Can You Help Me Solve This Linear Initial-Value Question?

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In summary, the conversation is about a question from a book that does not have any answers provided. The question involves finding the integral of xe^(2/x) and the conversation discusses different approaches and solutions, ultimately concluding that there is no elementary way to solve the problem. There is a suggestion to use the exponential integral if familiar with it, but it is suspected that there may be a typo in the original question.
  • #1
string_656
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Im doing questions out of a book, and I am stuck on one question, and it doesn't give any answers

dy/dx*(x) = 2y/x + X^2...when y(1) = 3 ... = ... dy/dx - 2y/x^2 = x
i have trouble intergrating...

I(x) = e ^ (intergral of) -2/x^2) = e^(2/x)

e^(2/x)*y = (intergral of) xe^(2/x)

im just having trouble with this ^^

can you help?
thanks
 
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  • #2
string_656 said:
Im doing questions out of a book, and I am stuck on one question, and it doesn't give any answers

dy/dx*(x) = 2y/x + X^2...when y(1) = 3 ... = ... dy/dx - 2y/x^2 = x
i have trouble intergrating...

I(x) = e ^ (intergral of) -2/x^2) = e^(2/x)

e^(2/x)*y = (intergral of) xe^(2/x)

im just having trouble with this ^^

can you help?
thanks

Your work looks OK. That integral can't be expressed in terms of the usual elementary functions. If it's a textbook exercise I would think there might be a typo somewhere.
 
  • #3
On the contrary, that has a simple polynomial solution.

Your difficulty is in your very first step: your equation is dy/dx= 2y/x+ x2 and then you rewrote it as dy/dx- 2y/x2= x!

It should be dy/dx- 2y/x= x2. Now the integrating factor is
[tex]e^{\int \frac{-2}{x}dx}= x^{-2}[/tex].
 
  • #4
string_656 said:
Im doing questions out of a book, and I am stuck on one question, and it doesn't give any answers

dy/dx*(x) = 2y/x + X^2...when y(1) = 3 ... = ... dy/dx - 2y/x^2 = x
i have trouble intergrating...

I(x) = e ^ (intergral of) -2/x^2) = e^(2/x)

e^(2/x)*y = (intergral of) xe^(2/x)

im just having trouble with this ^^

can you help?
thanks

HallsofIvy said:
On the contrary, that has a simple polynomial solution.

Your difficulty is in your very first step: your equation is dy/dx= 2y/x+ x2 and then you rewrote it as dy/dx- 2y/x2= x!

It should be dy/dx- 2y/x= x2. Now the integrating factor is
[tex]e^{\int \frac{-2}{x}dx}= x^{-2}[/tex].

But aren't you missing the x in the first term of the original DE: dy/dx*(x)?
I would have written it as x*dy/dx but anyway you must divide through by that before computing the integrating factor.
 
  • #5
im only having trouble intergrating xe^(2/x).. if i can do that i can solve the problem..
 
  • #6
string_656 said:
im only having trouble intergrating xe^(2/x).. if i can do that i can solve the problem..

In my previous reply I said:

"Your work looks OK. That integral can't be expressed in terms of the usual elementary functions. If it's a textbook exercise I would think there might be a typo somewhere."

There *is* no elementary anti-derivative for that function. None of your usual techniques can ever work.
 
  • #7
oh ok.. thanks. but how could i do it... is there a way of doing it?
 
  • #8
There is no elementary way. If you are familiar with the exponential integral:

http://en.wikipedia.org/wiki/Exponential_integral

there is a way to express the answer in terms of that. Like I said before, if this is a typical homework problem I suspect there is a misprint. As another poster has pointed out, if the x factor in the leading term was missing, it would be straightforward. I can give you the expression Maple gives for an answer if you want it.
 

What is a linear initial-value question?

A linear initial-value question is a type of mathematical problem that involves finding a solution to a linear differential equation, where the value of the unknown function and its derivatives are given at a specific initial point or condition.

What are the steps to solving a linear initial-value question?

The steps to solving a linear initial-value question are:

  1. Identify the given differential equation and initial condition.
  2. Check if the equation is linear and if the initial condition is consistent with the equation.
  3. Apply the appropriate method to solve the equation, such as separation of variables or integrating factor.
  4. Use the initial condition to find the value of the constant of integration.
  5. Write the final solution in the form of y = f(x).

What are the common applications of linear initial-value questions?

Linear initial-value questions are commonly used in various fields of science and engineering, such as physics, chemistry, economics, and biology, to model and analyze real-world phenomena. They are especially useful in understanding the behavior of systems that can be described by linear differential equations.

What are the differences between a linear initial-value question and a non-linear initial-value question?

The main difference between a linear initial-value question and a non-linear initial-value question is that linear equations have a constant rate of change, while non-linear equations have a varying rate of change. This means that the solution to a linear initial-value question will always be a straight line, while the solution to a non-linear initial-value question can be a curve or a more complex shape.

Can a linear initial-value question have multiple solutions?

No, a linear initial-value question can only have one unique solution that satisfies both the differential equation and the initial condition. This is because linear equations have a constant rate of change, so there can only be one possible outcome for a given set of initial conditions.

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