second order ode

The general solution of $\frac{d^2 y}{d x^2}-6 \frac{d y}{d x}+9 y=0$ is

The position of a particle y(t) is described by the differential equation : $${{{d^2}y} \over {d{t^2}}} = - {{dy} \over {dt}} - {{5y} \over 4}$$. The initial conditions are y(0) =...

The general solution of the differential equation $$\,\,{{{d^2}y} \over {d{x^2}}} + 2{{dy} \over {dx}} - 5y = 0\,\,\,$$ in terms of arbitrary constants $${K_1}$$ and $${K_2}$$ is

The particular solution of the initial value problem given below is $$\,\,{{{d^2}y} \over {d{x^2}}} + 12{{dy} \over {dx}} + 36y = 0\,\,$$ with $$\,y\left( 0 \right) = 3\,\,$$ and $...

The Solution of the differential equation $$\,{{{d^2}y} \over {d{t^2}}} + 2{{dy} \over {dt}} + y = 0\,\,$$ with $$\,y\left( 0 \right) = {y^1}\left( 0 \right) = 1\,\,$$ is

With initial values $$\,\,\,y\left( 0 \right) = y'\left( 0 \right) = 1,\,\,\,$$ the solution of the differential equation $$\,\,{{{d^2}y} \over {d{x^2}}} - 4{{dy} \over {dx}} + 4y...

If $$a$$ and $$b$$ are constants, the most general solution of the differential equation $$\,{{{d^2}x} \over {d{t^2}}} + 2{{dx} \over {dt}} + x = 0$$ is

A function $$n(x)$$ satisfies the differential equation $${{{d^2}n\left( x \right)} \over {d{x^2}}} - {{n\left( x \right)} \over {{L^2}}} = 0$$ where $$L$$ is a constant. The bound...

The solution of the differential equation $${k^2}{{{d^2}y} \over {d\,{x^2}}} = y - {y_2}\,\,$$ under the boundary conditions (i) $$y = {y_1}$$ at $$x=0$$ and (ii) $$y = {y_2}$$ at...

For the differential equation $${{{d^2}y} \over {d{x^2}}} + {k^2}y = 0,$$ the boundary conditions are (i) $$y=0$$ for $$x=0$$ and (ii) $$y=0$$ for $$x=a$$ The form of non-zero solu...

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

$$y = {e^{ - 2x}}$$ is a solution of the differential equation $$\,{y^{11}} + {y^1} - 2y = 0$$