Differential Equations MCQs Question and Answer in Engineering Maths

Given that \[{\rm{\ddot x}} + 3{\rm{x}} = 0,\] and x(0) = 1, \[{\rm{\dot x}}\left( 0 \right) = 0,\] what is x(1) ?

A) -0.99 B) -0.16 C) 0.16 D) 0.99

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Answer: Option B

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

A) $${\text{u}} = \frac{{{\text{Ux}}}}{{\text{L}}}$$ 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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Answer: Option B

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A solution of the following differential equation is given by $$\frac{{{{\text{d}}^2}{\text{y}}}}{{{\text{d}}{{\text{x}}^2}}} - 5\frac{{{\text{dy}}}}{{{\text{dx}}}} + 6{\text{y}} = 0$$

A) y = e^2x + e^-3x B) y = e^2x + e^3x C) y = e^-2x + e^3x D) y = e^-2x + e^-3x

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Answer: Option B

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