Introduction

A cricket ball is thrown vertically upwards with a velocity of v ft/sec. The retardation is $\propto v$ or $=kv$ . Find the maximum height reached (Y) and the time of flight to the vertex (T).

Prove that the Initial velocity u is given by: $u= - k\,Y + g\,T$

The acceleration = -kv - g

To find the time of Flight T $\frac{dv}{dt}\;= - (kv + g)$ $\therefore\;\;\;\;\;\;\frac{dv}{kv + g}\;= - dt$ thus $\frac{1}{k}Ln(kv + g)\;= - t + c$

When t = 0 v = u $\therefore\;\;\;\;\;\;\frac{1}{k}Ln(ku + g) = 0 + c$ $\therefore\;\;\;\;\;\;\frac{1}{k}Ln(kv + g)\;= - t + \frac{1}{k}Ln\,(ku + g)$ Thus $kt = Ln\left(\frac{ku + g}{kv + g} \right)$

At the vertex t = T and v = 0 so $kT = Ln\left(\frac{ku + g}{0 + g} \right)$ i.e. ${T = \frac{1}{k}\;Ln\left(\frac{ku}{g} + 1 \right)$

For Height Y $v\;\frac{dv}{dy}= - (kv + g)$ $v\;\frac{dv}{(kv + g)}= - dy$ $\int \frac{v}{kv + g}dv = -\int dy$ $\int \left[\frac{1}{k} - \frac{g/k}{k\,v + g} \right]dv= - y + D$ $\therefore\;\;\;\;\;\;\frac{1}{k}\;v - \frac{g}{k^2}\;Ln\,(kv + g)= - y + D$

But when y = 0, v = u so $\frac{1}{k}\;u - \frac{g}{k^2}\;Ln\,(ku + g) = 0 + D$ $\therefore\;\;\;\;\;\;\frac{1}{k}\;v - \frac{g}{k^2}\;Ln\,(kv + g)= - y + \frac{1}{k}u - \frac{g}{k^2}\,Ln(ku + g)$

At the Vertex v = 0 $\therefore\;\;\;\;\;0 - \frac{g}{k^2}\,Ln g= - Y + \frac{1}{k}u - \frac{g}{k^2}\,Ln(ku + g)$ $y = \frac{1}{k}u - \frac{g}{k^2}\,Ln\,\left(\frac{ku + g}{g} \right)$ $y = \frac{1}{k}u - \frac{g}{k^2}\,Ln\,\left(\frac{ku}{g} + 1 \right)$ $= \frac{1}{k}u - \frac{g}{k}\times \frac{1}{k}\,Ln\,\left(\frac{ku}{g} + 1 \right)$

The flight time is: ${T = \frac{1}{k}\;Ln\left(\frac{ku}{g} + 1 \right)$

Max Height is: $H= \frac{1}{k}u - \frac{g}{k}\times \frac{1}{k}\,Ln\,\left(\frac{ku}{g} + 1 \right)$

Basic trigonometrical examples

The reference page on Trigonometrical Formulae includes:-

Considering the first equation , this can be re-written as:-

Now if during the solution of a differential equation we arrive at :-

we can compare the right hand side with the right hand side of (5) and we can see that they are of the same form but $Cos\,\alpha$ has been replaced by "3" and $Sin\,\alpha$ by "4". Clearly this can not be correct as the Sine and Cosine can not have a value above unity but if we draw the following right angled triangle.

Values of Sine and Cosine $\alpha$ can be obtained which can be put into equation (5)

This can be re-arranged to satisfy the requirements of equation (6)

$\mathbf{\ddot{x} + 4\dot{x} + 3x = 15\;sin\,3t}$

To find the Particular Integral:-

$Let\;\;\;\;\;x = A\;cos\,3t + B\;sin\,3t$

$\therefore\;\;\;\;\;\dot{x} = -\,3A\;sin\,3t + 3B\;cos\,3t$

$\therefore\;\;\;\;\;\ddot{x} = -\,9A\;cos\,3t - 9B\;sin\,3t$

Substituting in the original equation.

$\,9A\;cos\,3t - 9B\;sin\,3t\;-12\,A\;sin\,3t + 12\,B\;cos\,3t + 3\,A\;cos\,3t + 3\,B\;sin\,3t = 5\;sin\,3t$

Equating the coefficients of sin 3t and cos 3t

$(12B + 3A - 9A)\;cos\,3t = 0$

$(3B - 9B - 12A)\;sin\,3t = 5\;sin\,3t$

$\;\;\;\;\;12\;B = 6\;A\;\;\;\;or\;\;\;\;A = 2B$

$\;\;\; - 6B - 12A = 5$

$\;\;\;\;\;x\;= - \frac{\sqrt{5}}{6}\;sin\,(3t + \alpha )$

To find the Complementary Function using the D factor:-

$\ddot{x} + 4\dot{x} + 3x = 0$

$\;\;\;\;D^2\;+4D + 3 = 0$

$\;\;\;D = -3\;\;\;\;or\;\;\;\;-1$

$A\;e^{-t} + B\,e^{-3t}$

And the General Solution is:-

$\mathbf{x\;= - \frac{\sqrt{5}}{6}\;sin\,(3t + \alpha ) + A\;e^{-t} + B\,e^{-3t}\;\;\;\;where\;\;tan\,\alpha = 2}$

Find the General Solution to the following equation.

To find the Particular Integral:-

$Let\;\;\;\;\;x = A\;cos\,4t + B\;sin\,4t$

$\therefore\;\;\;\;\;\dot{x} = -4A\;sin\,4t + 4B\;cos\,4t$

$\therefore\;\;\;\;\;\ddot{x} = -16A\;cos\,4t - 16B\;sin\,4t$

Re-writing equation (1)

$\therefore\;\;\;-\,16A\,cos\,4t - 16B\,sin\,4t + 9A\,cos\,4t + 9B\,sin\,4t = 15\,cos\,4t$

$\;\;\;\;(-16A + 9A)\,cos\,4t + (9B - 16B)\,sin\,4t = 15\,cos\,4t$

Equating the coefficients:-

-7A = 15 and -7B = 0

To find the complementary function:-

$D^2 + 9D = 0$

$\therefore\;\;\;\;\;\;D = 3$

$\;\;\;x = A\,cos\,3t + B\,sin\,3t$

And the General Solution is :

$\mathbf{x = -\,\frac{\sqrt{5}}{6}\;cos\,4t + A\,cos\,3t + B\,sin\,3t}$

To find the Particular Integral:

$Let\;\;\;\;\;y = A\,cos\,5t + B\,sin\,5t$

$\therefore\;\;\;\;\;\dot{y} = -\,5A\,sin\,5t + 5B\,cos\,5t$

$\therefore\;\;\;\;\;\ddot{y} = -\,25A\,cos\,5t - 25B\,sin\,5t$

Substituting the above values in equation (1)

$- 25\,A\,cos\,5t - 25\,B\,sin\,5t - 20\,\a\,sin\,5t + 20\,B\,cos\,5t + 40\,A\,cos\,5t + 40\,B\,sin\,5t = 500\,cos\,5t$

$(- 25A\;+20B + 40A)\,cos\,5t = 500\,cos\,5t$

$(-\,25B\;-\,20A\;+\,40B)\;sin\,5t = 0$

$\therefore\;\;\;\;15B - 20A = 0$

$And\;\;\;\;15A + 20B = 500$

from which B = 16 and A = 12

$\therefore\;\;\;\;y = 12\;cos\,5t + 16\;sin\,5t$

$= 20\;cos\,(5t - \alpha )\;\;\;\;\;where\;tan\,\alpha = \frac{4}{3}$

To find the Complementary Function:

$\ddot{y} + 4\dot{y} + 40\,y = 0$

$D^2 + 4D + 40 = 0$

$\therefore\;\;\;\;D = -\,2\,\pm \sqrt{-\,36} = -\,2\,\pm j\,6$

$\therefore\;\;\;\;y = e^{-2t}\;(A\,cos\,6t + B\,sin\,6t)$

Thus the General Solution is :

$\mathbf{y = 20\,cos\,(5t - \alpha ) + e^{-2t}\;(A\,cos\,6t + B\,sin\,6t)\;\;\;\;\;where\;tan\,\alpha = \frac{4}{3}}$

Solve $\ddot{x} + 4x = 4\,e^2t$

$\lambda = 2\;\;\;;\;\;\;k = 0\;\;\;;\;\;\;n = 2\;\;\;;\;\;\;a = 4$

Using the D operator

$D^2 + 4 = 0$

$\therefore\;\;\;\;D = \pm j2$

Therefore the General Solution is given by:-

$x = \frac{1}{2}\,e^{\frac{1}{2}t} + A\;cos\,2t + B\;sin\,2t$

$4\,\frac{d^2y}{dx^2} + y = 5\,e^{-\,0.25x}$

Re-write equation in the form:

$\frac{d^2y}{dx^2} + \frac{y}{4} = \frac{5}{4}\,e^{-0.25}$

Using equations (108) and (113)

$\lambda = -\,0.25\;\;;\;\;a = \frac{5}{4}\;\;;\;\;n^2 = \frac{1}{4}$

$\therefore\;\;\;\;\,A\;\left(\frac{1}{16} + \frac{4}{16} \right) = \frac{5}{4}$

$Thus\;\;\;\;A = \frac{5}{4}\times\frac{16}{5}$

Using the D operator:-

$D^2 + \frac{1}{4} = 0\;\;\;\;\;\;\therefore\;\;\;\;D = \pm j\,\frac{1}{2}$

$\mathbf{Hence\;\;\;\;\;y = 4\,e^{-\frac{1}{4}x} + A\;cos\,\frac{1}{2}x + B\;sin\,\frac{1}{2}x}$

Find the Complete Solution to:-

$Let\;\;\;\;\;y = A + B\,t + C\,t^2$

$\therefore\;\;\;\;\;\dot{y} = B + 2\,C\,t$

$\ddot{y} = 2\,C$

Substituting these values in equation (1)

$2C + 3(B + 2CT) + 2(A + BT + CT^2) = 1 + 12T + 2T^2$

$C = 1\;\;;\;\;B = 3\;\;;\;\;C = -\,5$

Therefore the Particular integral is:

$y = -\,5 + 3t + t^2$

Using the D Factor to find the Complimentary Function:-

From Equation (138)

$D^2 + 3D + 2 = 0$

Hence D = -1 or - 2

Therefore the Complete Solution is given by:

$\mathbf{y = -\,5 + 3t + t^2 + A_1e^{-t} + A_2e^{-2t}}$

Find the Complete Solution of:

$\frac{d^2y}{dx^2} - 4\,y = x^2$

$y = A + Bx + Cx^2$

$\dot{y} = B + 2\,Cx$

$\ddot{y} = 2\,C$

Substituting in equation (147)

$2\,C - 4\,A - 4\,Bx - 4\,Cx^2$

From which A = 0 ; B = -1/8 ; C = -1/4

$y = -\,\frac{x}{8} - \frac{1}{4}x^2$

Using the D factor to find the Complementary Function

$(D^2 - 4) = 0\;\;\;\;and\;\;\;therefore\;\;\;\;D = -\,2\;\;or\;\;+2$

$\mathbf{Hence\;\;\;y = A\,e^{2x} + B\,e^{-2x} - \frac{x}{8} - \frac{x^2}{4}}$