Differential Equations MCQs Question and Answer in Engineering Maths

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

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

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

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

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

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

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)

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

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

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