The general solution of the differential equation, d x 4 d 4 y − 2 d x 3 d 3 y + 2 d x 2 d 2 y − 2 dx dy + y = 0 is
A. y = ( C 1 − C 2 x ) e x + C 3 cos x + C 4 sin x
B. y = ( C 1 + C 2 x ) e x − C 2 cos x + C 4 sin x
C. y = ( C 1 + C 2 x ) e x + C 3 cos x + C 4 sin x
D. y = ( C 1 + C 2 x ) e x + C 3 cos x − C 4 sin x
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The solution of the differential equation,
y 1 − x 2 dy + x 1 − y 2 dx = 0 is
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The maximum value of the solution y(t) of the differential equation y ( t ) + y ¨ ( t ) = 0 with initial conditions y ˙ ( 0 ) = 1 and y(0) = 1, for t ≥ 0 is
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The order of the differential equation d t 2 d 2 y + ( dt dy ) 3 + y 4 = e − t is
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The order and degree of the differential equation
d x 3 d 3 y + 4 ( dx dy ) 3 + y 2 = 0 are respectively
A. 3 and 2
B. 2 and 3
C. 3 and 3
D. 3 and 1
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Consider the following second-order differential equation: y" - 4y' + 3y = 2t - 3t2
The particular solution of the differential equation is
A. -2 - 2t - t2
B. -2t - t2
C. 2t - t2
D. -2 - 2t - 3t2
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The differential equation [ 1 + ( dx dy ) 2 ] 3 = C 2 [ d x 2 d 2 y ] 2 is of
A. 2nd order and 3rd degree
B. 3rd order and 2nd degree
C. 2nd order and 2nd degree
D. 3rd order and 3rd degree
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Consider a system governed by the following equations:
dt d x 1 ( t ) = x 2 ( t ) − x 1 ( t ) ; dt d x 2 ( t ) = x 1 ( t ) − x 2 ( t )
The initial conditions are such that x 1 ( 0 ) < x 2 ( 0 ) < ∞. Let x 1 f = t → ∞ lim x 1 ( t ) and x 2 f = t → ∞ lim x 2 ( t ) . Which one of the following is true?
A. x 1 f < x 2 f < ∞
B. x 2 f < x 1 f < ∞
C. x 1 f = x 2 f < ∞
D. x 1 f = x 2 f = ∞
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The families of curves represented by the solution of the equation dx dy = − ( y x ) n for n = -1 and n = +1, respectively, are
A. Hyperbolas and Parabolas
B. Hyperbolas and Circles
C. Parabolas and Circles
D. Circles and Hyperbolas
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The solution of dx dy = y 2 with initial value y(0) = 1 bounded in the interval
A. − ∞ ⩽ x ⩽ ∞
B. − ∞ ⩽ x ⩽ 1
C. x < 1 , x > 1
D. − 2 ⩽ x ⩽ 2
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The respective expressions for complimentary function and particular integral part of the solution of the differential equation d x 4 d 4 y + 3 d x 2 d 2 y = 108 x 2 are
A. [ c 1 + c 2 x + c 3 sin 3 x + c 4 cos 3 x ] and [ 3 x 4 − 12 x 2 + c ] B. [ c 2 + c 3 sin 3 x + c 4 cos 3 x ] and [ 5 x 4 − 12 x 2 + c ] C. [ c 1 + c 3 sin 3 x + c 4 cos 3 x ] and [ 3 x 4 − 12 x 2 + c ] D. [ c 1 + c 2 x + c 3 sin 3 x + c 4 cos 3 x ] and [ 5 x 4 − 12 x 2 + c ] Select an option to see the answer and solution.
The solution of the differential equation d t 2 d 2 y + 2 dt dy + y = 0 with y(0) = y'(0) = 1 is
A. (2 - t)et
B. (1 + 2t)e-t
C. (2 + t)e-t
D. (1 - 2t)et
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Consider the differential equation ( t 2 − 81 ) dt dy + 5 ty = sin ( t ) with y(1) = 2π. There
exists a unique solution for this differential equation when t belongs to the interval
A. (-2, 2)
B. (-10, 10)
C. (-10, 2)
D. (0, 10)
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The solution to 6yy' - 25x = 0 represents a
A. family of circles
B. family of ellipses
C. family of parabolas
D. family of hyperbolas
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Which one of the following is the general solution of the first order differential equation dx dy = ( x + y − 1 ) 2 , where x, y are real?
A. y = 1 + x + tan-1 (x + c), where c is a constant
B. y = 1 + x + tan(x + c), where c is a constant
C. y = 1 - x + tan-1 (x + c), where c is a constant
D. y = 1 - x + tan(x + c), where c is a constant
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If roots of the auxiliary equation of d x 2 d 2 y + a dx dy + by = 0 are real and equal, the general solution of the differential equation is
A. y = c 1 e − 2 ax + c 2 e 2 ax
B. y = ( c 1 + c 2 x ) e − 2 ax
C. y = ( c 1 + c 2 ln x ) e − 2 ax
D. y = ( c 1 cos x + c 2 sin x ) e − 2 ax
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The solution of the ordinary differential equation dx dy + 2y = 0 for the boundary condition, y = 5 at x = 1 is
A. y = e-2x
B. y = 2e-2x
C. y = 10.95 e-2x
D. y = 36.95 e-2x
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If y is the solution of the differential equation
y 3 dx dy + x 3 = 0 , y ( 0 ) = 1
the value of y(-1) is
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The general solution of the differential equation dx dy = cos ( x + y ) , with c as a constant, is
A. y + sin ( x + y ) = x + c
B. tan ( 2 x + y ) = y + c
C. cos ( 2 x + y ) = x + c
D. tan ( 2 x + y ) = x + c
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The solution of the differential equation, d x 2 d 2 y − dx dy − 2 y = 3 e 2 x , where, y(0) = 0 and y'(0) = -2 is
A. y = e-x - e2x + xe2x
B. y = ex - e-2x - xe2x
C. y = e-x + e2x + xe2x
D. y = ex - e-2x + xe2x
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