Differential Equation

  • #1

Homework Statement


Solve the inital value first order linear differential equation

y'=y+x y(0) = 2



Homework Equations





The Attempt at a Solution



y'-y=x

That's as far as I got. I'm not sure how to approach this. I've looked through my notes and book, and I don't have any examples that are similar to this. I looked online and it said something about raising the coefficient of y to e, but I'm not sure what to do after.
 

Answers and Replies

  • #2
MathematicalPhysicist
Gold Member
4,594
311
You should multiply by an exponential factor.
y'-y=x
exp(-x)y'-exp(-x)y=exp(-x)x
(exp(-x)y)'=exp(-x)x
Usually the exponential factor is of the form:
if y'+f(x)y=g(x)
then the exponential factor is:
u=exp([tex]\int f(x)dx[/tex])
Take the integral of xexp(-x) by parts, and then your'e done.
 
  • #3
82
0
Hi; I tried using an integration factor, and using the initial value provided (y(0)=2), I arrived at a particular solution of [tex]y=x-1+3e^{-x}[/tex].

I just saw loop's reply above. The integration factor I obtained was [tex]e^x[/tex].
 
  • #4
HallsofIvy
Science Advisor
Homework Helper
41,847
967
For the linear equation y'+ f(x)y= g(x), and integrating factor is a function m(x) such that multiplying by it, m(x)y'+ m(x)f(x)y= f(x)g(x), reduces the left side to a single derivative: (m(x)y)'. Since, by the product rule, (m(x)y)'= my'+ m'y= my'+ mfy, we must have m'= m(x)f(x) which is, itself, a separable differential equation: dm/m= f(x)dx so [itex]ln(m)= \int f(x)dx[/itex] and so [itex]m(x)= e^{\int f(x)dx}[/itex]. For this particular problem f(x) is the constant -1 so your integrating factor is [itex]e^{-x}[/itex], not [itex]e^x[/itex]. Multiplying the equation by [itex]e^{-x}[/itex], we have [itex]e^{-x}y'- e^{-x}y= (e^{-x}y)'= xe^{-x}[/itex]. Integrating both sides of that (the left side by parts, as loop quantum gravity said) you get [itex]e^{-x}y= -(x+1)e^{-x}+ C[/itex] or [itex]y= -x- 1+ Ce^x[/itex] and, since y(0)= 2, 2= -1+ C and C= 3.
[itex]y= -x-1+ 3e^x[/itex]

Notice that if [itex]y= x- 1+ 3e^{-x}[/itex] then [itex]y'= 1- 3e^{-x}[/itex] while [itex]y+ x= x- 1+ 3e^{-x}+ x= 2x- 1+ 3e^{-x}[/itex] so your y does NOT satisfy the differential equation.
 
Last edited by a moderator:
  • #5
82
0
Thanks for pointing out that mistake Ivy...
 

Related Threads on Differential Equation

Replies
5
Views
751
  • Last Post
Replies
8
Views
847
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
2
Views
917
  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
0
Views
849
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
1
Views
906
  • Last Post
Replies
1
Views
1K
Top