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Practice Problems for the final:
Problem 1 Find the general solutions of the following DEs.
a) x 2 y   xy  y 2  0
solution:
y
y2
y   x  2  0 : homogeneous equation.
x
dy
y
 v  x dv .
v  x , y  vx, and
dx
dx
v  x dv  v  v 2 .
dx
x dv  v 2 : separable equation.
dx
dv  dx .
x
v2
 1v  ln|x|C.
 xy  ln|x|C.
y x .
ln|x|C
Note: You can consider the above equation as a Bernoulli equation with n  2.
b)x 2  1y   x  1y  1
solution:
y   x2  1 y  2 1
: linear equation.
x 1
x 1
y
 21 .
y 
x1
x 1
Integration factor x  e
 x11 dx  e lnx1  x  1.
d x  1y  x  1  1 .
x1
dx
x2  1
x  1y  ln|x  1|C.
y
ln|x  1|C
.
x1
1
c)y 2 y   2xy 3  6x
solution:
y   2xy  6xy 2 : Bernoulli equation with n  2.
dy
 1 v 2/3 dv .
3
dx
dx
v  y 12  y 3 , y  v 1/3 and
1 v 2/3 dv  2xv 1/3  6xv 2/3 .
3
dx
dv  6xv  18x : linear equation.
dx
Integrating factor x  e
 6xdx  e 3x 2 .
d e 3x 2 v  18xe 3x 2 .
dx
e 3x v  18  xe 3x dx  3e 3x  C, u  3x 2 and du  6xdx.
2
2
2
2
v  3  Ce 3x .
2
y  3  Ce 3x  1/3 .
d) y  
xy
sol) Let v  x  y, then y  v  x and
dv 
dx
dy
 dv  1.
dx
dx
dv  1 v .
dx
v  1 : separable equation.
dv
 dx.
v 1

Let w 
dv
 x  C.
v 1
v  1. Then
dw  1 v 1/2 dv
2
and
dv  2v 1/2 dw  2w  1dw.
Hence,

dv

v 1

2w  1
2 dw
dw  2  w
w
 2w  2 ln|w|C  2 v  1  2 ln v  1  C.
Thus,
x  2 x  y  1  2 ln x  y  1  C.
2
Problem 2 A hemispherical bowl (with top radius 4) shaped water tank is slowly losing
water at its lower end. As a result, the height of water in the tank, given by yt
satisfies
y
dy
 1
.
72 8y  y 2 
dt
a) Solve the DE for yt when the tank is initially full.
solution: Since the tank is initially full and the top radius is 4, y0  4.
By separating variables,
8y 1/2  y 3/2 dy   1 dt.
72
16 y 3/2  2 y 5/2   t  C.
5
72
3
y0  4, C  448 .
15
b) How long does it take for the tank to be empty?
solution: You need to find t for which yt  0.
t  72  148 .
15
Problem 3 Determine whether the given functions are linearly independent or not.
a) fx  e x sin x, gx  e x cos x.
solution:
Wf, g 
e x sin x
e x cos x
e 2x sin x  cos x e x cos x  sin x
 e 2x  sin 2 x  sin x cos x  e 2x sin x cos x  cos 2 x  e 2x  0.
Hence f and g are linearly independent.
b) fx  sin 2x, gx  sin x cos x and hx  e x .
solution: Note that sin 2x  2 sin x cos x. Hence
fx  2gx  0hx  0.
Since we can write 0 as a (nontrivial) linear combination of f, g and h, they are linearly
dependent.
Remark : Nontrivial combination means linear combination other 0f  0g  0h  0.
3
Problem 4 Find the unique solution to the initial value problem
y   7y   12y  x  2e 3x ,
y0  0, y  0  2.
solution: Characteristic equation
r 2  7r  12  r  3r  4  0,
r  3, 4.
y c  c 1 e 3x  c 2 e 4x .
Your first guess for y p might be y p  Ax  Be 3x . But since you have e 3x in y c , the
correct candidate should be
y p  xAx  Be 3x .
Hence,
y p  7y p  12y p  2Ax  2A  Be 3x  x  2e 3x .
 2A  1,
2A  B  1.
A   12 ,
B  3.
y  y c  y p  c 1 e 3x  c 2 e 4x  3xe 3x 
1
2
x 2 e 3x .
Now,
y0  c 1  c 2  0,
y  0  3c 1  4c 2  3  0.
So, c 1  5 and c 2  5.
Hence, the unique solution y is
y  5e 3x  5e 4x 
1
2
x 2 e 3x  3xe 3x .
4
Problem 5 Find the general solution of the differential equation
y   y   4y   6y  e 2x  2x.
solution: Characteristic equation is r 3  r 2  4r  6  0. We observe that r  1 is a
solution and using this, we factor
r 3  r 2  4r  6  r  1r 2  2r  6  0.
r  1,
1  5 i.
So, y c  c 1 e x  c 2 e x cos 5 x  c 3 e x sin 5 x. Find each particular solution for e 2x , 2x.
n
For e 2x : Try y p 1  Ae 2x . Note that y p 1  A2 n e 2x . So,
A8  4  4  2  6e 2x  e 2x .
A
1
18
1
18
y p1 
.
e 2x .
For 2x: Try y p 2  Bx  C. Note that y p 2  y  0  0 and y p 2  B. So,
6Bx  4B  6C  2x.
6B  2,
B
1
3
4B  6C  0.
C   29 .
,
y p2 
1
3
x
2
9
.
Therefore,
y  y c  y p 1  y p 2  c 1 e x  c 2 e x cos 5 x  c 3 e x sin 5 x 
1
18
e 2x 
1
3
x
2
9
.
5
Problem 6 Find the general solution to the differential equation
ex
y   2y   y  1x
2 .
solution: Characteristic equation: r 2  2r  1  0, r  1 2  0, r  1, 1. So, two
linearly independent solutions are y 1  e x , y 2  xe x .
ex
Since the right hand side is fx  1x
2 , we need to use the variation of parameter
method.
y p  u 1 y 1  u 2 y 2  u 1 e x  u 2 xe x ,
where
y f
y f
u1    2 , u2   1 .
W
W
W
u1   
u2 
ex
e
x
xe x 

xe x
x
e  xe
ex
1x 2
2x
e
ex 
e
x
 e 2x x  1  xe 2x  e 2x .
dx   
ex
1x 2
2x
dx 

x
  1 ln1  x 2 .
2
2
1x
1 dx  arctan x.
1  x2
Hence,
y p   1 e x ln1  x 2   xe x arctan x.
2
Thus, the general solution is
y  c 1 e x  c 2 xe x  12 e x ln1  x 2   xe x arctan x.
6
Problem 7 Find the Fourier series solution of the end point problem
x   2x  1
x0  0,
x1  0.
solution: According to the boundary data, we need to find the sine series of 1,
0  t  1.
4
n
1
b n  2  sin ntdt 
0,
0
So, 1   n odd
4
n
n odd
,
n even.
sin nt. Now, we are looking for the solution xt of
x   2x 

4
n sin nt.
n odd
Try
xt 
 B n sin nt.
n odd
This would satisfy the end point conditions. We need to determine B n .
x     n 2  2 B n sin nt.
n odd
Hence,
 B n 2   2 n 2  sin nt  
4
n sin nt.
n odd
nodd
It follow that
4 ,
B n 2   2 n 2   n
for n odd.
4
,
n2  n 2  2 
for n odd.
So, we have
Bn 
Therefore,
xt 

n odd
4
sin nt.
n2  n 2  2 
7
Problem 8Find a particular solution of the following equations.
a) x   2x  sin t
solution: Using undetermined coefficient method, set x p  A sin t  B cos t. Note here
that x c does not have either sin t or cos t. Since there is no x  term, we can try
x p  A sin t. We can easily deduce that A  1. So, x p  sin t.
b) x   2x   n odd
4
n
sin nt. (Find a formal Fourier series solution.)
solution: Since there is no x  term, we can try
x p t 
 B n sin nt.
n odd
Then,
x p t 
 n 2 B n sin nt.
n odd
Plugging in
x p , x p
into the equation, we get
 n 2  2B n sin nt  
n odd
4
n
sin nt.
n odd
So,
Bn 
4
.
n2  n 2 
Hence,
x p t 

n odd
4
sin nt.
n2  n 2 
8
Problem 9 Consider the following eigenvalue problem
X   X  0, 0  x  
X0  0, X    0.
Show that the eigenvalues  n and eigenfunctions X n are given by
2n  1 2
2n  1x
, X n  sin
, n  1, 2,  .
n 
2
2
4
You may use the following fact: cos x  0 if and only if x 
odd).
solution: i)   0: Xx  C  Dx. Since X0  0, C  0.
n
2
, n  1, 3, 5,  (i.e, n:
X  D  0.
Hence   0 is not an eigenvalue.
ii)   0: set    2 ,   0 Then, X    2 X  0. The zeros of its characteristic equation
is . So,
Xx  Ae x  Be x .
X0  A  B  0,
B  A.
X    Ae   Be   Ae   e    0.
 0, A  0 and so B  0. Hence, Xx  0. Thus, if   0, it is not an
Since e   e 
eigenvalue.
iii)   0: set    2 ,   0. Then, the general solution
Xx  A cos x  B sin x.
X0  A  0,
X    A sin x  B cos x  B cos x  0.
, n  1, 2,  .
Now, we do not want B  0. So, cos   0, which implies that   2n1
2
2n1 2
Thus the eigenvalues are  n  4 2 . The corresponding solutions (eigenfunctions)
2n1x
are X n x  sin 2 .
9
Problem 10 By separating the variables, solve the following wave equation
4u xx  u tt , 0  x  , t  0
u0, t  0, u, t  0, t  0.
ux, 0  1, u t x, 0  0,
0  x  .
solution: Set ux, t  XxTt.
4X  T  XT  .
X   T   .
X
4T
X   X  0.
So,  n 
n22
2
X0  0, X  0.
 n 2 and X n x  sin nx. Plug in  n  n 2 into T   4T  0. We get
T   4n 2  0.
Tt  c 1 cos 2nt  c 2 sin 2nt.
Since u t x, 0  0, T  0  0.
T  t  2nc 1 sin 2nt  2nc 2 cos 2nt, T  t  2nc 2  0,
Hence,
T n t  cos 2nt.
Now
c 2  0.

ux, t 
 c n sin nx cos 2nt.
n1

ux, 0 
 c n sin nx  1.
m1
Fourier sine series of 1 is

4
n
sin nx.
n odd
Hence,
cn  4
n,
if n is odd and otherwise, c n  0.
Thus,
ux, t 

4
n
sin nx cos 2nt.
n odd
10
Problem 11 Find the solution of the following problem. You are not required to
determine the coefficients.
u xx  u yy  0, 0  x  2, 0  y  2
ux, 0  ux, 2  0
u0, y  0.
solution:
u xx  u yy  0,
0  x  2, 0  y  2
1
ux, 0  ux, 2  0
2
Set ux, y  XxYy. Then, 1 tells us that
X  Y  XY   0.


 X  Y  .
X
Y
So, we have
Y   Y  0, 3
X   X  0.
4
From 2, we know Y0  Y2  0. Using these boundary conditions and the equation
3 above, we have a familiar eigenvalue problem
Y   Y  0
Y0  Y2  0.
2 2
Eigenvalues:  n  n 4
ny
Eigenfunctions: Y n y  sin 2 .
We find from u0, y  0 that X0  0. Now plug in  n 
consider
2 2
X   n x  0
n22
4
into the equation 4 and
X0  0.
We know that
Xx  c 1 e
nx
2
 c2e
nx
2
.
Since X0  c 1  c 2  0, c 2  c 1 .
Hence,
nx
nx
Xx  c 1 e 2  e  2 .
Thus,
X n x  e
nx
2
 e
nx
2
.
So,

ux, y 

 c n X n xY n y   c n e
n1
n1
nx
2
 e
nx
2
 sin
ny
2
.
11