Differential Equations MCQs Question and Answer in Engineering Maths Page-7 section-1

61.
The general solution of the differential equation $$\frac{{{{\text{d}}^2}{\text{y}}}}{{{\text{d}}{{\text{x}}^2}}} + 2\frac{{{\text{dy}}}}{{{\text{dx}}}} - 5{\text{y}} = 0$$ in terms of arbitrary constants K₁ and K₂ is

A. K₁e^{(-1 + √6)x} + K₂e^{(-1 - √6)x}

B. K₁e^{(-1 + √8)x} + K₂e^{(-1 - √8)x}

C. K₁e^{(-2 + √6)x} + K₂e^{(-2 - √6)x}

D. K₁e^{(-2 + √8)x} + K₂e^{(-2 - √8)x}

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62.
General solution of the Cauchy-Euler equation $${{\text{x}}^2}\frac{{{{\text{d}}^2}{\text{y}}}}{{{\text{d}}{{\text{x}}^2}}} - 7{\text{x}}\frac{{{\text{dy}}}}{{{\text{dx}}}} + 16{\text{y}} = 0$$ is

A. y = c₁x² + c₂x⁴

B. y = c₁x² + c₂x^-4

C. y = (c₁ + c₂ln x) x⁴

D. y = c₁x⁴ + c₂x^-4ln x

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63.
Biotransformation of an organic compound having concentration (x) can be modeled using an ordinary differential equation $$\frac{{{\text{dx}}}}{{{\text{dt}}}} + {\text{k}}{{\text{x}}^2} = 0,$$ where k is the reaction rate constant. If x = a at t = 0, the solution of the equation is

A. x = ae^-kt

B. $$\frac{1}{{\text{x}}} = \frac{1}{{\text{a}}} + {\text{kt}}$$

C. x = a(1 - e^-kt)

D. x = a + kt

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64.
The solution of initial value problem; $$\frac{{\partial {\text{u}}}}{{\partial {\text{x}}}} = 2\frac{{\partial {\text{u}}}}{{\partial {\text{t}}}} + {\text{u,}}$$ where u(x, 0) = 6e^-3x is

A. u = 6e^{-3x + t}

B. u = 6e^{-(2x + 2t)}

C. u = 6e^{-(3x + 2t)}

D. u = 6e^{-(3x + 2t)}

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65.
A function n(x) satisfies the differential equation $$\frac{{{{\text{d}}^2}{\text{n}}\left( {\text{x}} \right)}}{{{\text{d}}{{\text{x}}^2}}} - \frac{{{\text{n}}\left( {\text{x}} \right)}}{{{{\text{L}}^2}}} = 0$$ where L is a constant. The boundary conditions are: n(0) = K and n($$\infty $$) = 0. The solution to this equation is

A. n(x) = K exp(x/L)

B. n(x)= K exp(-x/√L)

C. n(x) = K² exp(-x/L)

D. n(x) = K exp(-x/L)

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66.
The solutions of differential equation $$\frac{{{{\text{d}}^2}{\text{y}}}}{{{\text{d}}{{\text{x}}^2}}} + \frac{{{\text{2dy}}}}{{{\text{dx}}}} + 2{\text{y}} = 0$$ are

A. e^{-(1 + i)x}, e^{-(1 - i)x}

B. e^{(1 + i)x}, e^{(1 - i)x}

C. e^{-(1 + i)x}, e^{(1 - i)x}

D. e^{(1 + i)x}, e^{-(1 - i)x}

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67.
It is given that y'' + 2y' + y = 0, y(0) = 0, y(1) = 0. What is y (0.5)?

A. 0

B. 0.37

C. 0.62

D. 1.13

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68.
A function y(t), such that y(0) = 1 and y(1) = 3e^-1, is a solution of the differential equation $$\frac{{{{\text{d}}^2}{\text{y}}}}{{{\text{d}}{{\text{t}}^2}}} + 2\frac{{{\text{dy}}}}{{{\text{dt}}}} + {\text{y}} = 0.$$ Then y(2) is

A. 5e^-1

B. 5e^-2

C. 7e^-1

D. 7e^-2

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69.
A curve passes through the point (x = 1, y = 0) and satisfies the differential equation $$\frac{{{\text{dy}}}}{{{\text{dx}}}} = \frac{{{{\text{x}}^2} + {{\text{y}}^2}}}{{2{\text{y}}}} + \frac{{\text{y}}}{{\text{x}}}.$$ The equation that describes the curve is

A. $$\ln \left( {1 + \frac{{{{\text{y}}^2}}}{{{{\text{x}}^2}}}} \right) = {\text{x}} - 1$$

B. $$\frac{1}{2}\ln \left( {1 + \frac{{{{\text{y}}^2}}}{{{{\text{x}}^2}}}} \right) = {\text{x}} - 1$$

C. $$\ln \left( {1 + \frac{{\text{y}}}{{\text{x}}}} \right) = {\text{x}} - 1$$

D. $$\frac{1}{2}\ln \left( {1 + \frac{{\text{y}}}{{\text{x}}}} \right) = {\text{x}} - 1$$

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70.
The differential equation $$\frac{{{\text{dy}}}}{{{\text{dx}}}} + 4{\text{y}} = 5$$ is valid in the domain 0 ≤ x ≤ 1 with y(0) = 2.25. The solution of differential equation is

A. y = e^-4x + 1.25

B. y = e^-4x + 5

C. y = e^4x + 5

D. y = e^4x + 1.25

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Differential Equations MCQs Question and Answer in Engineering Maths

61. The general solution of the differential equation $$\frac{{{{\text{d}}^2}{\text{y}}}}{{{\text{d}}{{\text{x}}^2}}} + 2\frac{{{\text{dy}}}}{{{\text{dx}}}} - 5{\text{y}} = 0$$ in terms of arbitrary constants K1 and K2 is

Answer & Solution

62. General solution of the Cauchy-Euler equation $${{\text{x}}^2}\frac{{{{\text{d}}^2}{\text{y}}}}{{{\text{d}}{{\text{x}}^2}}} - 7{\text{x}}\frac{{{\text{dy}}}}{{{\text{dx}}}} + 16{\text{y}} = 0$$ is

Answer & Solution

63. Biotransformation of an organic compound having concentration (x) can be modeled using an ordinary differential equation $$\frac{{{\text{dx}}}}{{{\text{dt}}}} + {\text{k}}{{\text{x}}^2} = 0,$$ where k is the reaction rate constant. If x = a at t = 0, the solution of the equation is

Answer & Solution

64. The solution of initial value problem; $$\frac{{\partial {\text{u}}}}{{\partial {\text{x}}}} = 2\frac{{\partial {\text{u}}}}{{\partial {\text{t}}}} + {\text{u,}}$$ where u(x, 0) = 6e-3x is

Answer & Solution

Answer & Solution

66. The solutions of differential equation $$\frac{{{{\text{d}}^2}{\text{y}}}}{{{\text{d}}{{\text{x}}^2}}} + \frac{{{\text{2dy}}}}{{{\text{dx}}}} + 2{\text{y}} = 0$$ are

Answer & Solution

67. It is given that y'' + 2y' + y = 0, y(0) = 0, y(1) = 0. What is y (0.5)?

Answer & Solution

68. A function y(t), such that y(0) = 1 and y(1) = 3e-1, is a solution of the differential equation $$\frac{{{{\text{d}}^2}{\text{y}}}}{{{\text{d}}{{\text{t}}^2}}} + 2\frac{{{\text{dy}}}}{{{\text{dt}}}} + {\text{y}} = 0.$$ Then y(2) is

Answer & Solution

69. A curve passes through the point (x = 1, y = 0) and satisfies the differential equation $$\frac{{{\text{dy}}}}{{{\text{dx}}}} = \frac{{{{\text{x}}^2} + {{\text{y}}^2}}}{{2{\text{y}}}} + \frac{{\text{y}}}{{\text{x}}}.$$ The equation that describes the curve is

Answer & Solution

70. The differential equation $$\frac{{{\text{dy}}}}{{{\text{dx}}}} + 4{\text{y}} = 5$$ is valid in the domain 0 ≤ x ≤ 1 with y(0) = 2.25. The solution of differential equation is

Answer & Solution

61.
The general solution of the differential equation $$\frac{{{{\text{d}}^2}{\text{y}}}}{{{\text{d}}{{\text{x}}^2}}} + 2\frac{{{\text{dy}}}}{{{\text{dx}}}} - 5{\text{y}} = 0$$ in terms of arbitrary constants K₁ and K₂ is

62.
General solution of the Cauchy-Euler equation $${{\text{x}}^2}\frac{{{{\text{d}}^2}{\text{y}}}}{{{\text{d}}{{\text{x}}^2}}} - 7{\text{x}}\frac{{{\text{dy}}}}{{{\text{dx}}}} + 16{\text{y}} = 0$$ is

63.
Biotransformation of an organic compound having concentration (x) can be modeled using an ordinary differential equation $$\frac{{{\text{dx}}}}{{{\text{dt}}}} + {\text{k}}{{\text{x}}^2} = 0,$$ where k is the reaction rate constant. If x = a at t = 0, the solution of the equation is

64.
The solution of initial value problem; $$\frac{{\partial {\text{u}}}}{{\partial {\text{x}}}} = 2\frac{{\partial {\text{u}}}}{{\partial {\text{t}}}} + {\text{u,}}$$ where u(x, 0) = 6e^-3x is

66.
The solutions of differential equation $$\frac{{{{\text{d}}^2}{\text{y}}}}{{{\text{d}}{{\text{x}}^2}}} + \frac{{{\text{2dy}}}}{{{\text{dx}}}} + 2{\text{y}} = 0$$ are

67.
It is given that y'' + 2y' + y = 0, y(0) = 0, y(1) = 0. What is y (0.5)?

68.
A function y(t), such that y(0) = 1 and y(1) = 3e^-1, is a solution of the differential equation $$\frac{{{{\text{d}}^2}{\text{y}}}}{{{\text{d}}{{\text{t}}^2}}} + 2\frac{{{\text{dy}}}}{{{\text{dt}}}} + {\text{y}} = 0.$$ Then y(2) is

69.
A curve passes through the point (x = 1, y = 0) and satisfies the differential equation $$\frac{{{\text{dy}}}}{{{\text{dx}}}} = \frac{{{{\text{x}}^2} + {{\text{y}}^2}}}{{2{\text{y}}}} + \frac{{\text{y}}}{{\text{x}}}.$$ The equation that describes the curve is

70.
The differential equation $$\frac{{{\text{dy}}}}{{{\text{dx}}}} + 4{\text{y}} = 5$$ is valid in the domain 0 ≤ x ≤ 1 with y(0) = 2.25. The solution of differential equation is