June 2011

Problem 03 | Separation of Variables

Problem 03
$xy \, y' = 1 + y^2$,   when   $x = 2$,   $y = 3$.
 

Solution 03
$xy \, y' = 1 + y^2$

$xy \dfrac{dy}{dx} = 1 + y^2$
 

Problem 06 | Separation of Variables

Problem 06
$2y \, dx = 3x \, dy$,   when   $x = 2$,   $y = -1$.
 

Solution 06
From Solution 04,
$\dfrac{x^2}{y^3} = c$
 

Problem 05 | Separation of Variables

Problem 05
$2y \, dx = 3x \, dy$,   when   $x = -2$,   $y = 1$.
 

Solution 05
From Solution 04,
$\dfrac{x^2}{y^3} = c$
 
 

Problem 07 | Separation of Variables

Problem 07
$y' = x \exp (y - x^2)$,   when   $x = 0$,   $y = 0$.
 

Solution 07

Problem 08 | Separation of Variables

Problem 08
$xy^2 \, dx + e^x \, dy = 0$,   when   $x \to \infty$,   $y \to \frac{1}{2}$.
 

Solution 08
$xy^2 \, dx + e^x \, dy = 0$

$\dfrac{xy^2 \, dx}{y^2 e^x} + \dfrac{e^x \, dy}{y^2 e^x} = 0$

$\dfrac{x \, dx}{e^x} + \dfrac{dy}{y^2} = 0$

$\displaystyle \int xe^{-x} \, dx + \int y^{-2} \, dy = 0$
 

For   $\displaystyle \int xe^{-x} \, dx$
Let
$u = x$,   $du = dx$

$dv = \int e^{-x} \, dx$,   $v = -e^{-x}$
 

Problem 09 | Separation of Variables

Problem 09
$(2a^2 - r^2) \, dr = r^3 \sin \theta \, d\theta$,   when   $\theta = 0$,   $r = a$.
 

Solution 09

Problem 10 | Separation of Variables

Problem 10
$v (dv / dx) = g$,   when   $x = x_o$,   $v = v_o$.
 

Solution 10

Problem 11 | Separation of Variables

Problem 11
$(1 - x)y' = y^2$
 

Solution 11

Problem 12 | Separation of Variables

Problem 12
$\sin x \sin y \, dx + \cos x \cos y \, dy = 0$
 

Solution 12
$\sin x \sin y \, dx + \cos x \cos y \, dy = 0$

$\dfrac{\sin x \sin y \, dx}{\sin y \cos x} + \dfrac{\cos x \cos y \, dy}{\sin y \cos x} = 0$

$\dfrac{\sin x \, dx}{\cos x} + \dfrac{\cos y \, dy}{\sin y} = 0$

$\displaystyle -\int \dfrac{-\sin x \, dx}{\cos x} + \int \dfrac{\cos y \, dy}{\sin y} = 0$

$-\ln (\cos x) + \ln (\sin y) = \ln c$

$\ln \dfrac{\sin y}{\cos x}= \ln c$

$\dfrac{\sin y}{\cos x}= c$

Problem 342 | Equilibrium of Parallel Force System

Problem 342
The wheel loads on a jeep are given in Fig. P-342. Determine the distance x so that the reaction of the beam at A is twice as great as the reaction at B.
 

Wheel loads of a jeep moving across the beam

 

Problem 343 | Equilibrium of Parallel Force System

Problem 343
The weight W of a traveling crane is 20 tons acting as shown in Fig. P-343. To prevent the crane from tipping to the right when carrying a load P of 20 tons, a counterweight Q is used. Determine the value and position of Q so that the crane will remain in equilibrium both when the maximum load P is applied and when the load P is removed.
 

Twenty tonner traveling crane

 

Problem 13 | Separation of Variables

Problem 13
$xy^3 \, dx + e^{x^2} \, dy = 0$
 

Solution 13

Problem 14 - 15 | Separation of Variables

Problem 14
$2y \, dx = 3x \, dy$

Solution 14

 

Problem 15
$my \, dx = nx \, dy$

Solution 15
$my \, dx = nx \, dy$

$m\dfrac{dx}{x} = n\dfrac{dy}{y}$

$m\ln x = n\ln y + \ln c$

$\ln x^m = \ln y^n + \ln c$

$\ln x^m = \ln cy^n$

Equilibrium of Parallel Force System

Conditions for Equilibrium of Parallel Forces
The sum of all the forces is zero.

$\Sigma F = 0$

 

The sum of moment at any point O is zero.

$\Sigma M_O = 0$

 

Problem 16 | Separation of Variables

Problem 16
$y' = xy^2$

Solution 16

Problem 17 | Separation of Variables

Problem 17
$\dfrac{dV}{dP} = -\dfrac{V}{P}$
 

Solution 17

Problem 18 | Separation of Variables

Problem 18
$ye^{2x} \, dx = (4 + e^{2x}) \, dy$
 

Solution 18
$ye^{2x} \, dx = (4 + e^{2x}) \, dy$

$\dfrac{ye^{2x} \, dx}{y(4 + e^{2x})} = \dfrac{(4 + e^{2x}) \, dy}{y(4 + e^{2x})}$

$\dfrac{e^{2x} \, dx}{4 + e^{2x}} = \dfrac{dy}{y}$

$\displaystyle \dfrac{1}{2} \int \dfrac{e^{2x} (2 \, dx)}{4 + e^{2x}} = \int \dfrac{dy}{y}$

$\frac{1}{2} \ln (4 + e^{2x}) = \ln y + \ln c$

$\frac{1}{2} \ln (4 + e^{2x}) = \ln cy$

$\ln (4 + e^{2x}) = 2\ln cy$

$\ln (4 + e^{2x}) = \ln (cy)^2$

$\ln (4 + e^{2x}) = \ln c^2y^2$

Problem 19 | Separation of Variables

Problem 19
$dr = b(\cos \theta \, dr + r \sin \theta \, d\theta)$
 

Solution 19
$dr = b(\cos \theta \, dr + r \sin \theta \, d\theta)$

$dr = b\cos \theta \, dr + br \sin \theta \, d\theta$

$dr - b\cos \theta \, dr = br \sin \theta \, d\theta$

$(1 - b\cos \theta) \, dr = br \sin \theta \, d\theta$

$\dfrac{(1 - b\cos \theta) \, dr}{r(1 - b\cos \theta)} = \dfrac{br \sin \theta \, d\theta}{r(1 - b\cos \theta)}$

$\dfrac{dr}{r} = \dfrac{b \sin \theta \, d\theta}{1 - b\cos \theta}$

Problem 20 | Separation of Variables

Problem 20
$xy \, dx - (x + 2) \, dy = 0$
 

Solution 20
$xy \, dx - (x + 2) \, dy = 0$

$\dfrac{xy \, dx}{y(x + 2)} - \dfrac{(x + 2) \, dy}{y(x + 2)} = 0$

$\dfrac{x \, dx}{x + 2} - \dfrac{dy}{y} = 0$

$\left( 1 - \dfrac{2}{x + 2} \right) \, dx - \dfrac{dy}{y} = 0$

$\displaystyle \int dx - 2 \int \dfrac{dx}{x + 2} - \int \dfrac{dy}{y} = 0$

$x - 2 \ln (x + 2) - \ln y = \ln c$

$x = \ln c + \ln y + 2 \ln (x + 2)$

$x = \ln c + \ln y + \ln (x + 2)^2$

$x = \ln cy(x + 2)^2$

$\ln e^x = \ln cy(x + 2)^2$

Problem 21 | Separation of Variables

Problem 21
$x^2 \, dx + y(x - 1) \, dy = 0$
 

Solution 21
$x^2 \, dx + y(x - 1) \, dy = 0$

$\dfrac{x^2 \, dx}{x - 1} + y \, dy = 0$
 

By long division
$\dfrac{x^2 \, dx}{x - 1} = x + 1 + \dfrac{1}{x - 1}$
 

Thus,
$\left[ (x + 1) + \dfrac{1}{x - 1} \right] \, dx + y \, dy = 0$

$\displaystyle \int (x + 1) \, dx + \int \dfrac{dx}{x - 1} + \int y \, dy = 0$

$\frac{1}{2}(x + 1)^2 + \ln |x - 1| + \frac{1}{2}y^2 + \ln c = 0$

$(x + 1)^2 + 2\ln |x - 1| + y^2 + 2\ln c = 0$

Problem 22 | Separation of Variables

Problem 22
$(xy + x) \, dx = (x^2y^2 + x^2 + y^2 + 1) \, dy = 0$
 

Solution 22
$(xy + x) \, dx = (x^2y^2 + x^2 + y^2 + 1) \, dy = 0$

$x(y + 1) \, dx = (x^2 + 1)(y^2 + 1) \, dy = 0$

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

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

By long division
$\dfrac{y^2 + 1}{y + 1} = y - 1 + \dfrac{2}{y + 1}$
 

Thus,
$\dfrac{x \, dx}{x^2 + 1} = \left( y - 1 + \dfrac{2}{y + 1} \right) \, dy = 0$

Problem 23 | Separation of Variables

Problem 23
$x \cos^2 y \, dx + \tan y \, dy = 0$
 

Solution 23

Equilibrium of Non-Concurrent Force System

There are three equilibrium conditions that can be used for non-concurrent, non-parallel force system.
 

The sum of all forces in the x-direction or horizontal is zero.

$\Sigma F_x = 0$   or   $\Sigma F_H = 0$

 

Problem 346 | Equilibrium of Non-Concurrent Force System

Problem 346
A boom AB is supported in a horizontal position by a hinge A and a cable which runs from C over a small pulley at D as shown in Fig. P-346. Compute the tension T in the cable and the horizontal and vertical components of the reaction at A. Neglect the size of the pulley at D.
 

Cable and boom structure

 

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