Differential Equations MCQs Question and Answer in Engineering Maths Page-6 section-1

51.
The solution of $$\frac{{{{\text{d}}^2}{\text{y}}}}{{{\text{d}}{{\text{x}}^2}}} + 2\frac{{{\text{dy}}}}{{{\text{dx}}}} + 17{\text{y}} = 0;$$ y(0) = 1, $$\frac{{{\text{dy}}}}{{{\text{dx}}}}\left( {\frac{\pi }{4}} \right) = 0$$ in the range 0 < x < $$\frac{\pi }{4}$$ is given by

A. $${{\text{e}}^{ - {\text{x}}}}\left( {\cos 4{\text{x}} + \frac{1}{4}\sin 4{\text{x}}} \right)$$

B. $${{\text{e}}^{\text{x}}}\left( {\cos 4{\text{x}} - \frac{1}{4}\sin 4{\text{x}}} \right)$$

C. $${{\text{e}}^{ - 4{\text{x}}}}\left( {\cos {\text{x}} - \frac{1}{4}\sin {\text{x}}} \right)$$

D. $${{\text{e}}^{ - 4{\text{x}}}}\left( {\cos 4{\text{x}} - \frac{1}{4}\sin 4{\text{x}}} \right)$$

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52.
The differential equation $$\frac{{{{\text{d}}^2}{\text{y}}}}{{{\text{d}}{{\text{x}}^2}}} + 16{\text{y}} = 0$$ for y(x) with the two boundary conditions $${\left. {\frac{{{\text{dy}}}}{{{\text{dx}}}}} \right|_{{\text{x}} = 0}} = 1$$ and $${\left. {\frac{{{\text{dy}}}}{{{\text{dx}}}}} \right|_{{\text{x}} = \frac{\pi }{2}}} = - 1$$ has

A. no solution

B. exactly two solutions

C. exactly one solution

D. infinitely many solutions

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53.
The solution of the differential equation $${{\text{x}}^2}\frac{{{{\text{d}}^2}{\text{y}}}}{{{\text{d}}{{\text{x}}^2}}} - {\text{x}}\frac{{{\text{dy}}}}{{{\text{dx}}}} + {\text{y}} = \log {\text{x}}$$ is

A. y = (c₁ + c₂x) log x + 2 log x + 3

B. y = (c₁ + c₂x²) log x + log x + 2

C. y = (c₁ + c₂x) log x + log x + 2

D. y = (c₁ + c₂log x) x + log x + 2

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54.
Solution of the differential equation $$3{\text{y}}\frac{{{\text{dy}}}}{{{\text{dx}}}} + 2{\text{x}} = 0$$ represents a family of

A. ellipses

B. circles

C. parabolas

D. hyperbolas

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55.
The solution of differential equation $$\frac{{{\text{dy}}}}{{{\text{dx}}}} - {{\text{y}}^2} = 1$$ satisfying condition y = 0 is

A. $${\text{y}} = {{\text{e}}^{{{\text{x}}^2}}}$$

B. $${\text{y}} = \sqrt {\text{x}} $$

C. $${\text{y}} = \cot \left( {{\text{x}} + \frac{\pi }{4}} \right)$$

D. $${\text{y}} = \tan \left( {{\text{x}} + \frac{\pi }{4}} \right)$$

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56.
The solution of the initial value problem $$\frac{{{\text{dy}}}}{{{\text{dx}}}} = - 2{\text{xy}};$$ y(0) = 2 is

A. 1 + e^-x²

B. 2e^-x²

C. 1 + e^x²

D. 2e^x²

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57.
An ordinary differential equation is given below.
$$\left( {\frac{{{\text{dy}}}}{{{\text{dx}}}}} \right)\left( {{\text{x}}\ln {\text{x}}} \right) = {\text{y}}$$
The solution for the above equation is
(Note: K denotes a constant in the options)

A. y = Kxe^x

B. y = Kxe^-x

C. y = Klnx

D. y = Kxlnx

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58.
The solution to the differential equation $$\frac{{{{\text{d}}^2} \cdot {\text{u}}}}{{{\text{d}}{{\text{x}}^2}}} - {\text{k}}\frac{{{\text{du}}}}{{{\text{dx}}}} = 0$$ is where k is constant, subjected to the boundary conditions u(0) = 0 and u(L) = U, is

A. $${\text{u}} = {\text{U}}\frac{{\text{x}}}{{\text{L}}}$$

B. $${\text{u}} = {\text{U}}\left( {\frac{{1 - {{\text{e}}^{{\text{kx}}}}}}{{1 - {{\text{e}}^{{\text{kL}}}}}}} \right)$$

C. $${\text{u}} = {\text{U}}\left( {\frac{{1 - {{\text{e}}^{ - {\text{kx}}}}}}{{1 - {{\text{e}}^{ - {\text{kL}}}}}}} \right)$$

D. $${\text{u}} = {\text{U}}\left( {\frac{{1 + {{\text{e}}^{{\text{kx}}}}}}{{1 + {{\text{e}}^{{\text{kL}}}}}}} \right)$$

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59.
The matrix form of the linear system $$\frac{{{\text{dx}}}}{{{\text{dt}}}} = 3{\text{x}} - 5{\text{y}}$$ and $$\frac{{{\text{dy}}}}{{{\text{dt}}}} = 4{\text{x}} + 8{\text{y}}$$ is

A. \[\frac{{\text{d}}}{{{\text{dt}}}}\left\{ {\begin{array}{*{20}{c}} {\text{x}} \\ {\text{y}} \end{array}} \right\} = \left[ {\begin{array}{*{20}{c}} 3&{ - 5} \\ 4&8 \end{array}} \right]\left\{ {\begin{array}{*{20}{c}} {\text{x}} \\ {\text{y}} \end{array}} \right\}\]

B. \[\frac{{\text{d}}}{{{\text{dt}}}}\left\{ {\begin{array}{*{20}{c}} {\text{x}} \\ {\text{y}} \end{array}} \right\} = \left[ {\begin{array}{*{20}{c}} 3&8 \\ 4&{ - 5} \end{array}} \right]\left\{ {\begin{array}{*{20}{c}} {\text{x}} \\ {\text{y}} \end{array}} \right\}\]

C. \[\frac{{\text{d}}}{{{\text{dt}}}}\left\{ {\begin{array}{*{20}{c}} {\text{x}} \\ {\text{y}} \end{array}} \right\} = \left[ {\begin{array}{*{20}{c}} 4&{ - 5} \\ 3&8 \end{array}} \right]\left\{ {\begin{array}{*{20}{c}} {\text{x}} \\ {\text{y}} \end{array}} \right\}\]

D. \[\frac{{\text{d}}}{{{\text{dt}}}}\left\{ {\begin{array}{*{20}{c}} {\text{x}} \\ {\text{y}} \end{array}} \right\} = \left[ {\begin{array}{*{20}{c}} 4&8 \\ 3&{ - 5} \end{array}} \right]\left\{ {\begin{array}{*{20}{c}} {\text{x}} \\ {\text{y}} \end{array}} \right\}\]

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60.
The solution of the equation $$\frac{{{\text{dQ}}}}{{{\text{dt}}}} + {\text{Q}} = 1$$ with Q = 0 at t = 0 is

A. Q(t) = e^-t - 1

B. Q(t) = 1 + e^-t

C. Q(t) = 1 - e^t

D. Q(t) = 1 - e^-t

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Differential Equations MCQs Question and Answer in Engineering Maths

51. The solution of $$\frac{{{{\text{d}}^2}{\text{y}}}}{{{\text{d}}{{\text{x}}^2}}} + 2\frac{{{\text{dy}}}}{{{\text{dx}}}} + 17{\text{y}} = 0;$$ y(0) = 1, $$\frac{{{\text{dy}}}}{{{\text{dx}}}}\left( {\frac{\pi }{4}} \right) = 0$$ in the range 0 < x < $$\frac{\pi }{4}$$ is given by

Answer & Solution

Answer & Solution

53. The solution of the differential equation $${{\text{x}}^2}\frac{{{{\text{d}}^2}{\text{y}}}}{{{\text{d}}{{\text{x}}^2}}} - {\text{x}}\frac{{{\text{dy}}}}{{{\text{dx}}}} + {\text{y}} = \log {\text{x}}$$ is

Answer & Solution

54. Solution of the differential equation $$3{\text{y}}\frac{{{\text{dy}}}}{{{\text{dx}}}} + 2{\text{x}} = 0$$ represents a family of

Answer & Solution

55. The solution of differential equation $$\frac{{{\text{dy}}}}{{{\text{dx}}}} - {{\text{y}}^2} = 1$$ satisfying condition y = 0 is

Answer & Solution

56. The solution of the initial value problem $$\frac{{{\text{dy}}}}{{{\text{dx}}}} = - 2{\text{xy}};$$ y(0) = 2 is

Answer & Solution

57. An ordinary differential equation is given below. $$\left( {\frac{{{\text{dy}}}}{{{\text{dx}}}}} \right)\left( {{\text{x}}\ln {\text{x}}} \right) = {\text{y}}$$ The solution for the above equation is (Note: K denotes a constant in the options)

Answer & Solution

58. The solution to the differential equation $$\frac{{{{\text{d}}^2} \cdot {\text{u}}}}{{{\text{d}}{{\text{x}}^2}}} - {\text{k}}\frac{{{\text{du}}}}{{{\text{dx}}}} = 0$$ is where k is constant, subjected to the boundary conditions u(0) = 0 and u(L) = U, is

Answer & Solution

59. The matrix form of the linear system $$\frac{{{\text{dx}}}}{{{\text{dt}}}} = 3{\text{x}} - 5{\text{y}}$$ and $$\frac{{{\text{dy}}}}{{{\text{dt}}}} = 4{\text{x}} + 8{\text{y}}$$ is

Answer & Solution

60. The solution of the equation $$\frac{{{\text{dQ}}}}{{{\text{dt}}}} + {\text{Q}} = 1$$ with Q = 0 at t = 0 is

Answer & Solution

51.
The solution of $$\frac{{{{\text{d}}^2}{\text{y}}}}{{{\text{d}}{{\text{x}}^2}}} + 2\frac{{{\text{dy}}}}{{{\text{dx}}}} + 17{\text{y}} = 0;$$ y(0) = 1, $$\frac{{{\text{dy}}}}{{{\text{dx}}}}\left( {\frac{\pi }{4}} \right) = 0$$ in the range 0 < x < $$\frac{\pi }{4}$$ is given by

53.
The solution of the differential equation $${{\text{x}}^2}\frac{{{{\text{d}}^2}{\text{y}}}}{{{\text{d}}{{\text{x}}^2}}} - {\text{x}}\frac{{{\text{dy}}}}{{{\text{dx}}}} + {\text{y}} = \log {\text{x}}$$ is

54.
Solution of the differential equation $$3{\text{y}}\frac{{{\text{dy}}}}{{{\text{dx}}}} + 2{\text{x}} = 0$$ represents a family of

55.
The solution of differential equation $$\frac{{{\text{dy}}}}{{{\text{dx}}}} - {{\text{y}}^2} = 1$$ satisfying condition y = 0 is

56.
The solution of the initial value problem $$\frac{{{\text{dy}}}}{{{\text{dx}}}} = - 2{\text{xy}};$$ y(0) = 2 is

57.
An ordinary differential equation is given below.
$$\left( {\frac{{{\text{dy}}}}{{{\text{dx}}}}} \right)\left( {{\text{x}}\ln {\text{x}}} \right) = {\text{y}}$$
The solution for the above equation is
(Note: K denotes a constant in the options)

58.
The solution to the differential equation $$\frac{{{{\text{d}}^2} \cdot {\text{u}}}}{{{\text{d}}{{\text{x}}^2}}} - {\text{k}}\frac{{{\text{du}}}}{{{\text{dx}}}} = 0$$ is where k is constant, subjected to the boundary conditions u(0) = 0 and u(L) = U, is

59.
The matrix form of the linear system $$\frac{{{\text{dx}}}}{{{\text{dt}}}} = 3{\text{x}} - 5{\text{y}}$$ and $$\frac{{{\text{dy}}}}{{{\text{dt}}}} = 4{\text{x}} + 8{\text{y}}$$ is

60.
The solution of the equation $$\frac{{{\text{dQ}}}}{{{\text{dt}}}} + {\text{Q}} = 1$$ with Q = 0 at t = 0 is