MCQ Unit I-1
MCQ Unit I-1
Differential Equations
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                              𝑑2𝑦
((Q))1_// The solution of D.E. 𝑑𝑥 2 + 9𝑦 = 0 is
((A)) 𝑐1 cos2𝑥 + 𝑐2 sin2𝑥
((B)) (𝑐1 𝑥 + 𝑐2 )𝑒 −3𝑥
((C)) 𝑐1 𝑒 3𝑥 + 𝑐2 𝑒 −3𝑥
((D)) 𝑐1 cos3𝑥 + 𝑐2 sin3𝑥
((E))D
((F))
                               𝑑2𝑦   𝑑𝑦
((Q))1_// The solution of D.E. 𝑑𝑥 2 − 𝑑𝑥 − 2𝑦 = 0 is
((A)) 𝑐1 𝑒 2𝑥 + 𝑐2 𝑒 𝑥
((B)) 𝑐1 𝑒 2𝑥 + 𝑐2 𝑒 −𝑥
((C)) 𝑐1 𝑒 −2𝑥 + 𝑐2 𝑒 𝑥
((D)) 𝑐1 𝑒 −2𝑥 + 𝑐2 𝑒 −𝑥
((E))B
((F))
((Q))1_// If 𝑚1 = 𝛼 + 𝑖𝛽 ((A))nd 𝑚2 = 𝛼 − 𝑖𝛽 are two complex roots of auxiliary equation of second order
DE 𝜙(𝐷)𝑦 = 0 then it’s solution is
((A)) 𝑒 𝛽𝑥 [𝑐1 cos𝛼𝑥 + 𝑐2 sin𝛼𝑥]
((B)) 𝑒 𝛼𝑥 [(𝑐1 𝑥 + 𝑐2 )cos𝛽𝑥 + (𝑐3 𝑥 + 𝑐4 )sin𝛽𝑥]
((C)) 𝑐1 𝑒 𝛼𝑥 + 𝑐2 𝑒 𝛽𝑥
((D)) 𝑒 𝛼𝑥 [𝑐1 cos𝛽𝑥 + 𝑐2 sin𝛽𝑥]
((E))D
((F))
                              𝑑2𝑦      𝑑𝑦
((Q))1_// The solution of D.E.𝑑𝑥 2 − 5 𝑑𝑥 + 6𝑦 = 0 is
((A)) 𝑐1 𝑒 2𝑥 + 𝑐2 𝑒 −3𝑥
((B)) 𝑐1 𝑒 −2𝑥 + 𝑐2 𝑒 3𝑥
((C)) 𝑐1 𝑒 −2𝑥 + 𝑐2 𝑒 −3𝑥
((D)) 𝑐1 𝑒 2𝑥 + 𝑐2 𝑒 3𝑥
((E))D
((F))
                                𝑑2𝑦       𝑑𝑦
((Q))1_// The solution of DE           − 5 𝑑𝑥 − 6𝑦 = 0 is
                                𝑑𝑥 2
((A)) 𝑐1 𝑒 −𝑥 + 𝑐2 𝑒 6𝑥
((B)) 𝑐1 𝑒 −2𝑥 + 𝑐2 𝑒 − 3𝑥
((C)) 𝑐1 𝑒 3𝑥 + 𝑐2 𝑒 2𝑥
((D)) 𝑐1 𝑒 −3𝑥 + 𝑐2 𝑒 − 2𝑥
((E))A
((F))
                                  𝑑2𝑦      𝑑𝑦
((Q))1_// The solution of D.E. 2 𝑑𝑥 2 − 𝑑𝑥 − 10𝑦 = 0 is
                     5
((A)) 𝑐1 𝑒 −2𝑥 + 𝑐2 𝑒 ( ⁄2)𝑥
                          5
((B)) 𝑐1 𝑒 −2𝑥 + 𝑐2 𝑒 −( ⁄2)𝑥
                      5
((C)) 𝑐1 𝑒 2𝑥 + 𝑐2 𝑒 ( ⁄2)𝑥
                        3
((D)) 𝑐1 𝑒 −2𝑥 + 𝑐2 𝑒 ( ⁄2)𝑥
((E))A
((F))
                                𝑑2𝑦
((Q))1_// The solution of D.E.𝑑𝑥 2 − 4𝑦 = 0 is
((A)) (𝑐1 𝑥 + 𝑐2 )𝑒 2𝑥
((B)) 𝑐1 𝑒 4𝑥 + 𝑐2 𝑒 −4𝑥
((C)) 𝑐1 𝑒 2𝑥 + 𝑐2 𝑒 −2𝑥
((D)) 𝑐1 cos2𝑥 + 𝑐2 sin2𝑥
((E))C
((F))
                                𝑑2𝑦        𝑑𝑦
((Q))1_// The solution of D.E.𝑑𝑥 2 + 2 𝑑𝑥 + 𝑦 = 0 is
((A)) 𝑐1 𝑒 2𝑥 + 𝑐2 𝑒 𝑥
((B)) 𝑐1 𝑒 𝑥 + 𝑐2 𝑒 −𝑥
((C)) (𝑐1 𝑥 + 𝑐2 )𝑒 −𝑥
((D)) (𝑐1 𝑥 + 𝑐2 )𝑒 𝑥
((E))C
((F))
                                𝑑2𝑦        𝑑𝑦
((Q))1_// The solution of D.E.𝑑𝑥 2 − 4 𝑑𝑥 + 4𝑦 = 0 is
((A)) (𝑐1 𝑥 + 𝑐2 )𝑒 2𝑥
((B)) (𝑐1 𝑥 + 𝑐2 )𝑒 −2𝑥
((C)) 𝑐1 𝑒 4𝑥 + 𝑐2 𝑒 −4𝑥
((D)) 𝑐1 𝑒 2𝑥 + 𝑐2 𝑒 −2𝑥
((E))A
((F))
                              𝑑3𝑦     𝑑𝑦
((Q))1_// The solution of D.E.𝑑𝑥 3 − 4 𝑑𝑥 = 0 is
((A)) 𝑐1 𝑒 2𝑥 + 𝑐2 𝑒 −2𝑥
((B)) 𝑐1 + 𝑐2 cos2𝑥 + 𝑐3 sin2𝑥
((C)) 𝑐1 𝑒 𝑥 + 𝑐2 𝑒 −2𝑥 + 𝑐3 𝑒 −3𝑥
((D)) 𝑐1 + 𝑐2 𝑒 2𝑥 + 𝑐3 𝑒 −2𝑥
((E))D
((F))
((Q))1_// The roots 𝑚1 , 𝑚2 , 𝑚3 ⋯ ⋯ , 𝑚𝑛 of auxiliary equation 𝜙(𝐷) = 0 are real. If three of these roots are
repeated, s((A))y, 𝑚1 = 𝑚2 = 𝑚3 and the remaining roots 𝑚4 , 𝑚5 ⋯ ⋯ 𝑚𝑛 are distinct then solution of
𝜙(𝐷)𝑦 = 0 is
((A)) 𝑐1 𝑒 𝑚1𝑥 + 𝑐2 𝑒 𝑚2𝑥 + ⋯ ⋯ + 𝑐𝑛 𝑒 𝑚𝑛𝑥
((B)) (𝑐1 𝑥 2 + 𝑐2 𝑥 + 𝑐3 )𝑒 𝑚1𝑥 + 𝑐4 𝑒 𝑚4𝑥 + ⋯ ⋯ + 𝑐𝑛 𝑒 𝑚𝑛𝑥
((C)) (𝑐1 𝑥 2 + 𝑐2 𝑥 + 𝑐3 )cos𝑚1 𝑥 + 𝑐4 cos𝑚4 𝑥 + ⋯ ⋯ + 𝑐𝑛 cos𝑚𝑛 𝑥
((D)) (𝑐1 𝑥 2 + 𝑐2 𝑥 + 𝑐3 )sin𝑚1 𝑥 + 𝑐4 sin𝑚4 𝑥 + ⋯ ⋯ + 𝑐𝑛 sin𝑚𝑛 𝑥
((E))B
((F))
                               𝑑2𝑦      𝑑𝑦
 ((Q))1_// The solution of D.E.𝑑𝑥 2 + 6 𝑑𝑥 + 9𝑦 = 0 is
((A)) 𝑐1 𝑒 −6𝑥 + 𝑐2 𝑒 −9𝑥
((B)) (𝑐1 𝑥 + 𝑐2 )𝑒 −3𝑥
((C)) (𝑐1 𝑥 + 𝑐2 )𝑒 3𝑥
((D)) 𝑐1 𝑒 3𝑥 + 𝑐2 𝑒 2𝑥
((E))B
((F))
                               𝑑2𝑦
((Q))1_// The solution of D.E.𝑑𝑥 2 + 𝑦 = 0 is
((A)) 𝑐1 𝑒 𝑥 + 𝑐2 𝑒 −𝑥
((B)) 𝑐1 cos𝑥 + 𝑐2 sin𝑥
((C)) (𝑐1 𝑥 + 𝑐2 )𝑒 −𝑥
((D)) 𝑒 𝑥 (𝑐1 cos𝑥 + 𝑐2 sin𝑥)
((E))B
((F))
                              𝑑2𝑦     𝑑𝑦
((Q))1_// The solution of D.E.𝑑𝑥 2 + 6 𝑑𝑥 + 10𝑦 = 0 is
((A)) 𝑒 𝑥 (𝑐1 cos𝑥 + 𝑐2 sin𝑥)
((B)) 𝑒 𝑥 (𝑐1 cos3𝑥 + 𝑐2 sin3𝑥)
((C)) 𝑐1 𝑒 5𝑥 + 𝑐2 𝑒 2𝑥
((D)) 𝑒 −3𝑥 (𝑐1cos𝑥 + 𝑐2 sin𝑥)
((E))D
((F))
                                     𝑑2𝑦    𝑑𝑦
 ((Q))1_// The solution of D.E.𝑑𝑥 2 + 𝑑𝑥 + 𝑦 = 0 is
 ((A)) 𝑒 𝑥 (𝑐1 cos𝑥 + 𝑐2 sin𝑥)
         x⁄           3               3
 ((B)) 𝑒   2 [𝑐1 cos ( ) 𝑥 + 𝑐2 sin ( ) 𝑥]
                      2               2
         −𝑥⁄2           √3              √3
 ((C)) 𝑒      [𝑐1 cos ( 2 ) 𝑥 + 𝑐2 sin ( 2 ) 𝑥]
 ((D)) 𝑐1 𝑒 𝑥 + 𝑐2 𝑒 −𝑥
 ((E))C
 ((F))
                                𝑑3𝑦        𝑑2𝑦    𝑑𝑦
((Q))1_// A solution of D.E. 𝑑𝑥 3 − 2 𝑑𝑥 2 + 4 𝑑𝑥 − 8y = 0 is
((A)) 𝑦 = 𝑐1 𝑒 𝑥 + 𝑐2 cos(2x) + 𝑐3 sin(2x)
((B)) 𝑦 = 𝑐1 𝑒 −2x + 𝑐2 cos(2x) + 𝑐3 sin(2x)
((C)) 𝑦 = (𝑐1 + 𝑐2 𝑥)𝑒 2x + 𝑐3 𝑒 −2x
((D)) 𝑦 = (𝑐1 + 𝑐2 𝑥)𝑒 −2x + 𝑐3 𝑒 2x
((E))A
((F))
                                𝑑3𝑦        𝑑𝑦
((Q))1_// A solution of D.E. 𝑑𝑥 3 − 2 𝑑𝑥 + 4y = 0 is
((A)) 𝑦 = 𝑐1 𝑒 2x + 𝑒 𝑥 [𝑐2 cos(𝑥) + 𝑐3 sin(𝑥)]
((B)) 𝑦 = 𝑐1 𝑒 2x + 𝑒 −𝑥 [𝑐2 cos(𝑥) + 𝑐3 sin(𝑥)]
((C)) 𝑦 = 𝑐1 𝑒 2x + 𝑐2 𝑒 −2x + 𝑐3 𝑒 𝑥
((D)) 𝑦 = 𝑐1 𝑒 2x + 𝑐2 𝑒 −𝑥 + 𝑐3 𝑒 𝑥
((E))B
((F))
                               𝑑4𝑦         𝑑2𝑦
((Q))1_// A solution of D.E. 𝑑𝑥 4 + 5 𝑑𝑥 2 + 4y = 0 is
 ((A)) 𝑦 = 𝑐1 cos(𝑥) + 𝑐2 sin(𝑥) + 𝑐3 𝑒 2x + 𝑐4 𝑒 −2x
 ((B)) 𝑦 = 𝑐1 𝑒 𝑥 + 𝑐2 𝑒 −𝑥 + 𝑐3 𝑒 2x + 𝑐4 𝑒 −2x
 ((C)) 𝑦 = 𝑐1 cos(𝑥) + 𝑐2 sin(𝑥) + 𝑐3 cos(2x) + 𝑐4 sin(2x)
 ((D)) 𝑦 = 𝑐1 𝑒 𝑥 + 𝑐2 𝑒 −𝑥 + 𝑐3 cos(2x) + 𝑐4 sin(2x)
 ((E))C
 ((F))
                                𝑑3𝑦
((Q))1_// A solution of D.E. 𝑑𝑥 3 + 𝑦 = 0 is
((A)) 𝑦 = 𝑐1 𝑒 −𝑥 + 𝑒 𝑥 [𝑐2 cos(√3𝑥) + 𝑐3 sin(√3𝑥)]
((B)) 𝑦 = 𝑐1 𝑒 −𝑥 + 𝑐2 𝑒 𝑥/2 + 𝑐3 𝑒 √3𝑥/2
                            √3𝑥            √3𝑥
((C)) 𝑦 = 𝑐1 𝑒 −𝑥 + 𝑒 −𝑥/2 [𝑐2 cos(
                                ) + 𝑐3 sin( )]
                              2             2
                −𝑥  𝑥/2     √3𝑥            √3𝑥
((D)) 𝑦 =   𝑐1 𝑒 + 𝑒 [𝑐2cos( 2 ) + 𝑐3 sin( 2 )]
((E))D
((F))
                               𝑑3𝑦         𝑑𝑦
((Q))1_// A solution of D.E.4 𝑑𝑥 3 − 7 𝑑𝑥 + 3y = 0 is
((A)) 𝑦 = 𝑐1 𝑒 2x + 𝑐2 𝑒 𝑥/2 + 𝑐3 𝑒 −𝑥/2
((B)) 𝑦 = 𝑐1 𝑒 𝑥 + 𝑐2 𝑒 𝑥/2 + 𝑐3 𝑒 −3x/2
((C)) 𝑦 = 𝑐1 + 𝑐2 𝑒 −𝑥 + 𝑐3 𝑒 −3x/4
((D)) 𝑦 = 𝑐1 𝑒 −𝑥 + 𝑐2 𝑒 𝑥/2 + 𝑐3 𝑒 −3x/2
((E))B
((F))
((Q))1_// If the roots 𝑚1 , 𝑚2 , 𝑚3 ⋯ ⋯ , 𝑚𝑛 of Auxiliary equation 𝜙(𝐷) = 0 Are real And distinct, then solution
of 𝜙(𝐷)𝑦 = 0 is
((A)) 𝑐1 𝑒 𝑚1𝑥 + 𝑐2 𝑒 𝑚2𝑥 + ⋯ ⋯ + 𝑐𝑛 𝑒 𝑚𝑛𝑥
((B)) 𝑐1 cos𝑚1 𝑥 + 𝑐2 cos𝑚2 𝑥 + ⋯ ⋯ + 𝑐𝑛 cos𝑚𝑛 𝑥
((C)) 𝑚1 𝑒 𝑐1𝑥 + 𝑚2 𝑒 𝑐2𝑥 + ⋯ ⋯ + 𝑚𝑛 𝑒 𝑐𝑛𝑥
((D)) 𝑐1 sin𝑚1 𝑥 + 𝑐2 sin𝑚2 𝑥 + ⋯ ⋯ + 𝑐𝑛 sin𝑚𝑛 𝑥
((E))A
((F))
                             𝑑3𝑦      𝑑𝑦
((Q))1_// A solution of D.E. 𝑑𝑥 3 − 3 𝑑𝑥 + 2y = 0 is
((A)) 𝑦 = (𝑐1 + 𝑐2 𝑥)𝑒 𝑥 + 𝑐3 𝑒 −2x
((B)) 𝑦 = (𝑐1 + 𝑐2 𝑥)𝑒 −𝑥 + 𝑐3 𝑒 2x
((C)) 𝑦 = (𝑐1 + 𝑐2 𝑥)𝑒 2x + 𝑐3 𝑒 −2x
((D)) 𝑦 = 𝑐1 𝑒 𝑥 + 𝑐2 𝑒 2x + 𝑐3 𝑒 3x
((E))A
((F))
                             𝑑3𝑦     𝑑2𝑦        𝑑𝑦
((Q))1_// A solution of D.E. 𝑑𝑥 3 − 𝑑𝑥 2 + 3 𝑑𝑥 + 5y = 0 is
((A)) 𝑦 = 𝑐1 𝑒 −𝑥 + 𝑒 𝑥 [𝑐2 cos(2x) + 𝑐3 sin(2x)]
((B)) 𝑦 = 𝑐1 𝑒 𝑥 + 𝑒 −𝑥 [𝑐2 cos(2x) + 𝑐3 sin(2x)]
((C)) 𝑦 = 𝑐1 𝑒 −𝑥 + 𝑒 −𝑥 [𝑐2 cos(2x) + 𝑐3 sin(2x)]
((D)) 𝑦 = 𝑐1 𝑒 𝑥 + 𝑒 𝑥 [𝑐2 cos(2x) + 𝑐3 sin(2x)]
((E))A
((F))
                             𝑑3𝑦      𝑑2𝑦        𝑑𝑦
((Q))2_// A solution of D.E. 𝑑𝑥 3 − 5 𝑑𝑥 2 + 7 𝑑𝑥 − 3y = 0 is
((A)) 𝑦 = 𝑐1 𝑒 𝑥 + 𝑐2 cos(2x) + 𝑐3 sin(2x)
((B)) 𝑦 = (𝑐1 + 𝑐2 𝑥)𝑒 𝑥 + 𝑐3 𝑒 3x
((C)) 𝑦 = (𝑐1 + 𝑐2 𝑥)𝑒 −𝑥 + 𝑐3 𝑒 −3x
((D)) 𝑦 = (𝑐1 + 𝑐2 𝑥)𝑒 −3x + 𝑐3 𝑒 𝑥
((E)) B
((F))
                             𝑑3𝑦      𝑑2𝑦             𝑑𝑦
((Q))2_// A solution of D.E. 𝑑𝑥 3 − 6 𝑑𝑥 2 + 12 𝑑𝑥 − 8y = 0 is
((A)) 𝑦 = (𝑐1 + 𝑐2 𝑥 + 𝑐3 𝑥 2 )𝑒 2x
((B)) 𝑦 = 𝑐1 𝑒 −2x + 𝑐2 cos(2x) + 𝑐3 sin(2x)
((C)) 𝑦 = (𝑐1 + 𝑐2 𝑥)𝑒 2x + 𝑐3 𝑒 −2x
((D)) 𝑦 = (𝑐1 + 𝑐2 𝑥 + 𝑐3 𝑥 2 )𝑒 −2x
((E)) A
((F))
                             𝑑4𝑦          𝑑2𝑦
((Q))2_// A solution of D.E. 𝑑𝑥 4 + 18 𝑑𝑥 2 + 81y = 0 is
((A)) 𝑦 = 𝑐1 𝑒 3x + 𝑐2 𝑒 −3x + 𝑐3 cos(3x) + 𝑐4 sin(3x)
((B)) 𝑦 = (𝑐1 + 𝑐2 𝑥)𝑒 −3x + 𝑐3 cos(2x) + 𝑐4 sin(2x)
((C)) 𝑦 = (𝑐1 + 𝑐2 𝑥)𝑒 3x + (𝑐3 + 𝑐4 𝑥)𝑒 −3x
((D)) 𝑦 = (𝑐1 + 𝑐2 𝑥)cos(3x) + (𝑐3 + 𝑐4 𝑥)sin(3x)
((E)) D
((F))
                             𝑑3𝑦          𝑑𝑦
((Q))2_// A solution of D.E. 𝑑𝑥 3 − 12 𝑑𝑥 + 16y = 0 is
((A)) 𝑦 = 𝑐1 𝑒 4x + 𝑐2 cos(2x) + 𝑐3 sin(2x)
((B)) 𝑦 = 𝑐1 𝑒 −2x + 𝑐2 cos(2x) + 𝑐3 sin(2x)
((C)) 𝑦 = (𝑐1 + 𝑐2 𝑥)𝑒 2x + 𝑐3 𝑒 −4x
((D)) 𝑦 = (𝑐1 + 𝑐2 𝑥)𝑒 −4x + 𝑐3 𝑒 2x
((E)) C
((F))
                            𝑑3𝑦      𝑑2𝑦             𝑑𝑦
((Q))1_// Solution of D.E.𝑑𝑥 3 + 6 𝑑𝑥 2 + 11 𝑑𝑥 + 6𝑦 = 0 is
((A)) 𝑐1 𝑒 𝑥 + 𝑐2 𝑒 2𝑥 + 𝑐3 𝑒 3𝑥
((B)) 𝑐1 𝑒 −𝑥 + 𝑐2 𝑒 2𝑥 + 𝑐3 𝑒 −3𝑥
((C)) 𝑐1 𝑒 −𝑥 + 𝑐2 𝑒 −2𝑥 + 𝑐3 𝑒 −3𝑥
((D)) 𝑐1 𝑒 𝑥 + 𝑐2 𝑒 −2𝑥 + 𝑐3 𝑒 3𝑥
((E)) C
((F))
                               𝑑3𝑦          𝑑𝑦
((Q))1_// Solution of D.E.𝑑𝑥 3 − 7 𝑑𝑥 − 6𝑦 = 0 is
((A)) 𝑐1 𝑒 𝑥 + 𝑐2 𝑒 2𝑥 + 𝑐3 𝑒 3𝑥
((B)) 𝑐1 𝑒 −𝑥 + 𝑐2 𝑒 −2𝑥 + 𝑐3 𝑒 6𝑥
((C)) 𝑐1 𝑒 −𝑥 + 𝑐2 𝑒 2𝑥 + 𝑐3 𝑒 𝑥
((D)) 𝑐1 𝑒 −𝑥 + 𝑐2 𝑒 −2𝑥 + 𝑐3 𝑒 3𝑥
((E)) D
((F))
                               𝑑3𝑦          𝑑2𝑦    𝑑𝑦
((Q))1_// Solution of D.E.𝑑𝑥 3 + 2 𝑑𝑥 2 + 𝑑𝑥 = 0 is
((A)) 𝑐1 + 𝑐2 𝑒 𝑥 + 𝑐3 𝑒 −𝑥
((B)) 𝑐1 + 𝑒 𝑥 (𝑐2 𝑥 + 𝑐3 )
((C)) 𝑒 −𝑥 (𝑐2 𝑥 + 𝑐3 )
((D)) 𝑐1 + 𝑒 −𝑥 (𝑐2 𝑥 + 𝑐3 )
((E)) D
((F))
                               𝑑3𝑦          𝑑2𝑦        𝑑𝑦
((Q))1_// Solution of D.E.𝑑𝑥 3 − 5 𝑑𝑥 2 + 8 𝑑𝑥 − 4𝑦 = 0 is
((A)) 𝑐1 𝑒 𝑥 + (𝑐2 𝑥 + 𝑐3 )𝑒 2𝑥
((B)) 𝑐1 𝑒 𝑥 + 𝑐2 𝑒 2𝑥 + 𝑐3 𝑒 3𝑥
((C)) (𝑐2 𝑥 + 𝑐3 )𝑒 2𝑥
((D)) 𝑐1 𝑒 −𝑥 + (𝑐2 𝑥 + 𝑐3 )𝑒 −2𝑥
((E)) A
((F))
                                      𝑑3𝑦
((Q))2_// The solution of D.E.𝑑𝑥 3 + 𝑦 = 0 is
                             √3                √3
((A)) 𝑐1 𝑒 𝑥 + 𝑒 𝑥 (𝑐2 cos      𝑥   + 𝑐3 sin      𝑥)
                             2                 2
                  1
                                1               1
((B)) 𝑐1 𝑒 − 𝑥 + 𝑒 2 𝑥 (𝑐2 cos 2 𝑥 + 𝑐3 sin 2 𝑥)
                 1
                               √3                 √3
((C)) 𝑐1 𝑒 − 𝑥 + 𝑒 2 𝑥 (𝑐2 cos 2 𝑥 + 𝑐3 sin        2
                                                     𝑥)
((D)) (𝑐1 + 𝑐2 𝑥 + 𝑐3 𝑥 2 )𝑒 − 𝑥
((E)) C
((F))
                                      𝑑3𝑦         𝑑𝑦
((Q))1_// The solution of D.E.𝑑𝑥 3 + 3 𝑑𝑥 = 0 is
((A)) 𝑐1 + 𝑐2 cos𝑥 + 𝑐3 sin𝑥
((B)) 𝑐1 cos𝑥 + 𝑐2 sin𝑥
((C)) 𝑐1 + 𝑐2 𝑒 √3 𝑥 + 𝑐3 𝑒 −√3 𝑥
((D)) 𝑐1 + 𝑐2 cos√3𝑥 + 𝑐3 sin√3𝑥
((E)) D
((F))
                           𝑑3𝑦    𝑑2𝑦    𝑑𝑦
((Q))1_// Solution of D.E.𝑑𝑥 3 + 𝑑𝑥 2 − 2 𝑑𝑥 + 12𝑦 = 0 is
((A)) 𝑐1 𝑒 −3𝑥 + 𝑒 𝑥 (𝑐2 cos√3𝑥 + 𝑐3 sin√3𝑥)
((B)) 𝑐1 𝑒 −3𝑥 + (𝑐2 cos3𝑥 + 𝑐3 sin3𝑥)
((C)) 𝑐1 𝑒 3𝑥 + 𝑒 −𝑥 (𝑐2cos√3𝑥 + 𝑐3 sin√3𝑥)
((D)) 𝑐1 𝑒 −𝑥 + 𝑐2 𝑒 −√3 𝑥 + 𝑐3 𝑒 √3 𝑥
((E)) A
((F))
                                 𝑑4𝑦
((Q))2_// The solution of D.E.𝑑𝑥 4 − 𝑦 = 0 is
((A)) (𝑐1 𝑥 + 𝑐2 )𝑒 −𝑥 + 𝑐3 cos𝑥 + 𝑐4 sin𝑥
((B)) (𝑐1 𝑥 + 𝑐2 )cos𝑥 + (𝑐3 𝑥 + 𝑐4 )sin𝑥
((C)) 𝑐1 𝑒 𝑥 + 𝑐2 𝑒 − 𝑥 + 𝑐3 cos𝑥 + 𝑐4 sin𝑥
((D)) (𝑐1 + 𝑐2 𝑥 + 𝑐3 𝑥 2 + 𝑐4 𝑥 3 )𝑒 𝑥
((E)) C
((F))
                  1                      𝑑
((Q))1_// 𝐷+𝑚 𝑓(𝑥) , where 𝐷 ≡ 𝑑𝑥 and 𝑚 is constant, is equal to
((A)) 𝑒 −𝑚𝑥 ∫ 𝑒 𝑚𝑥 𝑑𝑥
((B)) ∫ 𝑒 𝑚𝑥 𝑓(𝑥) 𝑑𝑥
((C)) 𝑒 𝑚𝑥 ∫ 𝑒 −𝑚𝑥 𝑓(𝑥) 𝑑𝑥
((D)) 𝑒 −𝑚𝑥 ∫ 𝑒 𝑚𝑥 𝑓(𝑥) 𝑑𝑥
((E)) D
((F))
                                 1
((Q))1_//Particular Integral of 𝜙(𝐷2) sin(𝑎𝑥 + 𝑏),𝜙(−𝑎2 ) = 0 , 𝜙 ′ (−𝑎2 ) ≠ 0is
              1
((A)) 𝑥 𝜙′ (−𝑎2) cos(𝑎𝑥 + 𝑏)
              1
((B)) 𝑥 𝜙′(−𝑎2) sin(𝑎𝑥 + 𝑏)
          1
((C)) 𝜙(−𝑎2) sin(𝑎𝑥 + 𝑏)
          1
((D)) 𝜙′(−𝑎2) sin(𝑎𝑥 + 𝑏)
((E)) B
((F))
                                     1
((Q))1_// Particular Integral of 𝜙(𝐷) 𝑒 𝑎 𝑥 𝑉 where 𝑉 is any function of 𝑥 is
                  1
((A)) 𝑒 𝑎𝑥 𝜙(𝐷−𝑎) 𝑉
              1
((B)) 𝑒 𝑎𝑥 𝜙(𝑎) 𝑉
                  1
((C)) 𝑒 𝑎 𝑥            𝑉
              𝜙(𝐷+𝑎)
          1
((D)) 𝜙(𝐷+𝑎) 𝑉
((E)) C
((F))
((Q))2_// Particular Integral of D.E.(𝐷3 + 9𝐷)𝑦 = sin3𝑥 is
        𝑥
((A)) − 18 cos3𝑥
             𝑥
((B)) − 18 sin3𝑥
((C)) −𝑥sin3𝑥
         1
((D)) − 18 sin3𝑥
((E)) B
((F))
                                     1
((Q))1_// Particular Integral of (𝐷−𝑎)𝑟 𝑒 𝑎 𝑥 is
        1
((A)) 𝑟! 𝑒 𝑎𝑥
        𝑥𝑟
((B))        𝑒 𝑎𝑥
        𝑟
        𝑥𝑟
((C)) 𝑟! 𝑒 𝑎𝑥
((D)) 𝑥 𝑟 𝑒 𝑎𝑥
((E)) C
((F))
                                    1
((Q))1_// Particular Integral of 𝜙(𝐷2) sin(𝑎𝑥 + 𝑏), 𝜙(−𝑎2 ) ≠ 0 is
             1
((A)) 𝜙(−𝑎2) cos(𝑎𝑥 + 𝑏)
             1
((B)) 𝜙(−𝑎2) sin(𝑎𝑥 + 𝑏)
                 1
((C)) 𝑥 𝜙′(−𝑎2) sin(𝑎𝑥 + 𝑏)
             1
((D)) 𝜙(𝑎2) sin(𝑎𝑥 + 𝑏)
((E)) B
((F))
                                    1
((Q))1_// Particular Integral of 𝜙(𝐷2) cos(𝑎𝑥 + 𝑏), 𝜙(−𝑎2 ) ≠ 0 is
             1
((A)) 𝜙(−𝑎2) cos(𝑎𝑥 + 𝑏)
             1
((B))                sin(𝑎𝑥 + 𝑏)
        𝜙(−𝑎2 )
            1
((C)) 𝑥 𝜙′(−𝑎2) cos(𝑎𝑥 + 𝑏)
             1
((D)) 𝜙(𝑎2) cos(𝑎𝑥 + 𝑏)
((E)) A
((F))
                                                  𝑑2𝑦   𝑑𝑦
((Q))2_// Particular Integral of D.E.𝑑𝑥 2 − 2 𝑑𝑥 + 5𝑦 = 10sin𝑥 is
        8
((A)) 3 sin𝑥
((B)) sin𝑥 − 2cos𝑥
((C)) 4sin𝑥 + 2cos𝑥
((D)) 2sin𝑥 + cos𝑥
((E)) D
((F))
                      1                       𝑑
((Q))1_// 𝐷−𝑚 𝑓(𝑥) , where 𝐷 ≡ 𝑑𝑥 And 𝑚 is constant, is
((A)) 𝑒 𝑚𝑥 ∫ 𝑒 −𝑚𝑥 𝑑𝑥
((B)) ∫ 𝑒 −𝑚𝑥 𝑓(𝑥) 𝑑𝑥
((C)) 𝑒 𝑚𝑥 ∫ 𝑒 −𝑚𝑥 𝑓(𝑥) 𝑑𝑥
((D)) 𝑒 −𝑚𝑥 ∫ 𝑒 𝑚𝑥 𝑓(𝑥) 𝑑𝑥
((E)) C
((F))
                                              1
((Q))1_// Particular Integral of 𝜙(𝐷) 𝑒 𝑎 𝑥 , 𝜙(𝑎) ≠ 0 is
            1
((A)) 𝜙(−𝑎) 𝑒 𝑎𝑥
                1
((B)) 𝑥 𝜙′(𝑎) 𝑒 𝑎𝑥
            1
((C)) 𝜙(𝑎2) 𝑒 𝑎𝑥
            1
((D))               𝑒 𝑎𝑥
        𝜙(𝑎)
((E)) D
((F))
                                              1
((Q))1_// Particular Integral for 𝜙(𝐷) 𝑥𝑉 , where 𝑉 is ((A)) function of 𝑥,is
                      1           1
((A)) [𝑥 −                 ]              𝑉
                    𝜙(𝐷) 𝜙(𝐷)
                    𝜙 ′ (𝐷)
((B)) [𝑥 −                  ] 𝜙(𝐷)𝑉
                    𝜙(𝐷)
                    𝜙 ′ (𝐷)
((C)) [𝑥 +                  ] 𝑉
                    𝜙(𝐷)
                    𝜙 ′ (𝐷)        1
((D)) [𝑥 −                    ]           𝑉
                     𝜙(𝐷)         𝜙(𝐷)
((E)) D
((F))
                                  1
((Q))1_// Particular Integral of 𝜙(𝐷2) cos(𝑎𝑥 + 𝑏), 𝜙(−𝑎2 ) = 0, 𝜙′ (−𝑎2 ) ≠ 0 is
              1
((A)) 𝜙′ (−𝑎2) cos(𝑎𝑥 + 𝑏)
          1
((B)) 𝜙′(−𝑎2) cos(𝑎𝑥 + 𝑏)
                  1
((C)) 𝑥 𝜙′(−𝑎2) sin(𝑎𝑥 + 𝑏)
                  1
((D)) 𝑥 𝜙′(−𝑎2) cos(𝑎𝑥 + 𝑏)
((E)) D
((F))
                                  1
((Q))1_// Particular Integral of 𝜙(𝐷2) sinh(𝑎𝑥 + 𝑏), 𝜙(𝑎2 ) ≠ 0 is
          1
((A)) 𝜙(𝑎2) cosh(𝑎𝑥 + 𝑏)
                  1
((B)) 𝑥 𝜙′(𝑎2) sinh(𝑎𝑥 + 𝑏)
          1
((C)) 𝜙(𝑎2) sinh(𝑎𝑥 + 𝑏)
          1
((D)) 𝜙(−𝑎2) sinh(𝑎𝑥 + 𝑏)
((E)) C
((F))
                                  1
((Q))1_// Particular Integral of 𝜙(𝐷2) cosh(𝑎𝑥 + 𝑏),𝜙(𝑎2 ) ≠ 0 , is
          1
((A)) 𝜙(𝑎2) cosh(𝑎𝑥 + 𝑏)
                  1
((B)) 𝑥 𝜙′(𝑎2) cosh(𝑎𝑥 + 𝑏)
          1
((C)) 𝜙(𝑎2) sinh(𝑎𝑥 + 𝑏)
          1
((D)) 𝜙(−𝑎2) cosh(𝑎𝑥 + 𝑏)
((E)) A
((F))
                              1       𝑥          𝑑
((Q))1_// Particular Integral𝐷+1 𝑒 𝑒 where 𝐷 ≡ 𝑑𝑥 is
                      𝑥
((A)) 𝑒 −𝑥 𝑒 𝑒
          𝑥
((B)) 𝑒 𝑒
              𝑥
((C)) 𝑒 𝑥 𝑒 𝑒
                 𝑥
((D)) 𝑒 − 2𝑥 𝑒 𝑒
((E)) A
((F))
                              1        𝑥             𝑑
((Q))2_// Particular Integral𝐷+2 𝑒 −𝑥 𝑒 𝑒 where 𝐷 ≡ 𝑑𝑥 is
                 𝑥
((A)) 𝑒 2𝑥 𝑒 𝑒
                 𝑥
((B)) 𝑒 − 2𝑥 𝑒 𝑒
          𝑥
((C)) 𝑒 𝑒
                𝑥
((D)) 𝑒 − 𝑥 𝑒 𝑒
((E)) B
((F))
                              1                          𝑑
((Q))2_// Particular Integral𝐷+2 𝑒 −𝑥 cos𝑒 𝑥where 𝐷 ≡ 𝑑𝑥 is
((A)) 𝑒 −𝑥 cos𝑒 𝑥
((B)) 𝑒 −𝑥 sin𝑒 𝑥
((C)) 𝑒 −2𝑥 cos𝑒 𝑥
((D)) 𝑒 −2𝑥 sin𝑒 𝑥
((E)) D
((F))
                              1                                          𝑑
((Q))2_// Particular Integral𝐷+2 𝑒 −2𝑥 sec 2 𝑥(1 + 2t((A))n𝑥) where 𝐷 ≡ 𝑑𝑥 is
((A)) 𝑒 −2𝑥 (1 + 2t((A))n2 𝑥)
((B)) 𝑒 −2𝑥 (t((A))n𝑥 + t((A))n2 𝑥)
((C)) 𝑒 2𝑥 (t((A))n𝑥 + 2t((A))n2 𝑥)
((D)) 𝑒 −2𝑥 (t((A))n𝑥 + sec𝑥)
((E)) B
((F))
                              1     1                 𝑑
((Q))2_// Particular Integral𝐷+1 (1+𝑒 𝑥)where 𝐷 ≡ 𝑑𝑥 is
((A)) 𝑒 𝑥 log(1 − 𝑒 𝑥 )
((B)) log(1 + 𝑒 𝑥 )
((C)) 𝑒 𝑥 log(1 + 𝑒 𝑥 )
((D)) 𝑒 −𝑥 log(1 + 𝑒 𝑥 )
((E)) D
((F))
                                         𝑑2𝑦     𝑑𝑦
((Q))2_// The Particular Integral of D.E.𝑑𝑥 2 − 7 𝑑𝑥 + 6𝑦 = 𝑒 2𝑥 is
             𝑥𝑒 2𝑥
((A)) −         3
             𝑒 2𝑥
((B)) −        4
        𝑒 2𝑥
((C))    4
        𝑒 2𝑥
((D))
       24
((E)) B
((F))
                               1                      𝑑
((Q))2_// Particular Integral     sin𝑒 𝑥 where 𝐷 ≡ is
                              𝐷+1                 𝑑𝑥
((A)) − 𝑒 −𝑥 sin𝑒 𝑥
((B)) 𝑒 𝑥 cos𝑒 𝑥
((C)) − 𝑒 −𝑥 cos𝑒 𝑥
((D)) 𝑒 −𝑥 cos𝑒 𝑥
((E)) C
((F))
                                        𝑑3𝑦     𝑑𝑦
((Q))2_// The Particular Integral of D.E.𝑑𝑥 3 − 4 𝑑𝑥 = 2cosh2𝑥 is
        1
((A)) 4 cosh2𝑥
        𝑥
((B)) 8 cosh2𝑥
        𝑥
((C)) 4 cosh2𝑥
        𝑥
((D)) 4 sinh2𝑥
((E)) C
((F))
                                        𝑑3𝑦
((Q))2_// The Particular Integral of D.E.𝑑𝑥 3 + 8𝑦 = 𝑥 4 + 2𝑥 + 1 is
        1
((A)) 8 (𝑥 4 + 5𝑥 + 1)
        1
((B)) 8 (𝑥 3 − 3𝑥 2 + 1)
((C)) 𝑥 4 − 𝑥 + 1
      1
((D)) 8 (𝑥 4 − 𝑥 + 1)
((E)) D
((F))
                                         𝑑2𝑦      𝑑𝑦
((Q))2_// The Particular Integral of D.E.𝑑𝑥 2 + 2 𝑑𝑥 + 𝑦 = 𝑒 − 𝑥 cos𝑥 is
((A)) 𝑒 𝑥 cos𝑥
((B)) −𝑒 − 𝑥 sin𝑥
((C)) −𝑒 − 𝑥 cos𝑥
((D)) (𝑐1 𝑥 + 𝑐2 )𝑒 − 𝑥
((E)) C
((F))
                                                              𝑑2𝑦   𝑑𝑦
((Q))2_// The Particular Integral of D.E.𝑑𝑥 2 − 2                        + 𝑦 = 𝑥𝑒 𝑥 sin𝑥 is
                                                                    𝑑𝑥
((A)) − 𝑒 𝑥 (𝑥sin𝑥 + 2cos𝑥)
((B)) 𝑒 𝑥 (𝑥sin𝑥 − 2cos𝑥)
((C)) (𝑥sin𝑥 + 2cos𝑥)
((D)) − 𝑒 𝑥 (𝑥cos𝑥 + 2sin𝑥)
((E)) A
((F))
                                        𝑑2𝑦      𝑑𝑦
((Q))2_// Solution of D.E.𝑑𝑥 2 + 𝑑𝑥 + 𝑦 = 𝑒 2𝑥 is
                       √3                      √3         1
((A)) 𝑒 𝑥 (𝑐1 cos         𝑥         + 𝑐2 sin      𝑥   ) − 7 𝑒 2𝑥
                        2                       2
             1
                                  √3                        √3              1
((B)) 𝑒 2𝑥 (𝑐1 cos                    𝑥   + 𝑐2 sin             𝑥     ) + 5 𝑒 2𝑥
                                   2                         2
                 1
                                    1                       1           1
((C)) 𝑒 − 2𝑥 (𝑐1 cos 2 𝑥 + 𝑐2 sin 2 𝑥 ) + 7 𝑒 𝑥
                 1
                                   √3                        √3                 1
((D)) 𝑒 − 2𝑥 (𝑐1 cos                  𝑥       + 𝑐2 sin          𝑥     ) + 7 𝑒 2𝑥
                                    2                         2
((E)) D
((F))
                                                                                                                          −1                 −2
                                                                                     𝑑𝑛𝑦                          𝑑𝑛           𝑦        𝑑𝑛        𝑦
((Q))1_// Cauchy’s Linear equation 𝑎0 𝑥 𝑛 𝑑𝑥 𝑛 + 𝑎1 𝑥 𝑛−1 𝑑𝑥 𝑛−1 + 𝑎2 𝑥 𝑛−2 𝑑𝑥 𝑛−2 + ⋯ ⋯ + 𝑎𝑛 𝑦 = 𝑓(𝑥) is reduced
to Linear D.E. with constant coefficient by using substitution
((A)) 𝑥 = 𝑒 𝑧
((B)) 𝑦 = 𝑒 𝑧
((C)) 𝑥 = log𝑧
              2
((D)) 𝑥 = 𝑒 𝑧
((E)) A
((F))
                                                                                           −1                     −2
                                                            𝑑𝑛𝑦                       𝑑𝑛        𝑦            𝑑𝑛        𝑦
((Q))1_//     Legendre’s Linear differential 𝑎0 (𝑎𝑥 + 𝑏)𝑛 𝑑𝑥 𝑛 + 𝑎1 (𝑎𝑥 + 𝑏)𝑛−1 𝑑𝑥 𝑛−1 + 𝑎2 (𝑎𝑥 + 𝑏)𝑛−2 𝑑𝑥 𝑛−2 +
⋯ ⋯ + 𝑎𝑛 𝑦 = 𝑓(𝑥) is reduced to Linear D.E. with constant coefficient by using substitution
((A)) 𝑥 = 𝑒 𝑧
((B)) 𝑎𝑥 + 𝑏 = 𝑒 𝑧
((C)) 𝑎𝑥 + 𝑏 = log𝑧
                   2
((D)) 𝑎𝑥 + 𝑏 = 𝑒 𝑧
((E)) C
((F))
                                                          𝑑2𝑦       𝑑𝑦
((Q))1_// On putting 𝑥 = 𝑒 𝑧 the transformed D.E. of 𝑥 2 𝑑𝑥 2 + 𝑥 𝑑𝑥 + 𝑦 = 𝑥is
     𝑑2𝑦    𝑑𝑦
((A)) 𝑑𝑧 2 + 𝑑𝑧 + 𝑦 = 𝑒 𝑧
     𝑑2𝑦
((B)) 𝑑𝑧 2 + 𝑦 = 𝑒 𝑧
     𝑑2𝑦
((C)) 𝑑𝑧 2 + 𝑦 = 𝑒 𝑥
     𝑑2𝑦
((D)) 𝑑𝑧 2 − 𝑦 = 𝑒 𝑧
((E)) B
((F))
                                                           𝑑2𝑦         𝑑𝑦
((Q))1_// On putting 𝑥 = 𝑒 𝑧 the transformed D.E. of 𝑥 2          −𝑥        + 4𝑦 = cos(log𝑥) + 𝑥sin(log𝑥) using 𝐷 ≡
                                                           𝑑𝑥 2        𝑑𝑥
𝑑
    is
𝑑𝑧
((A)) (𝐷2 − 𝐷 + 4)𝑦 = sin𝑧 + 𝑒 𝑧 cos𝑧
((B)) (𝐷2 − 2𝐷 + 4)𝑦 = cos(log𝑥) + 𝑥sin(log𝑥)
((C)) (𝐷2 + 2𝐷 + 4)𝑦 = cos𝑧 + 𝑒 −𝑧 sin𝑧
((D)) (𝐷2 − 2𝐷 + 4)𝑦 = cos𝑧 + 𝑒 𝑧 sin𝑧
((E)) D
((F))
                                                         𝑑2𝑦           𝑑𝑦                                𝑑
((Q))1_// On putting 𝑥 = 𝑒 𝑧 the transformed D.E. of 𝑥 2 2 − 3𝑥 + 5𝑦 = 𝑥 2 sin(log𝑥 ) using 𝐷 ≡ is
                                                        𝑑𝑥     𝑑𝑥                              𝑑𝑧
((A)) (𝐷2 − 4𝐷 + 5)𝑦 = 𝑒 2𝑧 sin𝑧
((B)) (𝐷2 − 4𝐷 + 5)𝑦 = 𝑥 2 sin(log𝑥)
((C)) (𝐷2 − 4𝐷 − 4)𝑦 = 𝑒 𝑧 sin𝑧
                           2
((D)) (𝐷2 − 3𝐷 + 5)𝑦 = 𝑒 𝑧 sin𝑧
((E)) A
((F))
                                                         𝑑2𝑦       𝑑𝑦            𝑥3                 𝑑
((Q))1_// On putting 𝑥 = 𝑒 𝑧 the transformed D.E. of 𝑥 2 𝑑𝑥 2 + 𝑥 𝑑𝑥 − 𝑦 = 1+𝑥 2 using 𝐷 ≡ 𝑑𝑧 is
                        𝑥3
((A)) (𝐷2 − 1)𝑦 = 1+𝑥 2
                                         𝑒 3𝑧
((B)) (𝐷2 − 2𝐷 − 1)𝑦 = 1+𝑒 2𝑧
                        𝑒 3𝑧
((C)) (𝐷2 − 1)𝑦 = 1+𝑒 2𝑧
                             3
                        𝑒𝑧
((D)) (𝐷2 − 1)𝑦 =                2
                    1+𝑒 𝑧
((E)) C
((F))
                                                                 𝑑2𝑦           𝑑𝑦                               𝑑
((Q))1_// On putting 𝑥 = 𝑒 𝑧 the transformed D.E .of 𝑥 2                − 5𝑥        + 5𝑦 = 𝑥 2 log𝑥 using 𝐷 ≡        is
                                                                 𝑑𝑥 2          𝑑𝑥                               𝑑𝑧
                                              𝑧2
((A)) (𝐷2 − 5𝐷 + 5)𝑦 = 𝑧 𝑒
((B)) (𝐷2 − 5𝐷 − 5)𝑦 = 𝑒 2𝑧 𝑧
((C)) (𝐷2 − 6𝐷 + 5)𝑦 = 𝑥 2 log𝑥
((D)) (𝐷2 − 6𝐷 + 5)𝑦 = 𝑧 𝑒 2𝑧
((E)) D
((F))
                                                                                      𝑑2𝑦             𝑑𝑦
((Q))1_// On putting 2𝑥 + 1 = 𝑒 𝑧 the transformed D.E. of (2𝑥 + 1)2 𝑑𝑥 2 − 2(2𝑥 + 1) 𝑑𝑥 + 5𝑦 = 6𝑥 using 𝐷 ≡
𝑑
     is
𝑑𝑧
                                     3
((A)) (𝐷2 − 2𝐷 − 3)𝑦 = 4 (𝑒 𝑧 − 1)
((B)) (𝐷2 + 2𝐷 + 3)𝑦 = 3(𝑒 𝑧 − 1)
                         3
((C)) (𝐷2 + 2𝐷 − 12)𝑦 = 4 (𝑒 𝑧 − 1)
((D)) (𝐷2 − 2𝐷 − 3)𝑦 = 6𝑥
((E)) A
((F))
                                                                                      𝑑2𝑦             𝑑𝑦
((Q))1_// On putting 3𝑥 + 2 = 𝑒 𝑧 the transformed D.E. of (3𝑥 + 2)2 𝑑𝑥 2 + 3(3𝑥 + 2) 𝑑𝑥 − 36𝑦 = 3𝑥 2 + 4𝑥 +
             𝑑
1 using 𝐷 ≡ 𝑑𝑧 is
                                          1
((A)) (𝐷2 + 3𝐷 − 36)𝑦 = 27 (𝑒 2𝑧 − 1)
                    1
((B)) (𝐷2 + 4)𝑦 = 9 (𝑒 2𝑧 − 1)
                    1
((C)) (𝐷2 − 4)𝑦 = (𝑒 2𝑧 − 1)
                  27
((D)) (𝐷2 − 9)𝑦 = (𝑒 2𝑧 − 1)
((E)) C
((F))
                                                   𝑑𝑥                     𝑑𝑦                                              𝑑
((Q))2_// For the simultaneous Linear DE           𝑑𝑡
                                                        + 2𝑥 − 3𝑦 = 𝑡,     𝑑𝑡
                                                                                − 3𝑥 + 2𝑦 = 𝑒 2𝑡 solution of 𝑦 using 𝐷 ≡ 𝑑𝑡 is
obtain from
((A)) (𝐷2 + 4𝐷 − 5)𝑥 = 1 + 2𝑡 + 3𝑒 2𝑡
((B)) (𝐷2 − 4𝐷 − 5)𝑦 = 𝑡 − 4𝑒 2𝑡
((C)) (𝐷2 + 4𝐷 + 5)𝑦 = 3𝑡 + 2𝑒 2𝑡
((D)) (𝐷2 + 4𝐷 − 5)𝑦 = 3𝑡 + 4𝑒 2𝑡
((E)) D
((F))
                                           𝑑𝑥                       𝑑𝑦                                              𝑑
((Q))2_// For the simultaneous Linear DE + 2𝑥 − 3𝑦 = 𝑡,                  − 3𝑥 + 2𝑦 = 𝑒 2𝑡 solution of 𝑥 using 𝐷 ≡        is
                                        𝑑𝑡                          𝑑𝑡                                              𝑑𝑡
obt((A))in from
((A)) (𝐷2 + 4𝐷 − 5)𝑥 = 1 + 2𝑡 + 3𝑒 2𝑡
((B)) (𝐷2 − 4𝐷 − 5)𝑥 = 1 + 2𝑡 − 3𝑒 2𝑡
((C)) (𝐷2 + 4𝐷 − 5)𝑥 = 3𝑡 + 3𝑒 2𝑡
((D)) (𝐷2 + 4𝐷 − 5)𝑦 = 3𝑡 + 4𝑒 2𝑡
((E)) A
((F))
                                           𝑑𝑥                        𝑑𝑦     𝑑𝑥                                      𝑑
((Q))2_// For the simultaneous Linear DE        − 3𝑥 − 6𝑦 = 𝑡 2 ,         + 𝑑𝑡 − 3𝑦 = 𝑒 𝑡 solution of 𝑥 using 𝐷 ≡ 𝑑𝑡 is
                                           𝑑𝑡                        𝑑𝑡
obt((A))in from
((A)) (𝐷2 + 9)𝑥 = 6𝑒 𝑡 − 3𝑡 2 + 2𝑡
((B)) (𝐷2 + 9)𝑦 = −2𝑒 𝑡 − 2𝑡
((C)) (𝐷2 − 9)𝑥 = 6𝑒 𝑡 − 3𝑡 2
((D)) (𝐷2 + 12𝐷 + 9)𝑥 = 6𝑒 𝑡 + 3𝑡 2 + 2𝑡
((E)) A
((F))
                                           𝑑𝑢                  𝑑𝑣                                          𝑑
((Q))2_// For the simultaneous Linear DE 𝑑𝑥 + 𝑣 = sin𝑥,             + 𝑢 = cos𝑥 solution of 𝑢 using 𝐷 ≡ 𝑑𝑥 is obtain
                                                               𝑑𝑥
from
((A)) (𝐷2 + 1)𝑢 = 2cos𝑥
((B)) (𝐷2 − 1)𝑢 = 0
((C)) (𝐷2 − 1)𝑢 = sin𝑥 − cos𝑥
((D)) (𝐷2 − 1)𝑣 = −2sin𝑥
((E)) B
((F))
                                           𝑑𝑢                  𝑑𝑣                                          𝑑
((Q))2_// For the simultaneous Linear DE 𝑑𝑥 + 𝑣 = sin𝑥,             + 𝑢 = cos𝑥 solution of 𝑣 using 𝐷 ≡ 𝑑𝑥 is obtain
                                                               𝑑𝑥
from
((A)) (𝐷2 + 1)𝑣 = 0
((B)) (𝐷2 − 1)𝑢 = 0
((C)) (𝐷2 − 1)𝑣 = −2sin𝑥
((D)) (𝐷2 + 1)𝑣 = sin𝑥 + cos𝑥
((E)) C
((F))
                                                    𝑑𝑥                                𝑑𝑦
((Q))2_//   For the simultaneous Linear DE 𝐿 𝑑𝑡 + 𝑅𝑥 + 𝑅(𝑥 − 𝑦) = 𝐸, 𝐿                     + 𝑅𝑦 − 𝑅(𝑥 − 𝑦) = 0 where
                                                                                      𝑑𝑡
                                                     𝑑
𝐿 , 𝑅 and 𝐸are constants, solution of 𝑥 using 𝐷 ≡ 𝑑𝑡 is obtain from
((A)) (𝐿2 𝐷2 + 4𝑅𝐿𝐷 + 5𝑅2 )𝑥 = 2𝑅𝐸 + 2𝑅
((B)) (𝐿2 𝐷2 + 4𝑅𝐿𝐷 + 3𝑅2 )𝑦 = 𝑅𝐸
((C)) (𝐿2 𝐷2 + 4𝑅𝐿𝐷 + 3𝑅2 )𝑥 = 2𝑅𝐸
((D)) (𝐿2 𝐷2 + 2𝑅𝐿𝐷 + 5𝑅2 )𝑥 = 2𝑅𝐸
((E)) C
((F))
                                                      𝑑𝑥                               𝑑𝑦
((Q))2_//   For the simultaneous Linear DE 𝐿 𝑑𝑡 + 𝑅𝑥 + 𝑅(𝑥 − 𝑦) = 𝐸, 𝐿                      + 𝑅𝑦 − 𝑅(𝑥 − 𝑦) = 0 where
                                                                                       𝑑𝑡
                                                        𝑑
𝐿 , 𝑅 and 𝐸 are constants, solution of 𝑦 using 𝐷 ≡ 𝑑𝑡 is obtain from
((A)) (𝐿2 𝐷2 + 4𝑅𝐿𝐷 + 5𝑅2 )𝑦 = 𝑅𝐸 + 2𝑅
((B)) (𝐿2 𝐷2 + 4𝑅𝐿𝐷 + 3𝑅2 )𝑦 = 𝑅𝐸
((C)) (𝐿2 𝐷2 + 4𝑅𝐿𝐷 + 3𝑅2 )𝑥 = 2𝑅𝐸
((D)) (𝐿2 𝐷2 + 2𝑅𝐿𝐷 + 5𝑅2 )𝑦 = 2𝑅𝐸
((E)) B
((F))
                                           𝑑𝑥                𝑑𝑦                                        𝑑
((Q))2_// For the simultaneous Linear DE 𝑑𝑡 + 𝑦 = 𝑒 𝑡 ,           + 𝑥 = 𝑒 −𝑡 solution of 𝑥 using 𝐷 ≡ 𝑑𝑡 is obtain from
                                                             𝑑𝑡
                                            𝑑𝑥                         𝑑𝑦                                           𝑑
((Q))2_// For the simultaneous Linear DE            + 5𝑥 − 2𝑦 = 𝑡,          + 2𝑥 + 𝑦 = 0, solution of 𝑦 using 𝐷 ≡ 𝑑𝑡 is
                                               𝑑𝑡                      𝑑𝑡
obtain from
((A)) (𝐷2 − 6𝐷 − 9)𝑦 = 2𝑡
((B)) (𝐷2 + 6𝐷 + 9)𝑥 = 1 + 𝑡
((C)) (𝐷2 + 6𝐷 + 1)𝑦 = 𝑡
((D)) (𝐷2 + 6𝐷 + 9)𝑦 = 2𝑡
((E)) B
((F))
                                          𝑑𝑥                𝑑𝑦                                         𝑑
((Q))2_// For the simultaneous Linear DE 𝑑𝑡 + 𝑦 = 𝑒 𝑡 ,           + 𝑥 = 𝑒 −𝑡 , solution of 𝑦 using 𝐷 ≡ 𝑑𝑡 is obtain from
                                                            𝑑𝑡
((A)) (𝐷2 − 1)𝑦 = 2𝑒 𝑡
((B)) (𝐷2 − 1)𝑦 = −𝑒 𝑡 − 𝑒 −𝑡
((C)) (𝐷2 + 1)𝑦 = 𝑒 −𝑡 + 𝑒 𝑡
((D)) (𝐷2 − 1)𝑥 = 𝑒 𝑡 − 𝑒 −𝑡
((E)) B
((F))
                                                                             𝑑𝑥                         𝑑𝑦                                                 𝑑
((Q))2_// For the simultaneous Linear DE                                             + 5𝑥 − 2𝑦 = 𝑡,          + 2𝑥 + 𝑦 = 0, solution of 𝑥 using 𝐷 ≡ 𝑑𝑡 is
                                                                             𝑑𝑡                         𝑑𝑡
obtain from
((A)) (𝐷2 + 6𝐷 + 9)𝑥 = 1 + 𝑡
((B)) (𝐷2 − 6𝐷 + 9)𝑥 = 2𝑡
((C)) (𝐷2 + 6𝐷 + 1)𝑥 = 𝑡
((D)) (𝐷2 + 6𝐷 + 9)𝑦 = 2𝑡
((E)) A
((F))
                                      𝑑𝑢                          𝑑𝑣
((Q))1_// For the D.E. + 𝑣 = sin 𝑥, + 𝑢 = cos 𝑥, the auxiliary equation for 𝑢 is
                      𝑑𝑥           𝑑𝑥
((A)) 𝐷2 + 1 = 0
((B)) 𝐷2 + 2 = 0
((C)) 𝐷2 − 1 = 0
((D)) 𝐷2 − 2 = 0
((E)) C
((F))
                                                                                                                            𝑑𝑥        𝑑𝑦        𝑑𝑧
((Q))1_// Considering the first two ratio of the symmetrical simultaneous DE                                                        = −𝑥𝑦 = 𝑥(𝑧−2𝑦), one of the
                                                                                                                            𝑦2
relation in the solution of DE is
((A)) 𝑥 2 + 𝑦 2 = 𝑐
((B)) 𝑥 3 + 𝑦 3 = 𝑐
             𝑥2     𝑦3
((C)) − 2 = 3 + 𝑐
((D)) 𝑥 2 − 𝑦 2 = 𝑐
((E)) A
((F))
                                                                                                                            𝑑𝑥             𝑑𝑦        𝑑𝑧
((Q))1_// Which of the following are the sets of multipliers for the solution of                                                     = 4𝑥−2𝑧 = 2𝑦−3𝑥
                                                                                                                       3𝑥−4𝑦
((A)) 𝑥, 𝑦, 𝑧
((B)) 4, 3, 2
      1 1 1
((C)) 𝑥 , 𝑦 , 𝑧
        1 1 1
((D)) 4 , 3 , 2
((E)) A
((F))
                                                                                                 𝑥𝑑𝑥        𝑑𝑦         𝑑𝑧
((Q))1_//        Considering the first & third ratio of the symmetrical simultaneous DE                = 𝑥 2𝑧 = 𝑦 3, one of the
                                                                                                 𝑦3𝑧
relation in the solution of DE is
((A)) 𝑥 2 − 𝑧 2 = 𝑐
((B)) 𝑥 4 − 𝑦 4 = 𝑐
((C)) 𝑥 3 − 𝑧 3 = 𝑐
((D)) 𝑥 − 𝑧 = 𝑐
((E)) A
((F))
                                                                          𝑥𝑑𝑥         𝑑𝑦    𝑑𝑧
((Q))1_// Using a set of multiplier as 1, 𝑦, 𝑧 the solution of DE 𝑧 2−2𝑦𝑧−𝑦2 = 𝑦+𝑧 = 𝑦−𝑧 is
((A)) 𝑥 2 + 𝑦 2 + 𝑧 2 = 𝑐
            𝑦2     𝑧2
((B)) 𝑥 + 2 + 2 = 𝑐
((C)) 𝑥 + 𝑦 + 𝑧 = 𝑐
((D)) 𝑥 + 𝑦 2 + 𝑧 2 = 𝑐
((E)) A
((F))
                                                                                                       𝑑𝑥              𝑑𝑦   𝑑𝑧
((Q))1_// Considering the second & third ratio of the symmetrical simultaneous DE                                 = 2𝑥𝑦 = 2𝑥𝑧, one
                                                                                                  𝑥 2 −𝑦 2 −𝑧 2
of the relation in the solution of DE is
      1      1
((A)) 𝑦 2 − 𝑧 2 = 𝑐
((B)) 𝑦 2 − 𝑧 2 = 𝑐
((C)) 𝑦 = 𝑐 𝑧
((D)) 𝑥 − 𝑧 = 𝑐
((E)) C
((F))
                                                                     𝑑𝑥     𝑑𝑦        𝑑𝑧
((Q))1_// Using a set of multiplier as 1,1,1 the solution of DE 𝑦−𝑧 = 𝑧−𝑥 = 𝑥−𝑦 is
((A)) 𝑥 2 + 𝑦 2 + 𝑧 2 = 𝑐
((B)) 𝑥 − 𝑦 − 𝑧 = 𝑐
((C)) 𝑥 + 𝑦 + 𝑧 = 𝑐
((D)) −𝑥 + 𝑦 − 𝑧 = 𝑐
((E)) C
((F))
                                                                                 𝑑𝑥          𝑑𝑦                   𝑑𝑧
((Q))1_// Using a set of multiplier as 𝑥 3 , 𝑦 3 , 𝑧 3 the solution of DE 𝑥(2𝑦 4−𝑧 4) = 𝑦(𝑧 4−2𝑥 4) = 𝑧(𝑥 4−𝑦 4) is
((A)) 𝑥 3 + 𝑦 3 + 𝑧 3 = 𝑐
((B)) 𝑥 4 + 𝑦 4 + 𝑧 4 = 𝑐
((C)) 𝑥 + 𝑦 + 𝑧 = 𝑐
((D)) 𝑥 𝑦𝑧 = 𝑐
((E)) B
((F))
                                                                                                 𝑑𝑥           𝑑𝑦         𝑑𝑧
((Q))1_//    Considering the first two ratio of the symmetrical simultaneous DE                          =           =        , one of the
                                                                                                 𝑦2𝑧          𝑥 2𝑧       𝑦2𝑥
relation in the solution of DE is
((A)) 𝑥 2 − 𝑦 2 = 𝑐
((B)) 𝑥 − 𝑦 = 𝑐
((C)) 𝑥 3 − 𝑦 3 = 𝑐
((D)) 𝑥 3 + 𝑦 3 = 𝑐
((E)) C
((F))
                                                                     𝑑𝑥       𝑑𝑦      𝑑𝑧
((Q))1_// Using a set of multiplier as 3, 2, 1 the solution of DE         = −𝑥 = 2𝑥−3𝑦 is
                                                                     𝑦
((A)) 3𝑥 2 + 2𝑦 2 + 𝑧 2 = 𝑐
      3    2  1
((B)) 𝑥 + 𝑦 + 𝑧 = 𝑐
((C)) 3𝑥 − 2𝑦 − 𝑧 = 𝑐
((D)) 3𝑥 + 2𝑦 + 𝑧 = 𝑐
((E)) D
((F))
                                                                         𝑑𝑥         𝑑𝑦          𝑑𝑧
((Q))1_// Using a set of multiplier as 𝑥 , 𝑦, 𝑧 the solution of DE             =           =             is
                                                                     3𝑧−4𝑦         4𝑥−2𝑧       2𝑦−3𝑥
       3     3        3
((A)) 𝑥 + 𝑦 + 𝑧 = 𝑐
      1   1   1
((B)) 𝑥 + 𝑦 + 𝑧 = 𝑐
((C)) 𝑥 + 𝑦 + 𝑧 = 𝑐
((D)) 𝑥 2 + 𝑦 2 + 𝑧 2 = 𝑐
((E)) D
((F))
                                                                                               𝑑𝑥        𝑑𝑦          𝑑𝑧
((Q))1_// Considering the first two ratio of the symmetrical simultaneous DE                         =        = 𝑥 2𝑦 2𝑧 2, one of the
                                                                                               𝑦2        𝑥2
relation in the solution of DE is
      1   1
((A)) 𝑥 − 𝑦 = 𝑐
((B)) 𝑥 − 𝑦 = 𝑐
((C)) 𝑥 2 − 𝑦 2 = 𝑐
((D)) 𝑥 3 − 𝑦 3 = 𝑐
((E)) D
((F))