2nd Midsession Exam MATH 336 Differential Equations

Wednesday 20/11/2024 -- 1 hour

Name: ________________________________________ Student ID: ______________

Question 1 (A): (K1+K2) Choose the correct answer [5 Marks]

[1]. If 𝒚𝟏 (𝒙) and 𝒚𝟐 (𝒙) are solutions of 𝒚′′ + 𝒚 = 𝟎, 𝒙 ∈ ℝ, such that

𝒚𝟏 (𝟎) = 𝟏, 𝒚′𝟏 (𝟎) = 𝟎, 𝒚𝟐 (𝟎) = 𝟎, 𝒚′𝟐 (𝟎) = 𝟏, then the general solution is written as:
(a) 𝑦(𝑥) = 𝑐1 cos(𝑥) + 𝑐2 sin(𝑥) (c) 𝑦(𝑥) = 𝑐1 sin(𝑥) + 𝑐2 cos(𝑥)
(b) 𝑦(𝑥) = 𝑐1 𝑒 𝑥 + 𝑐2 𝑒 −𝑥 (d) 𝑦(𝑥) = 𝑐1 𝑥 + 𝑐2 𝑥 −1

[2]. The complementary function C.F. (𝒚𝒄 ) of 𝒙𝟐 𝒚′′ − 𝟑𝒙𝒚′ + 𝟒𝒚 = 𝟎 is

(a) 𝑦𝑐 = (𝑐1 + 𝑐2 𝑥)𝑒 2𝑥 (b) 𝑦𝑐 = (𝑐1 + 𝑐2 ln 𝑥)𝑥 2 (c) 𝑦𝑐 = (𝑐1 + 𝑐2 𝑥)𝑥 2 (d) None

[3]. The trial particular solution (𝑦𝑝 ) of the DE 𝒚′′ + 𝒚 = 𝟐 sin(𝑥) is

(a) 𝑦𝑝 = 𝐴 cos 𝑥 + 𝐵 sin 𝑥 (b) 𝑦𝑝 = 𝐴𝑥 sin 𝑥 (c) 𝑦𝑝 = 𝐴𝑥 cos 𝑥 + 𝐵𝑥 sin 𝑥 (d) None

[4]. The general solution of the DE 𝒚′′′ − 𝟓𝒚′′ + 𝟖𝒚′ − 𝟒𝒚 = 𝟎 is

(a) 𝑦 = (𝑐1 + 𝑐2 𝑥 + 𝑐3 𝑥 2 )𝑒 𝑥 (c) 𝑦 = 𝑐1 𝑒 𝑥 + (𝑐2 + 𝑐3 𝑥)𝑒 2𝑥
(b) 𝑦 = (𝑐1 + 𝑐2 𝑥 + 𝑐3 𝑥 2 )𝑒 2𝑥 (d) 𝑦 = 𝑐1 𝑒 2𝑥 + (𝑐2 + 𝑐3 𝑥)𝑒 𝑥

[5]. The PDE uxx  2 xuxy  xu yy  2 yux  xu y  u  0 is Elliptic if

(b) 𝑥 < 1 (b) 0 < 𝑥 < 1 (c) 𝑥 > 1 (d) None

Question 1 (B): (K3) Solve the following differential equation [3 Marks]

𝒚′′ − 𝟗𝒚′ + 𝟗𝒚 = 𝟒𝒆−𝟒𝒙
Solution:
93 5 93 5
The auxiliary equation is 𝑚2 − 9𝑚 + 9 = 0  m  , .
2 2
9 3 5 x 93 5 x
yc  c1e 2  c2e 2 . 1.5 Marks
Let y p  Ae
4 x
 yp  4 Ae 4 x and yp  16 Ae 4 x , 0.5 Marks
Substituting in the given DE, we have
16 Ae 4 x  36 Ae 4 x  9 Ae 4 x  4e 4 x
4
 A .
61
4 4 x
Hence, y p  e . 0.5 Marks
61
9 3 5 93 5
x x 4 4 x
The general solution is y  c1e 2
 c2e 2
 e . 0.5 Marks
61

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 2nd Midsession Exam MATH 336 Differential Equations
Wednesday 20/11/2024 -- 1 hour
Name: ________________________________________ Student ID: ______________

Question 2: (S1+S2): Write the answer of the following [Marks 𝟐 × 𝟑 = 𝟔]

[1]. Find the general solution of the differential equation:
𝒙𝟐 𝒚′′ − 𝟒𝒙𝒚′ + 𝟏𝟑𝒚 = (𝐥𝐧 𝒙)𝟐 + 𝟐.
[2]. Apply the method of variation of parameters to solve the following differential equation:
𝒚′′ − 𝟒𝒚′ + 𝟒𝒚 = (𝒙𝟐 − 𝟏)𝒆𝟐𝒙 .
Let x  e or t  ln x , then
t
Solution [1]:
2
dy dy 2 d y d 2 y dy
Substituting x  and x   in the given DE, we get
dx dt dx 2 dt 2 dt
d2y dy
2
 5  13 y  t 2  2 . (1)
dt dt
5 3 3
The auxiliary equation is m2  5m  13  0  m   i,
2 2
5
t  3 3   3 3 
yc  e2 c1 cos  t   c2 sin  t   . 1.5 Marks
  2   2  
y p  At 2  Bt  C  yp  2 At  B and yp  2 A .
Substituting in (1), we have
13 At 2  (10 A  13B)t  (2 A  5B  13C )  t 2  2
1 10 262
 13 A  1 i.e. A  , 10 A  13B  0  B  2
, and 2 A  5B  13C  2  C  .
13 (13) (13)3
1 10 262
Hence y p  t 2  2
t . 1.5 Marks
13 (13) (13)3
5 
t 3 3   3 3  1 2 10 262
The general solution is y  e 2 c1 cos  t   c2 sin  t    t  2
t 3
.
  2   2  13 (13) (13)
Solution [2]: The auxiliary equation is m  4m  4  0  m  2,2
2

yc  c1e2 x  c2 xe2 x . 1 Marks

Let y1 ( x)  e 2 x and y2 ( x)  xe 2 x , then y p  Ay1 ( x)  B y2 ( x)
e2 x xe 2 x
W  e4 x . 0.5 Marks
2e 2x
e  2 xe
2x 2x

y2 ( x ) xe 2 x x4 x2
A   R( x)dx    4 x ( x 2  1)e 2 x dx    ( x 3  x)dx    . 0.5 Marks
W e 4 2
2x 3
y ( x) e x
B 1 R ( x)dx   4 x ( x 2  1)e 2 x dx   ( x 2  1)dx   x . 0.5 Marks
W e 3
 x 4
x 
2
x 3
 x 4
x 
2
y p      e 2 x    x  xe 2 x     e 2 x .
 4 2  3   12 2 
 x 4 x2 
Hence the general solution is y  c1e 2 x  c2 xe 2 x     e 2 x . 0.5 Marks
 12 2 

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 2nd Midsession Exam MATH 336 Differential Equations
Wednesday 20/11/2024 -- 1 hour
Name: ________________________________________ Student ID: ______________

Question 3: (S3+S4): Write the answer of the following [Marks 𝟐 × 𝟑 = 𝟔]

[1]. Use D’Alembert’s method to find the solution of the partial differential equation:
𝒖𝒕𝒕 = 𝒄𝟐 𝒖𝒙𝒙 ,
where −∞ < 𝒙 < ∞ , 𝒕 > 𝟎, with initial conditions 𝒖(𝒙, 𝟎) = 𝐜𝐨𝐬 𝒙 , 𝒖𝒕 (𝒙, 𝟎) = 𝐬𝐢𝐧 𝒙

[2]. Find the product solution of the following partial differential equation:
𝒖𝒙𝒙 − 𝟒𝒖𝒚𝒚 = 𝟎.
Solution [1]: Using D’Alembert’s method with u0 (x)  cos x and v0 (x)  sin x , we have
1 1 x ct
u(x ,t )  u0 (x - ct)  u0 (x  ct )  x-ct v0 (s)ds 1 Mark
2 2c
1 1 x ct
 cos(x  ct )  cos(x  ct )   sin s ds
2 2c x ct
1 1 x ct
 cos x cos ct  sin x sin ct  cos x cos ct  sin x sin ct     cos s x ct 1 Mark
2 2c
1
 cos x cos ct  cos(x  ct )  cos(x  ct )
2c
1
 cos x cos ct  (sin x sin ct ) . 1 Mark
c
Solution [2]: Given that 𝑢𝑥𝑥 − 4𝑢𝑦𝑦 = 0 (1)
Let u ( x, t )  XY then 𝑢𝑥𝑥 = 𝑋 ′′ 𝑌 𝑎𝑛𝑑 𝑢𝑦𝑦 = 𝑋𝑌 ′′
Substituting in (1), we have
𝑋 ′′ 𝑌= 4 𝑋𝑌 ′′
𝑋 ′′ 𝑌 ′′
This implies = = 𝜆 (𝑠𝑎𝑦) 0.5 Marks
4𝑋 𝑌
We need to solve 𝑋 − 4𝑋𝜆 = 0 and 𝑌 ′′ − 𝜆𝑌 = 0.
′′

Case 1: For 𝜆 = 0
𝑋 ′′ = 0 𝑌 ′′ = 0
𝑋 = 𝐴𝑥 + 𝐵 𝑌 = 𝐶𝑦 + 𝐷

The product solution 𝑢(𝑥, 𝑦) = (𝐴𝑥 + 𝐵)(𝐶𝑦 + 𝐷). 0.5 Marks

Case 2: For 𝜆 = 𝑘2 > 0
𝑌 ′′ − 𝑘 2 𝑌 = 0
𝑋 ′′ − 4𝑘 2 𝑋 = 0
𝑌 = 𝐶𝑒 𝑘𝑦 + 𝐵𝑒 −𝑘𝑦
𝑋 = 𝐴𝑒 2𝑘𝑥 + 𝐵𝑒 −2𝑘𝑥
The product solution 𝑢(𝑥, 𝑦) = (𝐴𝑒 2𝑘𝑥 + 𝐵𝑒 −2𝑘𝑥 )(𝐶𝑒 𝑘𝑦 + 𝐵𝑒 −𝑘𝑦 ). 1 Mark

Case 3: For 𝜆 = −𝑘2 > 0

𝑋 ′′ + 4𝑘 2 𝑋 = 0 𝑌 ′′ + 𝑘 2 𝑌 = 0
𝑋 = 𝐴𝑐𝑜𝑠(2𝑘𝑥) + 𝐵𝑠𝑖𝑛(2𝑘𝑥) 𝑌 = 𝐶𝑐𝑜𝑠(𝑘𝑦) + 𝐵𝑠𝑖𝑛(𝑘y)

The product solution 𝑢(𝑥, 𝑦) = (𝐴𝑐𝑜𝑠(2𝑘𝑥) + 𝐵𝑠𝑖𝑛(2𝑘𝑥))(𝐶𝑐𝑜𝑠(𝑘𝑦) + 𝐵𝑠𝑖𝑛(𝑘y). 1 Mark

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Wishing you all the best