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

EN
Answered by Ella N. Further Mathematics tutor

11220 Views

See similar Further Mathematics A Level tutors

Related Further Mathematics A Level answers

All answers ▸

Find the general solution to: d^(2)x/dt^(2) + 7 dx/dt + 12x = 2e^(-t)


What is the meaning of having a 3 by 3 matrix with determinent 0. Both geometrically and algebriaclly.


find an expression for the sum of the series of 1 + 1/2cosx + 1/4cos2x +1/8cos3x + ......


Prove that 27(23^n)+17(10^2n)+22n is divisible by 11 for n belongs to the natural numbers


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences