With K as a constant, the solution possible for the first order differential equation dx dy = e − 3 x is
A. − 3 1 e − 3 x + K
B. − 3 1 e 3 x + K
C. − 3 e − 3 x + K
D. − 3 e − x + K
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The solution of the first order differential equation x'(t) = -3x(t), x(0) = x0 is
A. x(t) = x0 e-3t
B. x(t) = x0 e-3
C. x(t) = x0 e − 3 1
D. x(t) = x0 e-1
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The solution of the equation x dx dy + y = 0 passing through the point (1, 1) is
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With initial condition x(1) = 0.5, the solution of the differential equation, t dt dx + x = t is
A. x = t − 2 1
B. x = t 2 − 2 1
C. x = 2 t 2
D. x = 2 t
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For the differential equation
d t 2 d 2 x + 6 dt dx + 8 x = 0 with initial conditions x(0) = 1 and
dy dx t = 0 = 0 , the solution is
A. x(t) = 2e-6t - e-2t
B. x(t) = 2e-2t - e-4t
C. x(t) = -e-6t + 2e-4t
D. x(t) = e-2t + 2e-4t
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Transformation to linear form by substituting v = y1 - n of the equation
dt dy + p ( t ) y = q ( t ) y n ; n > 0 will be
A. dt dv + ( 1 − n ) pv = ( 1 − n ) q
B. dt dv + ( 1 − n ) pv = ( 1 + n ) q
C. dt dv + ( 1 + n ) pv = ( 1 − n ) q
D. dt dv + ( 1 + n ) pv = ( 1 + n ) q
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Consider the differential equation y ¨ + 2 y ˙ + y = 0 with boundary conditions y(0) = 1, y(1) = 0. The value of y(2) is
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For the equation, dx dy + 7 x 2 y = 0 , if y ( 0 ) = 7 3 , then the value of y(1) is
A. 7 3 e − 3 7
B. 3 7 e − 3 7
C. 7 3 e − 7 3
D. 3 7 e − 7 3
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Consider two solutions x(t) = x
1 (t) and x(t) = x
2 (t) of the differential equation
d t 2 d 2 x ( t ) + x ( t ) = 0 , t > 0 , such that
x 2 = 0 , dt d x 2 ( t ) t = 0 = 1. The Wronskian
{\text{W}}\left( {\text{t}} \right) = \left| {\begin{array}{*{20}{c}}
{{{\text{x}}_1}\left( {\text{t}} \right)}&{{{\text{x}}_2}\left( {\text{t}} \right)} \\
{\frac{{{\text{d}}{{\text{x}}_1}\left( {\text{t}} \right)}}{{{\text{dt}}}}}&{\frac{{{\text{d}}{{\text{x}}_2}\left( {\text{t}} \right)}}{{{\text{dt}}}}}
\end{array}} \right| at
t = 2 π is
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If d t 2 d 2 y + y = 0 under the conditions y = 1, dt dy = 0 , when t = 0, then y is equal to
A. sin t
B. cos t
C. tan t
D. cot t
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The general solution of d x 2 d 2 y + y = 0 is
A. y = P cos x + Q sin x
B. y = P cos x
C. y = P sin x
D. y = P sin2 x
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The solution for the differential equation dx dy = x 2 y with the condition that y = 1 at x = 0 is
A. y = e 2 x 1
B. ln ( y ) = 3 x 3 + 4
C. ln ( y ) = 2 x 2
D. y = e 3 x 3
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The solution of the differential equation dx dy = ky , y ( 0 ) = c is
A. x = ce-ky
B. x = kecy
C. y = cekx
D. y = ce-kx
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The solution of the differential equation dx dy + y 2 = 0 is
A. y = x + c 1
B. y = 3 − x 3 + c
C. cex
D. unsolvable as equation is non-linear
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If y = 2x3 - 3x2 + 3x - 10, the value of ∆3 y will be (where, ∆ is forward differences operator)
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Match
List-I with
List-II and select the correct answer:
List-I
List-II
A. dx dy = x y
1. Circles
B. dx dy = − x y
2. Straight lines
C. dx dy = y x
3. Hyperbolas
D. dx dy = − y x
A. a-2, b-3, c-3, d-1
B. a-1, b-3, c-2, d-1
C. a-2, b-1, c-3, d-3
D. a-3, b-2, c-1, d-2
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If a and b are constants, the most general solution of the differential equation d t 2 d 2 x + 2 dt dx + x = 0 is
A. ae-t
B. ae-t + bte-t
C. aet + bte-t
D. ae-2t
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The solution of the differential equation, for t > 0, y''(t) + 2y'(t) + y(t) = 0 with initial conditions
y(0) = 0 and y'(0) = 1, is (u(t) denotes the unit step function),
A. te-t u(t)
B. (e-t - te-t )u(t)
C. (-e-t + te-t )u(t)
D. e-t u(t)
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A differential equation is given as
x 2 d x 2 d 2 y − 2 x dx dy + 2 y = 4
The solution of differential equation in terms of arbitrary constant C1 and C2 is
A. y = x 2 C 1 + C 2 x + 2
B. y = C 1 x 2 + C 2 x + 4
C. y = C 1 x 2 + C 2 x + 2
D. y = x 2 C 1 + C 2 x + 4
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The solution of the differential equation d x 2 d 2 y = 0 with boundary conditions
i . dx dy = 1 at x = 0 ; ii . dx dy = 1 at x = 1 is
A. y = 1
B. y = x
C. y = x + c where C is an arbitrary constant
D. y = C1 x + C2 where C1 and C2 are arbitrary
constants
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