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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) = eò 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) = eò P(x) dx
= eò -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 - 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 sustitute 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) = eò P(x) dx
= eò (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 = ò
u(x) Q(x) dx to obtain
x y = ò -2 xdx
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 sustitute 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) = eò P(x) dx
= eò (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 = ò
u(x) Q(x) dx to obtain
x y = ò - 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 sustitute 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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