Struggling with a Nonlinear Nonhomogenous ODE?

  • Thread starter Thread starter DreDD
  • Start date Start date
  • Tags Tags
    Linear Ode
Click For Summary

Homework Help Overview

The problem involves solving a first-order nonlinear nonhomogeneous ordinary differential equation (ODE) of the form y' + y = t^2, with initial conditions y(0) = 6 and y'(0) = -6. Participants are exploring methods to find both the homogeneous and particular solutions.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss attempts to solve the ODE using separation of variables and the method of undetermined coefficients. There is uncertainty about the necessity of using the second initial condition and whether the methods applied are appropriate for a first-order ODE.

Discussion Status

Some participants have provided feedback on the solutions presented, noting the consistency of the initial conditions with the ODE. There is acknowledgment of the need for clarification regarding the variable notation and suggestions for alternative methods of solving the equation, such as power series solutions.

Contextual Notes

There is a mention of confusion regarding the variable used in the equation, with participants noting that y should be expressed as a function of t rather than x. The discussion also touches on the implications of the initial conditions provided.

DreDD
Messages
4
Reaction score
0

Homework Statement



y' + y = t^2 , y(0) = 6, y'(0)= -6

Homework Equations





The Attempt at a Solution



first i tried to separate variables using y = ux but can't forward on and then i tried undetermined coeff. method. i found homogenous and particular solution but i am not sure about the solution because there is no need to use y'(0)=-6 and i really don't sure can i use this method for the first order ode.
 
Physics news on Phys.org
If you show me your homogeneous and particular solutions, I'll stand a better chance of telling you what is wrong with them than if I just wildly guess at what you may have done wrong! ;0)
 
x+1 = 0
x = -1

Yh = c1 e^-x

Yp= K2 x^2 + K1x + K0 and take Yp' and write the Yp and Yp' to the eq'n find K2, K1 and K0

Yp = x^2 - 2x +2

Y = c1 e^-x + x^2 - 2x + 2

there is only c1 and i don't need the second initial value
 
Shouldn't your y's be functions of t instead of x?!

and since y'(t)+y(t)=t^2, y'(0)+y(0)=0 => y'(0)=-y(0) which is consistent with your initial values, and so you can use either of them to find c1.

If on the other hand, you were given y(0)=6 and y'(0)=3, then there would be no solution since these initial conditions are inconsistent with your ODE.

Luckily, your initial conditions are consistent and so your method and solution are correct! :smile:
 
thanks for the help yes u r right it should be t . i don't like t :D
 
DreDD said:
thanks for the help yes u r right it should be t . i don't like t :D

t is good for you! :biggrin:

welcome to PF! :smile:
 
tiny-tim said:
t is good for you! :biggrin:

welcome to PF! :smile:

thx :D but it confuses my mind :D

by the way any other ways to solve this eq'n?
 
try "t" with ginseng...that should help your mind out :wink:

...as for other methods, I'm sure there are a few (such as power series solutions) but aside from plugging it into mathematica; this is the easiest way I know of.
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
7
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
Replies
7
Views
2K
Replies
9
Views
2K