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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

Ella N. GCSE Maths tutor, A Level Maths tutor, A Level Further Mathem...

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