Solving a First-order Differential Equation using the Integration Factor method

To solve a differential equation of the standard form below, we can make use of the Integration Factor method,

`(dy)/(dx)+F(x)y=G(x)`
Equation 1

We introduce a factor,

`I(x)`
Equation 2

Multiplying the differential equation by this ‘integrating factor’ we obtain,

`I(x) (dy)/(dx) +I(x)F(x)y=I(x)G(x)`
Equation 3

Which is now of the form of a derivative of a product of functions, Ie.,

`v(du)/(dx)+u(dv)/(dx)=(d(uv))/(dx)`
Equation 4

By inspection we obtain:

`v=I(x)`
Equation 5

`u'=(du)/(dx)=(dy)/(dx)`
Equation 6

`u=y`
Equation 7

`v'=(dv)/(dx)=I(x)F(x)`
Equation 8

And so,

`(dv)/(dx)=(dI(x))/(dx)=I(x)F(x)`
Equation 9

Which requires,

`(dI(x))/(I(x))=F(x)dx`
Equation 10

…and has the solution,

`lnI(x)=int F(x)dx`
Equation 11

Ie.,

`I(x)=exp( int F(x)dx)`
Equation 12

Combining the r.h.s of Equation 3 to that of Equation 4, we have

`(d(uv))/(dx)=I(x)G(x)`
Equation 13

Which integrates to

`uv=int I(x)G(x)dx + c`
Equation 14

From the definitions of u and v (equations 5 and 7), we have

`uv=I(x)y`
Equation 15

Which we know to be equal the r.h.s of equation 3, ie.,

`I(x)y=int I(x)G(x)dx + c`
Equation 16

By definition I and G are functions of x only, and so the integral can generally be found. Hence, we have an expression for y in terms of x.

`y=1/(I(x)) ( int I(x)G(x)dx + c)`
Equation 17