1
Part A  Questions & Answers
DECEMBER 2014 
  MA 2161
1.
Find the particular integral of Particular Integral = 2
 6 D  5  y  e 5 x
1
e 5 x
( D  6 D  5)
= 2
1
e 5 x
(25  30  5)
= x
= { put D = 5 }
1
e 5 x
2D  6
x
e 5 x { put D = 5 }
10  6
= 
2.
D
x 5 x
e
4
Transform the differential equation x2
coefficients.
dy
d2y
 x
+ 2y = 0 with constant 2
dx
dx
Put x = ez z = log x
xD = 
x2D2 =  (   1 ) in the given equation, we get [ (1)   + 2 ] y = 0
( 2  2 + 2 ) y = 0 where 

d
dz
DECEMBER 2014 
  MA 6251
1.
Find the particular integral of Particular Integral = D
2
 2 D  2  y  e  x sin 2 x
1
e  x sin 2 x
D  2D  2
2
Prepared by : R.Ravikumar, Department of Mathematics, Pallavan College of
Engineering
2
= e
Solve x2
1
sin 2 x
( D  1)  2( D  1)  2
2
= e  x
1
sin 2 x
D 1
= e  x
1
sin 2 x
4  1
= 2.
x
2
{ put D2 = 4 }
e x
sin 2 x
3
dy
d2y
+ x
= 0
dx
dx 2
Put x = ez z = log x
xD = 
x2D2 =  (   1 ) in the given equation, we get [ (1) +  ] y = 0
( 2 ) y = 0
The auxiliary equation is m2 = 0
 m = 0, 0
 the solution is y = ( Az  B)e 0 z where z = log x
JUNE 2014 
  MA 2161
1.
Solve the equation dy
d2 y
+ 2
+ y = 0
2
dx
dx
The auxiliary equation is m2 + 2m + 1 = 0 and hence m = 1, 1
 the solution is y = ( Ax + B ) e 
2.
x
Find the particular integral of the equation D
2
 9  y  33 x
Prepared by : R.Ravikumar, Department of Mathematics, Pallavan College of
Engineering
3
Particular Integral = 1
e 3 x
( D  9)
2
= 1
e 3 x
(9  9)
= x
1 3 x
e
2D
= 
x 3 x
e
6
{ put D = 3 }
DECEMBER 2013 
  MA 2161
1.
Solve ( D 2  4 ) y = 1
The auxiliary equation is m2  4 = 0
 m =  2
 the complementary function is Ae 2 x
Particular Integral = 1
e0 x
( D  4)
= 2
1
e0 x
(0  4)
= 
 the solution is y
2.
 Be 2 x
{ put D = 0 }
1
4
 Ae 2 x  Be 2 x 
Convert (3x2D2 + 5xD + 7)y = 1
2
2
log x into an equation with constant coefficients.
x
Put x = ez z = log x
xD = 
x2D2 =  (   1 ) in the given equation, we get [ 3(1) + 5 + 7 ] y = 2
z
ez
( 32 + 2 + 7 ) y = 2z e z
JUNE 2013 
  MA 2161
Prepared by : R.Ravikumar, Department of Mathematics, Pallavan College of
Engineering
4
1.
Find the particular integral of Partricular Integral = D
2
 2 D  1 y  cosh x
1 x x
e e 
2
D2  2D  1
= 1
ex
1
e x

2 D 2  2D  1 2 D 2  2D  1
= x ex
1 e x

2 2D  2 2 1  2  1
= x2 x
e x
e 
2
8
2.
Solve x2
dy
d2y
+ 4x
+ 2y = 0
dx
dx 2
Put x = ez z = log x
xD = 
x2D2 =  (   1 ) in the given equation, we get [ (1) + 4 + 2 ] y = 0
( 2 + 3 + 2 ) y = 0
The auxiliary equation is m2 + 3m + 2 = 0
 m = 1, 2
 the solution is Ae  z
 Be 2 z where z = log x
DECEMBER 2012 
  MA 2161
1.
Find the Wronskian of y1 , y2 of y   2
y   y  e x log x
The auxiliary equation is m2 – 2m + 1 = 0 and hence m = 1, 1
 the complementary function is y = ( Ax + B ) ex
Here Wronskian W = y1 y2'
 y2 y1' where y1 = xex , y2 = ex
Prepared by : R.Ravikumar, Department of Mathematics, Pallavan College of
Engineering
5
= x ex ex  ex { x ex + ex }
=  e2z
2.
Find the particular integral of ( D 2  4 D + 4 ) y = 2x
1
2x
2
( D  2)
Particular Integral = 1
e x (log 2)
( D  2) 2
= = 1
e x (log 2)
2
[(log 2)  2]
JUNE 2012 
  MA 2161
1.
Transform (2x+3)2 y  2 (2x+3) y + 2y = 6x into linear DE with constant coefficients.
Put 2x+3 = ez
z = log(2x+3)
(2x+3) D = 2 
(2x+3)2 D2 = 4  (   1 ) in the given equation, we get
( 42  8 + 2 )y = 3( ez  3 )
2.
Find the Particular Integral of (D1)2 y = ex sinx
Particular Integral = 1
e x sin x
( D  1) 2
= e
x
1
sin x
( D  1  1) 2
= e x
1
sin x
D2
=  ex sinx
DECEMBER 2011 
  MA 2161
1.
Transform the equation x 2 y  
x y   x into a linear differential equation with constant coefficients.
Given equation can be written as Put x = ez ( x 2 D 2 + x D ) y = x z = log x
Prepared by : R.Ravikumar, Department of Mathematics, Pallavan College of
Engineering
6
xD = 
x2D2 =  (   1 ) in the given equation, we get ( 2 ) y = e z
2.
Find the particular integral of (D2+4)y = sin 2x
Particular Integral = 1
sin 2 x
D 4
2
= 1
sin 2 x
4  4
= x
sin 2 x
2D
{ put D2 = 4 }
x
cos 2 x
4
= 
DECEMBER 2011 
  181202
1.
Find the particular integral of ( D 2  4 D + 4 ) y = x2 e 2x
Particular Integral = = e
2.
2x
1
x 2e 2 x
D  4D  4
2
1
x2
( D  2)  4( D  2)  4
2
= e 2 x
1
x2
2
D
= e 2 x
x4
12
Transform the differential equation (x2D2 + 4xD + 2)y = x 
1
to a differential x
equation with constant coefficients.
Put x = ez z = log x
xD = 
x2D2 =  (   1 ) in the given equation, we get [ (1) + 4 + 2 ] y = e z + e z
( 2 + 3 + 2 ) y = e z + e z
JUNE 2011 
  181202
Prepared by : R.Ravikumar, Department of Mathematics, Pallavan College of
Engineering
7
1.
Solve ( D3 + 1 ) y = 0
The auxiliary equation is m3 + 1 = 0
1
3
. Hence the solution is i
2
2
The roots are m = 1, 3
3

y  e  A cos
x  B sin   Ce  x
2
x

1
x
2
2.
Reduce the equation x 4 y  x 3 y   x 2 y   1 into linear equation with constant coefficients
Given equation is x
3
y   x 2 y   x y  
Put x = ez z = log x
xD = 
x2D2 =  (   1 ) 1
x
x3 D3 =  (   1 ) (   2 ) in the given equation, we get [ (1)(2)  (1) +  ] y = e z
( 3  32 + 2 ) y = e z
MAY 2011 
  MA 2161
1.
Transform the equation x 2 y  xy 
x into a linear differential equation with constant coefficients.
Refer Q. No.1 of December 2011  MA 2161
2.
Find the particular integral of (D2  4)y = cosh 2x 1
cosh 2 x
D 4
Particular Integral = 2
= 1  e 2 x  e 2 x

2 
D2  4 
1  e 2 x
1  e 2 x
= 2
  + 2


D  4  2
D 4 2
=  e 2 x
1  e 2 x
1
+ 

4  4  2 
(2) 2  4  2 
Prepared by : R.Ravikumar, Department of Mathematics, Pallavan College of
Engineering
8
1  e 
1  e 
+ x


2D  2 
2 D  2 
2 x
2x
= x
1  e 2 x
1  e 2 x
= x 
+ x
4  2 
4  2 
= x 2x
x
 e    e 2 x
8
8
JANUARY 2011 
  381101
1.
Solve ( D 2  3 D + 2 ) y = 0
The auxiliary equation is m2  3 m + 2 = 0  m = 1, 2
 the solution is y = A e x + B e2x 2.
Find the particular integral of ( D 2  2 ) y = e2x
Particular Integral = 1
( D  2)
2
e2 x
= 1
e2 x
(4  2)
= 1 2x
e
2
{ put D = 2 }
NOVEMBER 2010 MA 2161
1.
( x 2 D 2 + x D + 1 ) y = log x Reduce this into ODE with constant coefficients.
Put x = ez z = log x
xD = 
x2D2 =  (   1 ) in the given equation, we get ( 2 + 1 ) y = z
NOVEMBER 2010 MA 141
1.
Find the particular integral of ( D2  4 D + 4 ) y = 4 e3x
Prepared by : R.Ravikumar, Department of Mathematics, Pallavan College of
Engineering
9
Particular Integral = = 1
D  4 D  4)
2
4e3 x
1
4e3 x
(9  12  4)
{ put D = 3 }
4e3 x
= JUNE 2010 MA 2112
1.
Find the particular integral of ( D2 + 4 D + 5 ) y = e2x cosx
Particular Integral = = e
2.
2 x
1
e 2 x cos x
D  4D  5
2
1
cos x
( D  2)  4( D  2)  5
2
= e 2 x
1
cos x
D 1
= e 2 x
1
cos x
1  1
= e 2 x
x
cos x
2D
= e 2 x
x
sin x
2
2
{ put D2 = 1 }
Solve ( x 2 D 2 + 4x D + 2 ) y = 0
Put x = ez z = log x
xD = 
x2D2 =  (   1 ) in the given equation, we get ( 2 + 3  + 2 ) y = 0
The auxiliary equation is m2 + 3 m + 2 = 0  m = 1, 2
 the solution is y = A e z + B e2z y = Ax1 + B x2 MAY 2010 MA 2161
1.
x 2 y  + x y  = x. Transform this to ODE with constant coefficients.
Given equation is ( x2 D2 + xD )y = x
Prepared by : R.Ravikumar, Department of Mathematics, Pallavan College of
Engineering
10
Put x = ez z = log x
xD = 
x2D2 =  (   1 ) in the given equation, we get ( 2 ) y = ez
2.
Find the particular integral of ( D2 + 1 ) y = sinx
Particular Integral = 1
sin x
D 1
2
= 1
sin x
1  1
= x
sin x
2D
= 
{ put D2 = 1 }
x
cos x
2
APRIL 2010 MA 1101
1.
Solve dy
d3 y
d2 y
 3 + 3
 y = 0
3
2
dx
dx
dx
The auxiliary equation is m3  3m2 + 3 m  1 = 0  m = 1, 1, 1
 the solution is y = ( A x2 + Bx + c ) ex 2.
Reduce x2
dy
d2y
 x
+ y = log x into linear DE with constant coefficients.
2
dx
dx
Given equation is ( x2 D2  xD +1 )y = logx
Put x = ez z = log x
xD = 
x2D2 =  (   1 ) in the given equation, we get ( 2 2 + 1 ) y = z
APRIL 2010 MA 131
1.
Find the particular integral of ( D2 + 4 ) y = sin 2x
Particular Integral = = 1
sin 2 x
D 4
2
1
sin 2 x
4  4
Prepared by : R.Ravikumar, Department of Mathematics, Pallavan College of
Engineering
11
x
sin 2 x
2D
= = 
2.
x
cos 2 x
4
Solve y   5y + 6y = 0
The auxiliary equation is m2  5 m + 6 = 0  m = 2, 3
 the solution is y = A e2x + B e3x APRIL 2010 MA 1X01
1.
If y = 1 + et + sin t is the solution of dx
dy
 y  1  sin t ,
 x  cos t
dt
dt
find the solution x of the equation
Here x = cost  dy
dt
= cost  [  et + cos t ]
= et
APRIL 2010 MA 1102
1.
Find the particular integral of ( D  1 )2 y = x2
1
x2
2
( D  1)
Particular Integral = = 1
 1  D
= 2
 1  D
= 
2
x2
x2
1  2 D  3D 2  4 D 3 ......... x 2
= [ x2 + 4x + 6 ]
2.
Solve ( x 2 D 2 + 1 ) y = 0
Put x = ez z = log x
xD = 
x2D2 =  (   1 ) in the given equation, we get ( 2   + 1 ) y = 0
The auxiliary equation is m2  m + 1 = 0
 m = 1
3
i
2
2
Prepared by : R.Ravikumar, Department of Mathematics, Pallavan College of
Engineering
12
 the solution is y = e
1
z
2
{ A cos
3
3
z  B sin
z} where z = log x
2
2
APRIL 2010 MA 141
1.
Solve ( D 3 + 8 ) y = 0
The auxiliary equation is m3 + 8 = 0  m = 2, 1  i 3
 the solution is y = A e2x + ex { B cos 3 x + C sin 3 x } JANUARY 2010 MA 2112
1.
Find the particular integral of ( D + 1 )2 y = ex cosx
Particular Integral = 1
e  x cos x
( D  1) 2
= e
x
1
cos x
( D  1  1) 2
= e  x
1
cos x
D2
=  ex cosx
2.
Solve Dx  y = t; x + Dy = 1 where D = d/dt
Given equations are Dx  y = t ­­­­­­(1)
and
x + Dy = 1 ­­­­­(2)
Eliminating y between (1) and (2), we get ( D2 + 1 ) x = 2
The auxiliary equation is m2 + 1 = 0  m =  i
 the complementary function is x = A cost + B sint
Particular Integral = 1
2e0 x
( D  1)
= 2
1
2e0 x
(0  1)
{ put D = 0 }
= 2
 the solution is x = ( A cost + B sint ) + 2
From (1), y = Dx  t
= A sint + B cost  t
Prepared by : R.Ravikumar, Department of Mathematics, Pallavan College of
Engineering
13
Part B Questions
DECEMBER 2014 
  MA 2161
1.
Solve the differential equation y   a 2 y
2.
Solve the simultaneous differential equations 3.
Solve 
4.
Solve  x  1

D
3
2
 tan ax by variation of parameters method.
dx
dy
+ 2x  3y = t ;  3x + 2y = e 2t dt
dt
D 2  ( x  1) D  1 y  4 cos log( x  1)

 7 D  6  y  (1  x)e 2 x
DECEMBER 2014 
  MA 6251
1.
Solve ( D 2 + 2D + 5 ) y = e  x x2
2.
Solve 3.
Using the method of variation of parameters, solve 4.
Solve x D
2
2
 xD  1 y  sin(log x)
d2 y
+ 4 y =tan 2x
dx 2
dx
dy
+ y = e t ; x  = t dt
dt
JUNE 2014 
  MA 2161
1.
Solve ( D 2  4D + 3 ) y = cos2x + 2x2
2.
Solve d2 y
+ a2 y =tan ax using method of variation of parameters
2
dx
3.
Solve  x 2 D 2  xD  1 y  
4.
Solve the simultaneous equations  log x

 x 
2
dx
dy
+ 2y =  sint ;  2x = cos t dt
dt
Prepared by : R.Ravikumar, Department of Mathematics, Pallavan College of
Engineering
14
DECEMBER 2013 
  MA 2161
dy
d2 y
 2 + y = 8 xe x sin x
2
dx
dx
1.
Solve 2.
Solve by method of variation of parameters 2
3.
Solve x2
4.
Solve d2 y
+ 8y = tan 2x dx 2
dy
d2 y
log x
+ 4x + y = ee
2
dx
dx
dx
dy
+ 4x + 3y = t ; + 2x + 5y = e 2t dt
dt
JUNE 2013 
  MA 2161
dy
d2 y
e x
+ 2 + y = by the method of variation of parameters.
dx
x2
dx 2
1.
Solve 2.
Solve ( 3x + 2 )2 3.
Solve the simultaneous differential equations 4.
Solve x2
dy
d2 y
+ 3( 3x + 2 )  36 y = 3x2 + 4x + 1
2
dx
dx
dx
dy
+ 5x  2y = t ; + 2x + y = 0
dt
dt
dy
1
d2y
2
+ 4x
+ 2y = x  2
2
dx
x
dx
DECEMBER 2012 
  MA 2161
1.
Solve the equation ( D 2 + 5D + 4 ) y = e x sin2x
2.
Solve the equation 3.
Solve d2 y
+ y = cosec x by the method of variation of parameters.
dx 2
dx
dy
+ y = et, x  = t dt
dt
Prepared by : R.Ravikumar, Department of Mathematics, Pallavan College of
Engineering
15
4.
Solve the equation 1 dy
12 log x
d2 y
+ = 2
x dx
x2
dx
JUNE 2012 
  MA 2161
1.
Solve (D2+a2)y = sec ax using the method of variation of parameters
2.
Solve ( D 2  4D + 3 ) y = e x cos2x
3.
Solve ( x2 D2  x D + 4 ) y = x2 sin( log x ) 4.
Solve dx
dy
+ 2y = sin 2t,  2x = cos 2t dt
dt
DECEMBER 2011 
  MA 2161
1.
Solve the equation ( D 2  3D + 2 ) y = 2 cos(2x+3) + 2 ex
2.
Apply the method of variation of parameters to solve ( D 2 + 4 ) y = cot x
3.
Solve ( 1 + x )2 4.
Solve dy
d2 y
+ ( 1 + x ) + y = 4 cos[ log(1+x) ]
2
dx
dx
dx
dy
 y = t, + x = t2 given x(0) = y(0) = 2.
dt
dt
DECEMBER 2011 
  181202
1.
Solve the equation ( D 2 + 4D + 3 ) y = 6 e2x sin x sin 2x
2.
Using variation of parameters, solve ( 2D2  D  3 ) y = 25 e 2x
3.
Solve x2
4.
Solve dy
d2 y
 3x + 4y = x2 lnx
2
dx
dx
dx
dy
 3y + 2x = t,  3x + 2y = e2t
dt
dt
JUNE 2011 
  181202
1.
Solve ( D 2 + 3D + 2 ) y = sin x + x2 2.
Solve for x from the equations D2x + y = 3 e2t, Dx  Dy = 3 e2t 3.
Solve, by method of variation of parameters, the equation d2 y
+ a2 y = tan ax
2
dx
Prepared by : R.Ravikumar, Department of Mathematics, Pallavan College of
Engineering
16
MAY 2011 
  MA 2161
1.
Solve the equation 2.
Solve the equation 3.
Solve ( 1 + x )2 4.
Solve ( D 2 + 5D + 4 ) y = ex sin 2x d2 y
+ y = cosec x by the method of variation of parameters.
dx 2
dy
d2 y
+ ( 1 + x ) + y = 2 sin[ log(1+x) ] 2
dx
dx
dx
dy
 y = t and + x = t2
dt
dt
JANUARY 2011 381101
1.
Solve the equation ( D 2  4D + 3 ) y = sin 3x + x2
2.
Solve 3.
Solve x
4.
Solve ( D 2 + 5D + 4 ) y = ex sin 2x + 2
dx
dy
= 3y  2x + 5t, = 3x  2y + 2e2t dt
dt
2
y  4 xy  2 y  x 2 
1
x2
NOVEMBER 2010 MA 2161
1.
Solve ( D2 + 16 ) y = cos3 x
2.
Solve by method of variation of parameters, 3.
Solve ( x2 D2  3x D + 4 ) y = x2 cos( log x ) 4.
Solve d2 y
+ 4y = sec 2x
dx 2
dx
dy
+ 2y =  sin t,  2x = cos t given x = 1 and y = 0 at t = 0.
dt
dt
MAY 2010 MA 2161
1.
Solve the equation ( D 2 + 4D + 3 ) y = e x sin x
2.
Solve the equation ( D 2 + 1 ) y = x sin x by the method of variation of parameters
3.
Solve ( x2 D2  2x D  4 ) y = x2 + 2 log x 4.
Solve dx
dy
+ 2x + 3y = 2e2t, + 3x + 2y = 0
dt
dt
Prepared by : R.Ravikumar, Department of Mathematics, Pallavan College of
Engineering
17
MAY 2010 MA 1101
1.
Solve dy
d2 y
+ 4 + 3y = e x sin x + x e3x
2
dx
dx
2.
Solve dx
dy
+ 2x  3y = 5t,  3x + 2y = 2e2t
dt
dt
3.
Solve x2 4.
Solve by method of variation of parameters dy
d2 y
+ 2 x
 12y = x3 log x
2
dx
dx
dy
d2 y
­ 2 + 2y = e x sin x
2
dx
dx
MAY 2010 MA 131
1.
Solve ( D 2  5D + 6 ) y = e x cos 2x
2.
Solve ( x2 D 2  3xD + 4 ) y = x2 sin ( log x )
3.
Solve y  
4.
y = tanx by the method of variation of parameters
dx
dy
Solve + y = sin t, + x = cos t dt
dt
NOVEMBER 2009 MA 2161
1.
Solve the equation ( D 2  3D + 2 ) y = 2 cos (2x+3) + 2ex
2.
Apply the method of variation of parameters to solve ( D 2 + 4 ) y = cot 2x
3.
Solve the differential equation ( x2 D2  x D + 4 ) y = x2 sin ( log x ) 4.
Solve dx
dy
+ 2y = sin 2t,  2x = cos 2t given x = 1 and y = 0 at t = 0.
dt
dt
NOVEMBER 2009 MA 131
dy
d2 y
+ 4 + 4y = e  2x + e 3x sin x 2
dx
dx
1.
Solve 2.
Solve 3.
Solve ( D2  1 ) y = x2 + e 2x sin 3x 4.
Solve d2 y
+ y = sec x by the method of variation of parameters
dx 2
dx
dy
+ 2x + y = 0, + 5x  2y = t dt
dt
Prepared by : R.Ravikumar, Department of Mathematics, Pallavan College of
Engineering
18
NOVEMBER 2009 MA 1101
dx
dy
dy
  y = et, + x  y = e 2t dt
dt
dt
1.
Solve 2.
Solve x2 3.
Solve ( D2 + 4 )y = cos2 x + x2 + 1
4.
Using the method of variation of parameters, solve dy
d2 y
+ 8 x
+ 13y = log x
2
dx
dx
d2 y
+ y = cosec x
dx 2
MAY 2009 MA 2161
1.
solve ( D 2 + 4 ) y = x2 cos 2x
2.
Solve ( D 2 + a2 ) y = tan ax by the method of variation of parameters
3.
Solve the equation ( x2 D2 + x D + 5 ) y = x cos ( log x ) 4.
Solve dx
dy
+ y = sin t, + x = cos t given x = 2 and y = 0 at t = 0.
dt
dt
MAY 2009 MA 1101
dx
dy
+ y = et, x  = t dt
dt
1.
Solve 2.
Solve x2 3.
Solve the equation ( 3D 2 + D  14 ) y = cos 4x + 8e2x
4.
Solve by using method of variation of parameters UNIT
dy
d2 y
 3 x
 5y = sin ( log x )
2
dx
dx
No. of Question
Papers Verified
d2 y
+ 4y = tan x
dx 2
Questions Repeatedly Asked
Differential Equation with Variable Differential
Equation
Coefficient – 8 Marks
No. of Times
Asked
17
19
Method of Variation of Parameters – 8 Marks
18
Prepared by : R.Ravikumar, Department of Mathematics, Pallavan College of
Engineering
19
Prepared by : R.Ravikumar, Department of Mathematics, Pallavan College of
Engineering