

x = Ae^-3t + t/3 + 5/9

We put x. =. x₀ = 1, and find A = 4/9 so the solution for this initial condition is x = (4/9)e^-3t + t/3 + 5/9

in accord with the OP. Slightly more suggestive is when initial condition is x₀ = 0, and then we get

x = (5/9)e^-3t + t/3I'll write up another method tomorrow.

 

[ehild]

Slightly more suggestive is when initial condition is x₀ = 0, and then we get

x = (5/9)e^-3t + t/3I'll write up another method tomorrow.

It is nice that you wrote one method to solve a linear first order differential equation in detail, and promised to give the other ones. These methods (mainly the integrating factor method) are taught during the Calculus courses everywhere. The OP was familiar with one of them. He did not write the process as he had no problem with it. I also think it is useful for the visitors of PF when they can see full solutions, but full solutions are forbidden even then, when the OP has solved the problem already. I have got warnings and infraction for less.
By the way, the equation to be solved was dx/dt + 3x = 2 + t, but you wrote

x' + x =. t. +. 2

, and your last equation does not correspond to x(0)=0.

 

[Mark44]

It is nice that you wrote one method to solve a linear first order differential equation in detail, and promised to give the other ones. These methods (mainly the integrating factor method) are taught during the Calculus courses everywhere. The OP was familiar with one of them. He did not write the process as he had no problem with it. I also think it is useful for the visitors of PF when they can see full solutions, but full solutions are forbidden even then, when the OP has solved the problem already. I have got warnings and infraction for less.

If the OP has already done the problem and gotten a solution, I don't have a problem with giving a full solution in that case, and I don't believe the other mentors who are currently active would, either.

 

[ehild]

If the OP has already done the problem and gotten a solution, I don't have a problem with giving a full solution in that case, and I don't believe the other mentors who are currently active would, either.

It is a pleasure if that is allowed. Is it really? I am often inclined to show a simpler or nicer solution after the OP gave his one, but I have got warnings and even infraction for "almost" full solutions in the past.

 

[Mark44]

It is a pleasure if that is allowed. Is it really? I am often inclined to show a simpler or nicer solution after the OP gave his one, but I have got warnings and even infraction for "almost" full solutions in the past.

Here's the PF rule that applies:

Giving Full Answers: On helping with questions: Any and all assistance given to homework assignments or textbook style exercises should be given only after the questioner has shown some effort in solving the problem. If no attempt is made then the questioner should be asked to provide one before any assistance is given. Under no circumstances should complete solutions be provided to a questioner, whether or not an attempt has been made.

I believe the intent of the last sentence is that if the OP has provided at most only an attempt, a full solution should not be given. On the other hand, if the OP has shown a solution, it's OK for a poster to provide an alternative solution.

That's my take. I will bring it up with the mentors and see if they with me (or not). If they do, I'll look into changing that rule.

 

[ehild]

Here's the PF rule that applies:

I believe the intent of the last sentence is that if the OP has provided at most only an attempt, a full solution should not be given. On the other hand, if the OP has shown a solution, it's OK for a poster to provide an alternative solution.

That's my take. I will bring it up with the mentors and see if they with me (or not). If they do, I'll look into changing that rule.

Thank you Mark. It would be nice if a thread could be finished with a comprehensive full solution. As it is now, the threads usually end when the OP got enough hint to solve the problem, and the process of solution is not shown, even by the OP.

 

[epenguin]

Yes it seems the rules allow detailed solution and discussion of a solved problem.

Especially in this case where the boot was rather on the other foot - if the OP had not posted the correct solution I would probably have posted one of my incorrect ones!

I say there is a sound educational or learning principle involved.

And because of the Polya principle - when you've got an answer it isn't finished! They will miss concluding comments we might make. This solution is just an example of a wider principle, that technique can be applied to other things, there have been recent advances around the theme being treated... This question or subject matter is related to another.

This last is related thing could do a little to combat what is a vice made too prevalent just by our focussed question-and-answer thing that pervades education which misleads many students into seeing Science as just a fearful ritual in which you have to be able to recite the responses or quote the texts. Or see the world as divided into chapters and subjects, self-sufficient and unrelated.
...

Then we mustn't let happen what I have unfortunately seen a lot of in other worlds and situations - rules rigidly applied to the detriment of the forgotten objective they were invented for in the first place!

 

[epenguin]

Firstly probably the most important thing to for anyone to realize straightaway about this problem is that without solving the d.e. It should be obvious enough that long-term, that is at high t, the solution has to be close to x = t/3 , in fact x = y = z = t/3 .

Then about the solution, I invite comments about this not-quite-the-same reasoning or 'method'.

Denoting differentiation w.r.t. t by ' sign, the equation is

x' + 3x = 2 + t

Change variable to X = x - 2/3 , equation is

X' + 3X = t

Differentiating,

X'' + 3X' = 1

Changing variable again to Y = X' - 1/3, this is the homogeneous linear d.e.

Y' + 3Y = 0

Y'/Y = 3

Solution

Y = Ae^-3t

X' = Ae^-3t + 1/3

X = Ae^-3t + t/3 + K

x =. Ae^-3t + t/3 + K'

and the arbitrary constants are worked out as before.

Comments:

• in solving d.e.'s I've often enough found it useful to differentiate them (though usually thee are other methods too).
• This gives nicely the most important part of the solution - the long term t/3 term
• The virtue I see is that I/anyone does it from a starting point of greater ignorance, no need to know that if the 'forcing function' (RHS) is a polynomial you have to know or assume then'particular' solution is another polynomial
• I think it works for any polynomial - just have to differentiate it enough times (n) to make a constant.
• (I guess the first substitution was superfluous.)