The general form of the first order linear differential equation is as follows

** dy / dx + P(x) y = Q(x)**

where P(x) and Q(x) are functions of x.

If we multiply all terms in the differential equation given above by an unknown function u(x), the equation becomes

** u(x) dy / dx + u(x) P(x) y = u(x) Q(x)**

The left hand side in the above equation has a term u dy / dx, we might think of writing the whole left hand side of the equation as d (u y ) / dx. Using the product rule of derivatives we obtain

d (u y ) / dx = y du / dx + u dy / dx

For y du / dx + u dy / dx and u(x) dy / dx + u(x) P(x) y to be equal, we need to have

du / dx = u(x) P(x)

Which may be written as

(1/u) du / dx = P(x)

Integrate both sides to obtain

ln(u) = ò P(x) dx

Solve the above for u to obtain

u(x) = e^{ò P(x) dx}

u(x) is called the integrating factor. A solution for the unknown function u has been found. This will help in solving the differential equations.

d(uy) / dx = u(x) Q(x)

Integrate both sides to obtain

u(x) y = ò u(x) Q(x) dx

Finally solve for y to obtain

y = ( 1 / u(x) ) ò u(x) Q(x) dx

__Example 1:__ Solve the differential equation

**
dy / dx - 2 x y = x
**
__Solution to Example 1__

Comparing the given differential equation with the general first order differential equation, we have

P(x) = -2 x and Q(x) = x

Let us now find the integrating factor u(x)

u(x) = e^{ò P(x) dx}

= e^{ò -2 x dx}

= e^{ - x2}

We now substitute u(x)= e^{ - x2} and Q(x) = x in the equation u(x) y = ò
u(x) Q(x) dx to obtain

e^{ - x2} y = ò x e^{ - x2}dx

Integrate the right hand term to obtain

e^{ - x2} y = -(1/2) e^{-x2} + C , C is a constant of integration.

Solve the above for y to obtain

y = C e^{x2} - 1/2

As a practice, find dy / dx and substitute y and dy / dx in the given equation to check that the solution found is correct.

__Example 2:__ Solve the differential equation

**
dy / dx + y / x = - 2 for x > 0
**
__Solution to Example 2__

We first find P(x) and Q(x)

P(x) = 1 / x and Q(x) = - 2

The integrating factor u(x) is given by

u(x) = e^{ò P(x) dx}

= e^{ò (1 / x) dx}

= e^{ln |x|} = | x | = x since x > 0.

We now substitute u(x)= x and Q(x) = - 2 in the equation u(x) y = ò
u(x) Q(x) dx to obtain

x y = ò -2 x dx

Integrate the right hand term to obtain

x y = -x^{2} + C , C is a constant of integration.

Solve the above for y to obtain

y = C / x - x

As an exercise find dy / dx and substitute y and dy / dx in the given equation to check that the solution found is correct.

__Example 3:__ Solve the differential equation

**
x dy / dx + y = - x**^{3} for x > 0
__Solution to Example 3__

We first divide all terms of the equation by x to obtain

dy / dx + y / x = - x^{2}

We now find P(x) and Q(x)

P(x) = 1 / x and Q(x) = - x^{2}

The integrating factor u(x) is given by

u(x) = e^{ò P(x) dx}

= e^{ò (1 / x) dx}

= e^{ln |x|} = | x | = x since x > 0.

We now substitute u(x)= x and Q(x) = - x^{2}
in the equation u(x) y = ò
u(x) Q(x) dx to obtain

x y = ò - x^{3}
dx

Integration of the right hand term yields

x y = -x^{4} / 4 + C , C is a constant of integration.

Solve the above for y to obtain

y = C / x - x^{3} / 4

As an exercise find dy / dx and substitute y and dy / dx in the given equation to check that the solution found is correct.

NOTE: If you can "see" that the right hand side of the given equation

x dy / dx + y = - x^{3}

can be written as d(x y) / dx, the solution can be found easily as follows

d(x y) / dx = - x^{3}

Integrate both sides to obtain

x y = - x^{4} / 4 + C.

Then solve for y to obtain

y = - x^{3} / 4 + C / x

__Exercises:__ Solve the following differential equations.

1. dy / dx + y = 2x + 5

2. dy / dx + y = x^{4}

__Answers to Above Exercises__

1. y = 2x + 3 + C e^{-x} , C constant of integration.

2. y = x^{4} - 4x^{3} + 12x^{2} - 24 x + Ce^{-x} + 24, C a constant of integration.

More references on
Differential Equations

Differential Equations - Runge Kutta Method