Differential Equations MCQs Question and Answer in Engineering Maths

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21.
With K as a constant, the solution possible for the first order differential equation $$\frac{{{\text{dy}}}}{{{\text{dx}}}} = {{\text{e}}^{ - 3{\text{x}}}}$$   is

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24.
With initial condition x(1) = 0.5, the solution of the differential equation, $${\text{t}}\frac{{{\text{dx}}}}{{{\text{dt}}}} + {\text{x}} = {\text{t}}$$   is

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25.
For the differential equation $$\frac{{{{\text{d}}^2}{\text{x}}}}{{{\text{d}}{{\text{t}}^2}}} + 6\frac{{{\text{dx}}}}{{{\text{dt}}}} + 8{\text{x}} = 0$$     with initial conditions x(0) = 1 and $${\left. {\frac{{{\text{dx}}}}{{{\text{dy}}}}} \right|_{{\text{t}} = 0}} = 0,$$   the solution is

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26.
Transformation to linear form by substituting v = y1 - n of the equation
$$\frac{{{\text{dy}}}}{{{\text{dt}}}} + {\text{p}}\left( {\text{t}} \right){\text{y}} = {\text{q}}\left( {\text{t}} \right){{\text{y}}^{\text{n}}};\,{\text{n}} > 0$$      will be

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28.
For the equation, $$\frac{{{\text{dy}}}}{{{\text{dx}}}} + 7{{\text{x}}^2}{\text{y}} = 0,$$    if $${\text{y}}\left( {\text{0}} \right) = \frac{3}{7},$$   then the value of y(1) is

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29.
Consider two solutions x(t) = x1(t) and x(t) = x2(t) of the differential equation $$\frac{{{{\text{d}}^2}{\text{x}}\left( {\text{t}} \right)}}{{{\text{d}}{{\text{t}}^2}}} + {\text{x}}\left( {\text{t}} \right) = 0,\,{\text{t}} > 0,$$     such that $${{\text{x}}_2} = 0,\,{\left. {\frac{{{\text{d}}{{\text{x}}_2}\left( {\text{t}} \right)}}{{{\text{dt}}}}} \right|_{{\text{t}} = 0}} = 1.$$     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 $${\text{t}} = \frac{\pi }{2}$$  is

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30.
If $$\frac{{{{\text{d}}^2}{\text{y}}}}{{{\text{d}}{{\text{t}}^2}}} + {\text{y}} = 0$$   under the conditions y = 1, $$\frac{{{\text{dy}}}}{{{\text{dt}}}} = 0,$$  when t = 0, then y is equal to

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