Intergrating a Differential Equation using the intergrating factor method

  • Thread starter thomas49th
  • Start date
  • #1
655
0

Homework Statement



[tex]\frac{dy}{dx} + \frac{y}{2x} = -x^{\frac{1}{2}}[/tex]

Homework Equations




[tex]yT =\int{QT}dx + C[/tex]
where T is the intergrating factor

T = [tex]e^{\int{P}dx[/tex]
and P is the co-efficient of y from the differential equations


The Attempt at a Solution



well to find T we need to do:

[tex]e^{\int{\frac{1}{2x}}}dx[/tex]
[tex]e^{\frac{1}{2}\int{\frac{1}{x}}}dx[/tex]
[tex]e^{\frac{1}{2}ln|x|}[/tex]
[tex] = x^{\frac{1}{2}}[/tex]

so using [tex]yT =\int{QT}dx + C[/tex]

you get

[tex] yx^{\frac{1}{2}}= \int{x^{\frac{1}{2}x^-{\frac{1}{2}}}dx[/tex]
[tex] yx^{\frac{1}{2}}= \int{1}dx[/tex]
[tex] yx^{\frac{1}{2}}= x + c[/tex]

the answer in the back of the book says [tex]yx^{\frac{1}{2}} = \frac{1}{2}x^{2}[/tex]
Where have I gone wrong?

Thanks :)
 

Answers and Replies

  • #2
1,752
1
[tex]\frac{dy}{dx}-\left(-\frac{1}{2x}\right)y=-x^{\frac{1}{2}}[/tex]

Which makes your integrating factor [tex]x^\frac 1 2[/tex]

[tex]yx^\frac 1 2=-\int x^\frac 1 2 x^\frac 1 2dx=-\int xdx[/tex]

Watch your inputting.
 
  • #3
655
0
[tex]
yx^{\frac{1}{2}}= \int{x^{\frac{1}{2}}x^-{\frac{1}{2}}}dx
[/tex]

woops. above is what is ment to say
[tex]x^{\frac{1}{2}}x^{-\frac{1}{2}}[/tex]
simplifies to 1 right?

Thanks :)
 
Last edited:
  • #4
1,752
1
[tex]
yx^{\frac{1}{2}}= \int{x^{\frac{1}{2}}x^-{\frac{1}{2}}}dx
[/tex]

woops. above is what is ment to say
[tex]x^{\frac{1}{2}}x^-{\frac{1}{2}}[/tex]
simplifies to 1 right?

Thanks :)
No!!! Is the negative with x^1/2 or with just the 1/2?
 
  • #5
655
0
[tex]
x^{\frac{1}{2}}x^{-\frac{1}{2}}
[/tex]

is what i mean

thanks :)
 
  • #6
1,752
1
[tex]
x^{\frac{1}{2}}x^{-\frac{1}{2}}
[/tex]

is what i mean

thanks :)
I knew what you meant, but why would the negative be with the 1/2 making it [tex]x^{-\frac 1 2}[/tex]???

Isn't the right side [tex]-\int x^\frac 1 2dx[/tex] then multiplied by your integrating factor, [tex]-\int x^\frac 1 2x^\frac 1 2dx[/tex].
 
  • #7
655
0
your a genius :)
 

Related Threads on Intergrating a Differential Equation using the intergrating factor method

  • Last Post
Replies
4
Views
1K
Replies
1
Views
1K
  • Last Post
Replies
6
Views
1K
  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
4
Views
1K
Replies
7
Views
5K
  • Last Post
Replies
7
Views
3K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
5
Views
1K
Top