Which of the following is a solution of the differential equation $$\,{{{d^2}y} \over {d{x^2}}} + p{{dy} \over {dx}} + \left( {q + 1} \right)y = 0?$$ Where $$p=4, q=3$$

Find the solution of the differential equation $$\,{{{d^2}u} \over {d{t^2}}} + {\lambda ^2}y = \cos \left( {wt + k} \right)$$ with initial conditions $$\,y\left( 0 \right) = 0,\,\,{{dy\left( 0 \right)} \over {dt}} = 0.$$ Here $$\lambda ,$$ $$w$$ and $$k$$ are constants. Use either the method of undetermined coefficients (or) the operator $$\left( {D = {\raise0.5ex\hbox{$\scriptstyle d$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle {dt}$}}} \right)$$

The radial displacement in a rotating disc is governed by the differential equation $$\,\,{{{d^2}u} \over {d{x^2}}} + {1 \over x}{{du} \over {dx}} - {u \over {{x^2}}} = 8x.\,\,\,$$ where $$u$$ is the displacement and $$x$$ is the radius. If $$u=0$$ at $$x=0$$ and $$u=2$$ at $$x=1,$$ calculate the displacement at $$\,x = {1 \over {2.}}$$

Solve for $$y$$ if $${{{d^2}y} \over {d{t^2}}} + 2{{dy} \over {dt}} + y = 0$$ with $$y(0)=1$$ and $${y^1}\left( 0 \right) = 2$$