| Thread Closed |
differential equation and initial value |
Share Thread | Thread Tools |
| Sep23-09, 10:48 PM | #1 |
|
|
differential equation and initial value
1. The problem statement, all variables and given/known data
[img=http://img3.imageshack.us/img3/3417/questionec.th.jpg] 2. Relevant equations 3. The attempt at a solution I tried dividing both sides by (x^5+1), however the integration becomes really complex... can someone give me suggestions on how to do this? |
| Sep23-09, 10:54 PM | #2 |
|
Recognitions:
|
|
| Sep24-09, 12:04 AM | #3 |
|
|
that's what I did and then I need to integrate -10x^4/(x^5+1) dx right and then then to get the integrating factor is just e to the power of whatever the result of the integration is... however the integration is quite hard...
|
| Sep24-09, 03:06 PM | #4 |
|
Recognitions:
|
differential equation and initial value[tex]\int \frac{-10x^4}{x^5+1} dx[/tex] see how d/dx(x5+1)=5x4 ? Can you use a substitution to make this integral easier? |
| Sep24-09, 03:57 PM | #5 |
|
|
|
| Sep24-09, 05:00 PM | #6 |
|
Recognitions:
|
right so for y'+P(x)y=Q(x), when you multiply by an integrating factor 'u', the left side becomes d/dx(uy) . That's why we multiply by u in the first place. So you'll need to basically integrate uQ(x) w.r.t. x |
| Sep24-09, 06:23 PM | #7 |
|
|
and then divide that by the integrating factor right?
|
| Sep24-09, 06:25 PM | #8 |
|
Recognitions:
|
|
| Sep24-09, 07:17 PM | #9 |
|
|
what do you mean by if I need to? doesn't it always works like that?
|
| Sep24-09, 11:47 PM | #10 |
|
|
this now comes to:
[tex]\int \frac{x^2+5x-4}{x^5+1} e^{-2ln(x^5+1)} dx[/tex] I guess this can be simplified to: [tex]\int \frac{x^2+5x-4}{(x^5+1)^3} dx[/tex] is this true? how can I solve this such complex integration? |
| Sep25-09, 05:18 PM | #11 |
|
|
anyone please?
|
| Sep25-09, 09:24 PM | #12 |
|
Recognitions:
|
|
| Sep25-09, 11:34 PM | #13 |
|
|
Integrating [tex]\frac{10x^4}{(x^5+1)}[/tex] the result I got is [tex]2ln(x^5+1)[/tex] and so the integrating factor is [tex]e^{2ln(x^5+1)}[/tex] which simplifies to [tex](x^5+1)^2[/tex].
Then I do [tex] \int (x^2+5x-4)(x^5+1) [/tex] and the result of this integration I divide by [tex](x^5+1)^2[/tex] which is the integrating factor. Is this the correct step to find the solution? Please correct me if I am wrong. |
| Sep25-09, 11:43 PM | #14 |
|
Recognitions:
|
yes just integrate and divide, also don't forget the constant of integration |
| Sep26-09, 12:28 AM | #15 |
|
|
|
| Thread Closed |
| Thread Tools | |
Similar Threads for: differential equation and initial value
|
||||
| Thread | Forum | Replies | ||
| differential equation initial value prob | Calculus & Beyond Homework | 3 | ||
| Differential Equation - Initial Value Problem | Calculus & Beyond Homework | 14 | ||
| Initial Value Differential Equation Problem | Calculus & Beyond Homework | 9 | ||
| Need Help Solving Initial Value Differential Equation | Calculus & Beyond Homework | 1 | ||
| Differential in a initial value problem | Differential Equations | 1 | ||