First-Order Linear Differential Equations:
A First order linear differential equation is an equation of the form
y 0 + P (x)y = Q(x).
Where P and Q are functions of x. If the equation is written in
this form it is called standard form. The equation is called first
order because it only involves the function y and first derivatives
of y. We can solve this equation in general but it is better to
understand how to solve it than it is to just memorize the solution.
Integrating factors
We are trying to solve y 0 + P (x)y = Q(x). We make the following
observation (using the product rule):
R
R
d R P (x)dx
d
]
[y · e P (x)dx ] = y 0 · e P (x)dx + y ·
[e
dx
dx
Now
R
d R P (x)dx
] = e P (x)dx P (x) So we have
[e
dx
R
R
d
[y · e P (x)dx ] = e P (x)dx [y 0 + P (x)y]
dx
We define u(x) = e
the property that
R
P (x)dx
to be the integrating factor which has
d
[u(x)y] = u(x)[y 0 + P (x)y].
dx
Solving y 0 + P (x)y = Q(x) :
Now we start with our equation and multiply both sides by the
integrating factor:
d
[u(x)y] = u(x)[y 0 + P (x)y] = u(x)Q(x).
dx
Now we can integrate both sides and notice that the left-hand side
is already the derivative of something!
Z
Z
d
u(x)ydx = u(x)Q(x)dx.
dx
Z
u(x)y =
u(x)Q(x)dx ⇒ y =
1
u(x)
Z
u(x)Q(x)dx.
Solving y 0 + P (x)y = Q(x) :
The general solution to the first order linear differential equation is
given by
Z
1
y(x) =
u(x)Q(x)dx,
u(x)
with
R
u(x) = e
Now let’s do an example.
P (x)dx
.
Example: y 0 + 4xy = x
First we note that this is already in standard form with P (x) = 4x,
and Q(x) = x.
The first step is to find the integrating factor
u(x) = e
R
P (x)dx
=e
R
4xdx
2
= e2x .
Note that we do not need the general form of the integral until
the last step.
R
1
The next step is to find y = u(x)
u(x)Q(x)dx.
Recall that
1
e2x2
= e−2x
2.
Example: y 0 + 4xy = x continued
We have y =
solve
1
u(x)
R
2
u(x)Q(x)dx, with u(x) = e2x , so we have to
Z
2
−2x2
y=e
xe2x dx.
This integral can be done with the
z = 2x2 , then dz = 4xdx
Z
Z
1 z
2
xe2x dx =
e dz =
4
method of substitution, let
1 z
1 2
e + C = e2x + C.
4
4
Almost done, now let
1
2 1
2
2
y = e−2x [ e2x + C] = + Ce−2x .
4
4
Note how important the +C is in the integral, without it we lose
an entire part of the solution!
Example: y 0 + 4xy = x continued
We can have the utmost confidence in our solution if we check our
work. If we did everything right then we will be able to solve the
differential equation with our answer, i.e.
y=
Now
1
2
+ Ce−2x should satisfy y 0 + 4xy = x.
4
2
2
y 0 = Ce−2x (−4x) = −4Cxe−2x .
So
1
2
2
y 0 + 4xy = −4Cxe−2x + 4xCe−2x + 4x = x
4
and it works!
Is there another way to do that problem?
Yes, we can also solve that problem using separation of variables.
dy
dy
1
+ 4xy = x ⇒
= x(1 − 4y) ⇒
dy = xdx.
dx
dx
(1 − 4y)
Try solving it this way and see if you get the same answer (you
should!)
Another Example:
Find the solution to
xy 0 + y = x2 + 1
First we write
y0 +
1
1
1
1
y = x + ⇒ P (x) = and Q(x) = x + .
x
x
x
x
Therefore
u(x) = e
R
P (x)dx
= x,
and
1
y=
u(x)
Z
1
Q(x)u(x)dx =
x
Z
1
x2 + 1dx = x2 + 1 + Cx−1 .
3
Solve xy 0 + y = x2 + 1
Let’s check our answer y = 31 x2 + 1 + Cx−1 :
2
y 0 = x − Cx−2
3
2
1
⇒ xy 0 + y = x2 − Cx−1 + x2 + 1 + Cx−1
3
3
⇒ xy 0 + x = x2 + 1!