MAS201 Applied D.E. Fri 1pm Quiz 3 Class: ID: Name:
1
10 points
Solve a differential equation, that is, find a general solution.
[Hint: Consider order of terms and make a lucky guess.] x
2 y
00
− xy
0
+ y = x
2 ln x, x > 0
Solution.
Cauchy-Euler equation.
Its auxiliary equation with y = x m is m ( m − 1) − m + 1 = ( m − 1)
2
= 0 m = 1, so x and x ln x are fundamental set of solutions for homogeneous DE.
(no need to check linear independence by Wronskian as this method was already covered in textbook) ( + 3 points)
Now, enough to find a particular solution.
Let L be differential operator. Guess y = x
2 ln x , y
0
= 2 x ln x + x, y
00
= 2 ln x + 3
⇒ Ly = 3 x
2
+ 2 x
2 ln x − 2 x
2 ln x − x
2
+ x
2 ln x = 2 x
2
+ x
2 ln x
Choose y p
1
Ly p
2
= x 2 ln x and try y = x
= − 2 x
2 by linearity.
2
∴ y p
= y p
1
+ y p
2
= x
2 ln x − 2 x
2 to get Ly =
( + 5 points) x 2 . Choose y p
2
= − 2 x 2 (then
Thus, the general solution is y = ( c
1
+ c
2 ln x ) x + x
2 ln x − 2 x
2
( + 2 points)
29, Sep
Remark.
You can find a particular solution through variation of parameters, which essentially goes the same. The solution above is just the same way as undetermined coefficient method if you guess the right form of y p
.
This resembles the form y p
= P ( x ) e
λx with P ( x ) polynomial for inhomogeneous linear DE with constant coefficients.
This is because Cauchy-Euler equation basically is another representation of linear DE with constant coefficients. Thus the form has to be of P (ln x ) x m
. (Refer the remark at the end of sec 3.6 in your textbook) page 1 of 2

MAS201 Applied D.E. Fri 1pm Quiz 3 Class: ID:
2
2+8 points
Solve a differential equation.
y
00
− 2 y
0
+ y =


2 sin x x ≤ 0

0 x ≥ 0
(1) Find a fundamental set of solutions for homogeneous case.
Name:
(2) Find a particular solution and find the general solution.
[Hint: The solution must be of C
2
.]
Solution.
(1) It is e x and xe x
.
( + 2 points)
(2) To fit sin x , try cos x and it works, so y p
= cos x for x ≤ 0.
( + 2 points)
For x ≥ 0, choose y p a homogeneous solution such that it is C
2
. Note that enough to check C
1 since continuity of y
00 then follows from the DE.

 y (0) = 1
Then it is the same as solving DE with initial condition
 y
0
(0) = 0 from x ≤ 0.
( + 2 points)
Take y p
= Ae x
+ Bxe x
, then


 y (0) = A = 1 y
0
(0) = A + B = 0
⇒


A = 1

B = − 1

 cos x
∴ y p
=

(1 − x ) e x x ≤ 0 x ≥ 0
( + 2 points)
Thus, the general solution is y = c
1 e x
+ c
2 xe x

 cos x
+

(1 − x ) e x x ≤ 0 x ≥ 0
( + 2 points) (1)
29, Sep page 2 of 2