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Section 3.2 - Reduction of Order
Section 3.2
Copyright©Arunabha Biswas
127
Consider the 2nd order homogeneous linear DE:
a2 (x)y 00 + a1 (x)y 0 + a0 (x)y = 0
⇒ y 00 +
a1 (x) 0 a0 (x)
y +
y=0
a2 (x)
a2 (x)
(since a2 (x) 6= 0)
⇒ y 00 + P (x)y + Q(x)y = 0
Theorem (Reduction of Order)
If y1 (x) is a known solution of this DE then the other linearly
independent solution of this can be found as:
Z
y2 (x) = y1 (x)
Section 3.2
R
e− P (x) dx
dx
[y1 (x)]2
Copyright©Arunabha Biswas
128
Exercise: Use “reduction of order” to find a second linearly
independent solution of y 00 + 16y = 0 where y1 (x) = cos (4x) is a
known solution. Then write down the general solution.
Section 3.2
Copyright©Arunabha Biswas
129
Exercise: Use “reduction of order” to find a second linearly
independent solution of x2 y 00 − xy 0 + 2y = 0 where
y1 (x) = x sin (ln x) is a known solution. Then write down the
general solution.
Section 3.2
Copyright©Arunabha Biswas
130
Section 3.2
Copyright©Arunabha Biswas
131
Section 3.3 - Homogeneous Linear DE with Constant
Coefficients
Section 3.3
Copyright©Arunabha Biswas
132
Consider the homogeneous linear DE:
an y (n) + an−1 y (n−1) + · · · + a2 y 00 + a1 y 0 + a0 y = 0
where an , an−1 , · · · a2 , a1 , a0 are all real valued constants. Now
plug y = emx as a ”sample solution”. So we get:
emx (an mn + an−1 mn−1 + · · · a2 m2 + a1 m + a0 ) = 0
⇒ an mn + an−1 mn−1 + · · · a2 m2 + a1 m + a0 = 0
(So basically you replace all y (k) by mk )
This is a polynomial in m and known as “auxiliary equation” or
“characteristic equation”.
Solve it for m, and you will get n number of solutions.
Section 3.3
Copyright©Arunabha Biswas
133