2. THE USE OF
A KNOWN SOLUTION
OF AN EQUATION
TO FIND ANOTHER
3. Consider a general
second-order linear
equation
of the form
y"+p(x)y'+q(x)y=0….(1)
•Often, we can either guess
or elicit one solution to the
equation.
• But finding the second
independent solution is
more difficult.
•In this section we introduce
a method for finding that
second solution.
4. Here, we shall assume that we have found
the one solution y1
and
we shall suppose that the second solution
we seek is
y2= v · y1
for some undetermined function v.
Our job, then, is to find v.
5. Substituting y2= v · y1 into Equation (1), we get
or
Now we are assuming that y1 is a solution of the
differential equation (1), so the Expression in brackets
must vanish.
As a result,
We rearrange this equation to obtain
6. On integration, we get,
One more integration gives
We may rewrite this equation as,
7. Find the general solution of the differential equation
Consider the auxiliary equation m2-4m+4=0
Since m=2 is a root of this equation,
is a solution of the given equation.
Before applying the new methodology, we can observe that
8. According to the new formula, we can find a second solution
with
So, we have
𝑣 =
1
𝑒2𝑥 2
𝑒− −4 𝑑𝑥
𝑑𝑥 =
1
𝑒4𝑥
. 𝑒4𝑥
. 𝑑𝑥 = 1 𝑑𝑥 = 𝑥
The second solution to our differential equation is 𝑦2 = 𝑣. 𝑦1 = 𝑥. 𝑒2𝑥
The general solution is therefore 𝑦 = 𝐴. 𝑒2𝑥
+ 𝐵. 𝑥𝑒2𝑥
9. Verify that 𝑦1 = 𝑥2
is one solution of 𝑥2
𝑦” + 𝑥𝑦′
− 4𝑦 = 0 ,
and then find 𝑦2 and the general solution.
Verification:
𝑦1 = 𝑥2 ⇒ 𝑦1
′
= 2𝑥 & 𝑦1
′′
= 2
So,
𝑥2
𝑦1
′′
+ 𝑥𝑦1
′
− 4𝑦1 = 𝑥2
. 2 + 𝑥. 2𝑥 − 4. 𝑥2
= 0
Hence 𝑦1 = 𝑥2 is a solution of given equation.
10. Finding second solution:
The given equation can be expressed as 𝑦′′
+
1
𝑥
𝑦′
−
4
𝑥2 𝑦 = 0
Then 𝑝 𝑥 =
1
𝑥
and 𝑞 𝑥 = −
4
𝑥2
According to the new formula, we can find a second solution
with
So, we have
=
1
𝑥4
𝑒−
1
𝑥
𝑑𝑥
𝑑𝑥 =
1
𝑥4
𝑒− log 𝑥 𝑑𝑥
𝑑𝑥
11. Therefore,
V =
1
𝑥4
𝑒− log 𝑥 𝑑𝑥
𝑑𝑥 =
1
𝑥4
𝑒log 𝑥−1 𝑑𝑥
=
1
𝑥4
1
𝑥
𝑑𝑥 =
1
𝑥5
𝑑𝑥 = −
1
4
1
𝑥4
So, the second solution is
𝑦2 = 𝑣. 𝑦1 = −
1
4
1
𝑥4
𝑥2
= −
1
4𝑥2
= −
1
4
𝑥−2
Hence the general solution of the given equation is
𝑦 = 𝑐1 𝑥2
+ 𝑐2. −
1
4
𝑥−2
= 𝐴𝑥2
+ 𝐵. 𝑥−2
12. Assignment
Use the method of this section to find y2 and the general
solution of each of the following equations from the given
solution y1.
✍ 𝑦′′
+𝑦 = 0 , 𝑦1 𝑥 = sin 𝑥
✍ 𝑦′′
−𝑦 = 0, 𝑦1 𝑥 = 𝑒 𝑥
