Solve First Order Differential Equations

A tutorial on how to solve first order differential equations. Examples with detailed solutions are included.

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) =
P(x) dx

Solve the above for u to obtain

u(x) = e
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 =
u(x) Q(x) dx

Finally solve for y to obtain

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

= e -2 x dx

= e - x2

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

e - x2 y = x e - x2dx

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 ex2 - 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 P(x) dx

= e (1 / x) dx

= eln |x| = | x | = x since x > 0.

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

x y = -2 x dx

Integrate the right hand term to obtain

x y = -x2 + 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 = - x3 for x > 0

Solution to Example 3

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

dy / dx + y / x = - x2

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

P(x) = 1 / x and Q(x) = - x2

The integrating factor u(x) is given by

u(x) = e P(x) dx

= e (1 / x) dx

= eln |x| = | x | = x since x > 0.

We now substitute u(x)= x and Q(x) = - x2 in the equation u(x) y = u(x) Q(x) dx to obtain

x y = - x3 dx

Integration of the right hand term yields

x y = -x4 / 4 + C , C is a constant of integration.

Solve the above for y to obtain

y = C / x - x3 / 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 = - x3

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

d(x y) / dx = - x3

Integrate both sides to obtain

x y = - x4 / 4 + C.

Then solve for y to obtain

y = - x3 / 4 + C / x


Exercises: Solve the following differential equations.

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

2. dy / dx + y = x4

Answers to Above Exercises

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

2. y = x4 - 4x3 + 12x2 - 24 x + Ce-x + 24, C a constant of integration.

More references on Differential Equations

Differential Equations - Runge Kutta Method



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Updated: 2 April 2013

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