How to use the integrating factor?

Say you have a differential equation of the form:
dy/dx + Py = Q , where P and Q are functions of x.
Remember that even if your equation doesn't initially look like this, you may be able to rearrange it into this format!
The integrating factor is e^( ∫P dx).
If we multiply the whole equation by this integrating factor (not forgetting the right hand side!) then it will be in the form:
e^( ∫P dx) dy/dx + Pe^( ∫P dx) y = Qe^( ∫P dx)
This is useful to us because now the left hand side looks like something we might obtain from the product rule ( d/dx(uv)=u'v+v'u), with u=y and v=e^( ∫P dx).
So now we can write:
d/dx (ye^( ∫P dx)) = Qe^( ∫P dx)
Integrating both sides gives:
ye^( ∫P dx) = ∫ (Qe^( ∫P dx)) dx
This is the method, but it may be easier to understand with an example.
y = x dy/dx - 3x
This doesn't look like the form we want, but note that after a little rearranging, it can be expressed as:
dy/dx + y/x = 3
So here we have P=1/x, Q=3.
Our integrating factor is e^( ∫1/x dx) = e^(ln x) = x
Now we want to multiply the whole equation by this integrating factor, giving us:
x dy/dx + y = 3x
Noting that this looks like the product rule with u=y, v=x, we write:
d/dx (xy) = 3x
Integrating both sides gives
xy = ∫3x dx
xy = 1.5x^2 +c
Usually, you'll want the answer to be in the form y= something (although not always), so we divide through by x to give our final answer of:
y = 1.5x +c/x

Related Further Mathematics A Level answers

All answers ▸

A parabola with equation y^2=4ax for constant a is translated by the vector (2,3) to give the curve C. The curve C passes through the point (4,7), what is the value of a?


Unfortunately this box is to small to contain the question so please see the first paragraph of the answer box for the question.


find the sum of r from 0 to n of : 1/((r+1)(r+2)(r+3))


A 1kg ball is dropped of a 20m tall bridge onto tarmac. The ball experiences 2N of drag throughout its motion. The ground has a coefficient of restitution of 0.5. What is the maximum height the ball will reach after one bounce


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2024

Terms & Conditions|Privacy Policy