Solving xy' + 2y = sin(x) - Integrator Factor

  • Thread starter Thread starter paula17
  • Start date Start date
paula17
Messages
1
Reaction score
0
Hi guys, i need your help. I am not sure what is the integrator factor in this diferential equation to start solving it.

xy' + 2y = sin(x)

Thanks a lot
 
Physics news on Phys.org
You need to first write it as
y' + p(x)y = q(x)

Then evaluate
\mu = exp(\int p(x) dx)

Is that what you mean by integrating factor?
 
Or, if you would prefer to derive it yourself, an integrating factor is a function p(x) such that multiplying the equation by it, the left side becomes an "exact derivative":
\frac{d(p(x)xy)}{dx}= p(x)x\frac{dy}{dx}+ 2p(x)y
Since
\frac{d(p(x)xy)}{dx}= p(x)x\frac{dy}{dx}+ p(x)y+ p&#039;(x)xy[/itex] <br /> that means we must have xp&#039;+ p= 2p or xp&#039;= p, a simple separable differential equation for p. Solving it you get exactly what LeBrad said.
 

Similar threads

Replies
1
Views
1K
Replies
2
Views
2K
Replies
3
Views
3K
Replies
7
Views
3K
Back
Top