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Maths › Differential Equations ›

Differential Equations Worked Examples 1

This section contains worked examples of the type of differential equation which can be solved by integration

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Contents

Introduction
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Introduction

This page contains worked examples of the type of differential equation which can be solved by direct Integration. The questions are graded from fairly basic to some which were set as part of a London University Engineering degree. Most workings are hidden so that you can try the question before looking at the solution.

Example 1:

$\mathbf{\frac{dy}{dx} = \frac{1}{cos^2\,\frac{1}{2}x}}$

(1)

Workings

Thus

$y = 2 \tan(\frac{1}{2} x) + C$

(3)

Example 2:

$\mathbf{\frac{d^2y}{dx^2} = 2 + \frac{1}{x}}$

(4)

Workings

$\therefore\;\;\;y = x^2 + x\,ln\,x + Cx + K$

(6)

Example 3:

$\mathbf{\frac{dx}{dy} = 1 - y}$

(7)

$x = a (\tan (ax + c))$

(8)

Workings

$\therefore\;\;\;x = y - \frac{y^2}{2} + C$

(10)

Example 4:

$\mathbf{\frac{dy}{dx} = a^2 + y^2}$

(11)

Workings

$x = a\;(tan\,(ax + c))$

(13)

Example 5:

$\mathbf{\frac{dy}{dx} = \frac{x^2 - 1}{y}}$

(14)

Workings

$\therefore\;\;\frac{y^2}{2} = \frac{x^3}{3} - x + C$

(16)

Example 6:

$\mathbf{\frac{dy}{dx} = \frac{y}{x^2 - 1}}$

(17)

Workings

$\therefore\;\;y = C\;\sqrt{-\;\left(\frac{1 + x}{1 - x} \right)}$

(20)

Example 7:

$\mathbf{\sqrt{x - 1}\;\;dy + \sqrt{y - 1}\;\;dx = 0}$

(21)

Workings

$\therefore\;\; (y - 1) = 2\,(x - 1) + K\,(x - 1)^{\frac{1}{2}};+\;K^2\;\;where\;K = \frac{C}{2}$

(25)

Example 8:

$\mathbf{\frac{1}{r}\;\frac{dr}{d\theta } = cot\,\theta }$

(26)

Workings

$r = C\;sin\;\theta$

(29)

Example 9:

$\mathbf{\frac{d}{dr}\left(r\,\frac{dp}{dr} \right) = 0}}$

(30)

Workings

$\therefore\;\;p = C\,\ln\,r + K$

(33)

Example 10:

$\mathbf{x\,y\;dx - (1 + x^2)\;dy = 0}$

(34)

Workings

$thus\;\;\;\;y = K\;\sqrt{(1 + x^2)}$

(38)

Example 11:

$\mathbf{tan\,x\;cot\,y\;\frac{dy}{dx} = 2}$

(39)

Workings

$\therefore\;\;\;\;sin\,y = A\;sin^2\,x$

(42)

Example 12:

$\mathbf{x\,dy + \sqrt{y^2 + 4}\;\;dx = 0}$

(43)

Workings Then

$y = K\,x - \frac{1}{K\,x}$

(51)

Example 13:

$\mathbf{\frac{dy}{dx} = \frac{(1 + x)\;y^2}{x^3}}$

(52)

Workings

$y=\frac{2x^2}{2x + 1 - 2Cx^2}$

(56)

Example 14:

Find y if $\frac{d^4}{dx^4} = \frac{w}{EI}$ where w;E;and I are constants and $y = 0\;and\;\frac{d^2y}{dx^2} = 0$ at both x = 0 and x = l

Workings Thus

$y = \frac{w}{EI}\left(\frac{x^4}{24} - \frac{x^3l}{12} + \frac{l^3x}{24} \right)}$

(64)

Example 15:

Find y if

$\displaystyle \frac{d^4y}{dx^4} = \frac{w}{EI}$
y=0 at x=0 and x=l
$\displaystyle \frac{dy}{dx} = 0$ at x=0
$\displaystyle \frac{d^2 y}{dx^2} = 0$ at x = l

Workings Hence

$y = \frac{wx^2}{48EI}\left(2x^2 - 5xl + 3l^2 \right)$

(78)

Example 16:

Solve the following equation given that y=0 at x=0

$\mathbf{\frac{dy}{dx} = \sec\,y}$

(79)

Workings

$\therefore\;\;\;\;y = sin^{-1}x$

(82)

Example 17:

Solve the following equation given that x=1 at y=2

$\mathbf{(1 - x^2)\;y^2\;dx + dy = 0}$

(83)

Workings

$\frac{1}{y} = x - \frac{x^3}{3} - \frac{1}{6}$

(86)

Example 18:

Solve the following equation :-

$\mathbf{\frac{r\;tan\theta }{a^2 - r^2}\frar }{d\theta} = 1}$

(87)

given that

at $\theta = \frac{\pi }{4}$

Workings

$(a^2 - r^2) = \frac{a^3}{2}\;cosec^2\theta$

(94)

Example 19:

Solve the following equation given that v = o when $\phi = 0$

$\mathbf{\frac{dv}{d\phi } = e^{-v}\;sin\,\phi }}$

(95)

Workings

$v = \ln(2 - cos\,\phi)$

(99)

Example 20:

Show that the general solution to the equation;-

$\mathbf{(1 + y^2)\;dx + (1 + x^2)\;dy = 0}$

(100)

can be written in the form $\displaystyle y = \frac{a - x}{1 + ax}$ where a is an arbitrary constant. Find the particular solution which makes y = 2 when x = 1

Workings

$y = \left(\frac{-3 - x}{1-3x} \right) = \left(\frac{x+3}{3x-1} \right)$

(107)

Example 21:

$\mathbf{If\;\;(x-xy^2)dx + (y+yx^2)dy = 0\;\;\;Prove \;that\;y^2 = 1+C(1+x^2)}$

(108)

Workings So

(113)

Example 22:

If a and b are the radii of concentric spherical conductors at potentials of

respectively, then V is the potential at a distance r from the centre. Find the value of V if;-

$\mathbf{\frac{d}{dr}\left(r^2\;\frac{dV}{dr} \right) = 0}$

(114)

and

at r = a and V = 0 at r = b

Workings

$V = \left(\frac{1}{r} - \frac{1}{b} \right)\left(\frac{abV_1}{b-a} \right)$

(124)

Example 23:

For vertical motion of a particle under gravity, $\frac{d^2y}{dx^2} = g$ where y is the distance of the particle below some fixed point. Find the general expression for y in terms of t and also find the particular solution which makes y = 50 and $\frac{dy}{dt} = -\;20$ at t = 0. Take g as 32 ft/sec.sq.

Workings

Example 24:

A body which weighs 6 lbs, is acted upon by a force which diminishes uniformly with time from 1 lb weight to 1/2 lbs. wt. in 20 secs. If it starts from rest find its greatest velocity during the 20 secs. and how far it moves in this time.

Workings

Example 25:

A train of mass 300 tons travels along the level at a uniform speed of 80 ft/sec. against a resistance of 14 lb. wt/ton. It then climbs an incline of 1 in 100. Assuming that the horse power and the resistance remain constant, show that when the speed is v ft/sec.the retardation is (2v - 80)/5v ft/sec squared. Find the time taken for the speed to fall from 80 to 60 ft/sec. and how far the train has traveled in this time.(LU)

NOTE: One ton is 2240 lbs; One Horse power is 550 ft.lbs/sec; g = 32 ft/sec squared

Workings

Example 26:

The effective Horse Power of a ship of mass 10,000 tons is 6000 and its full speed is 20 mph. Assuming that the resistance to motion varies as the square of the speed and that the horse-power is constant, find the distance traveled from rest in attaining a speed of 16 mph (LU)

Note 60 mph is 88 ft/sec.

Workings

Example 27:

A body of weight W is shot upwards under gravity at a velocity of u and the air resistance is $\frac{W}{g}KV^2$ where K is a constant. Show that $\displaystyle V^2 = \left(u^2+\frac{g}{K} \right)e^{-2Kx}-\frac{g}{K}$ and the greatest height reached by the body is $\displaystyle \frac{1}{2K}\ln\left(1+\frac{Ku^2}{g} \right)$ . If the terminal velocity of the body when dropping under gravity is 1000 ft/sec. find the maximum height approximately when u = 2,500 ft./sec.

Workings

Example 28:

In a reservoir receiving flood water and discharging over a weir, if H ft. is the height of the water above the sill of weir at any time t mins it is known that $\frac{dt}{dH} = \frac{100}{10 - H^{\frac{3}{2}}}$ . By graphical integration or otherwise estimate the time required for the level to rise 3 ft. from the instant when overflow commences. (LU)

MISSING IMAGE!

13108/img_diff_eqn_100_1.jpg cannot be found in /users/13108/img_diff_eqn_100_1.jpg. Please contact the submission author.

Workings

Example 29:

A particle of mass m moves in a straight line in a medium whose resistance is $\displaystyle mk\,(v^3+\lambda ^2v)$ where v is the speed and k, $\lambda$ are positive constants, there being no other forces acting on the particle.If the particle moves through the origin with given velocity u, find its distance from the origin when its velocity is v and show that whatever the velocity u, the maximum distance from the origin is $\frac{\pi}{2k\lambda}$ (LU)

Workings

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Last Modified: 2009-01-16 12:46:32 Page Rendered: 2010-07-31 14:59:49

H	H^{1.5}	10-H^{1.5}	100/(10-H^{1.5}	Simpson Multiplier	Total
0.00	0.00	10.00	10.00	1	10.00
0.25	0.13	9.88	10.13	4	40.51
0.50	0.35	9.65	10.37	2	20.73
0.75	0.65	9.35	10.69	4	42.78
1.00	1.00	9.00	11.11	2	22.22
1.25	1.40	8.60	11.62	4	46.50
1.50	1.84	8.16	12.25	2	24.50
1.75	2.32	7.68	13.01	4	52.05
2.00	2.83	7.17	13.94	2	27.89
2.25	3.38	6.63	15.09	4	60.38
2.50	3.95	6.05	16.54	2	33.07
2.75	4.56	5.44	18.38	4	73.53
3.00	5.20	4.80	20.82	1	20.82
				Total:	474.98