August 2011

Problem 03 | Integrating Factors Found by Inspection

Problem 03
$(x^3y^3 + 1) \, dx + x^4y^2 \, dy = 0$
 

Solution 03
$(x^3y^3 + 1) \, dx + x^4y^2 \, dy = 0$

$x^3y^3 \, dx + dx + x^4y^2 \, dy = 0$

$(x^3y^3 \, dx + x^4y^2 \, dy) + dx = 0$

$x^3y^2(y \, dx + x \, dy) + dx = 0$

$x^3y^2 \, d(xy) + dx = 0$
 

Divide by x both sides
$x^2y^2 \, d(xy) + \dfrac{dx}{x} = 0$

$\displaystyle \int (xy)^2 \, d(xy) + \int \dfrac{dx}{x} = 0$

$\frac{1}{3}(xy)^3 + \ln x = -\ln c$

$x^3y^3 + 3\ln x = -3\ln c$

$x^3y^3 = -3\ln c - 3\ln x$

$x^3y^3 = -3(\ln c + \ln x)$

Problem 04 | Integrating Factors Found by Inspection

Problem 04
$2t \, ds + s(2 + s^2t) \, dt = 0$
 

Solution 04
$2t \, ds + s(2 + s^2t) \, dt = 0$

$2t \, ds + 2s \, dt + s^3t \, dt = 0$

$(2t \, ds + 2s \, dt) + s^3t \, dt = 0$

$2(t \, ds + s \, dt) + s^3t \, dt = 0$

$2 \, d(st) + s^3t \, dt = 0$

$\dfrac{2 \, d(st)}{s^3t^3} + \dfrac{s^3t \, dt}{s^3t^3} = 0$

$\dfrac{2 \, d(st)}{(st)^3} + \dfrac{dt}{t^2} = 0$

$2 (st)^{-3} \, d(st) + t^{-2} \, dt = 0$

$\displaystyle 2\int (st)^{-3} \, d(st) + \int t^{-2} \, dt = 0$

$-(st)^{-2} - t^{-1} = -c$

Annuities and Capitalized Cost

Annuity
An annuity is a series of equal payments made at equal intervals of time. Financial activities like installment payments, monthly rentals, life-insurance premium, monthly retirement benefits, are familiar examples of annuity.
 

Annuity can be certain or uncertain. In annuity certain, the specific amount of payments are set to begin and end at a specific length of time. A good example of annuity certain is the monthly payments of a car loan where the amount and number of payments are known. In annuity uncertain, the annuitant may be paid according to certain event. Example of annuity uncertain is life and accident insurance. In this example, the start of payment is not known and the amount of payment is dependent to which event.
 

Types of Annuities

Types of Simple Annuities
In engineering economy, annuities are classified into four categories. These are: (1) ordinary annuity, (2) annuity due, (3) deferred annuity, and (4) perpetuity. These four are actually simple annuities described in the previous page.
 

Derivation of Formula for the Future Amount of Ordinary Annuity

The sum of ordinary annuity is given by
 

$F = \dfrac{A[ \, (1 + i)^n - 1 \, ]}{i}$

 

To learn more about annuity, see this page: ordinary annuity, deferred annuity, annuity due, and perpetuity.
 

Derivation

Figure for Derivation of Sum of Ordinary Annuity

 

$F = \text{ Sum}$

$F = A + F_1 + F_2 + F_3 + \cdots + F_{n-1} + F_n$

$F = A + A(1 + i) + A(1 + i)^2 + A(1 + i)^3 + \cdots + A(1 + i)^{n-1} + A(1 + i)^n$
 

Interest and Discount

Interest
The amount of money earned for the use of borrowed capital is called interest. From the borrower’s point of view, interest is the amount of money paid for the capital. For the lender, interest is the income generated by the capital which he has lent.
 

There are two types of interest, simple interest and compound interest.
 

Simple Interest

Simple Interest
In simple interest, only the original principal bears interest and the interest to be paid varies directly with time.
 

The formula for simple interest is given by

$I = Prt$

 

The future amount is

Compound Interest

In compound interest, the interest earned by the principal at the end of each interest period (compounding period) is added to the principal. The sum (principal + interest) will earn another interest in the next compounding period.
 

Consider \$1000 invested in an account of 10% per year for 3 years. The figures below shows the contrast between simple interest and compound interest.
 

Problem 05 | Integrating Factors Found by Inspection

Problem 05
$y(x^4 - y^2) \, dx + x(x^4 + y^2) \, dy = 0$
 

Problem 05
$y(x^4 - y^2) \, dx + x(x^4 + y^2) \, dy = 0$

$x^4y \, dx - y^3 \, dx + x^5 \, dy + xy^2 \, dy = 0$

$(x^4y \, dx + x^5 \, dy) + (xy^2 \, dy - y^3 \, dx) = 0$

$x^4(y \, dx + x \, dy) + y^2(x \, dy - y \, dx) = 0$

$(y \, dx + x \, dy) + \dfrac{y^2(x \, dy - y \, dx)}{x^4} = 0$

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

Problem 06 - 07 | Integrating Factors Found by Inspection

Problem 06
$y(y^2 + 1) \, dx + x(y^2 - 1) \, dy = 0$

Solution 06
$y(y^2 + 1) \, dx + x(y^2 - 1) \, dy = 0$

$y^3 \, dx + y \, dx + xy^2 \, dy - x \, dy = 0$

$(xy^2 \, dy + y^3 \, dx) + (y \, dx - x \, dy) = 0$

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

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

$d(xy) + d\left( \dfrac{x}{y} \right) = 0$

$\displaystyle \int d(xy) + \int d\left( \dfrac{x}{y} \right) = 0$

$xy + \dfrac{x}{y} = c$

$xy^2 + x = cy$

Problem 11 | Integrating Factors Found by Inspection

Problem 11
$y(x^2 + y^2 - 1) \, dx + x(x^2 + y^2 + 1) \, dy = 0$
 

Solution 11
$y(x^2 + y^2 - 1) \, dx + x(x^2 + y^2 + 1) \, dy = 0$

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

$[ \, y(x^2 + y^2) \, dx + x(x^2 + y^2) \, dy \, ] - (y \, dx - x \, dy) = 0$

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

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

$d(xy) - d [ \, \arctan (y/x) \, ] = 0$

$\displaystyle \int d(xy) - \int d[ \, \arctan (y/x) \, ] = 0$

Laplace Transform of Derivatives

For first-order derivative:
$\mathcal{L} \left\{ f'(t) \right\} = s \, \mathcal{L} \left\{ f(t) \right\} - f(0)$
 

For second-order derivative:
$\mathcal{L} \left\{ f''(t) \right\} = s^2 \mathcal{L} \left\{ f(t) \right\} - s \, f(0) - f'(0)$
 

For third-order derivative:
$\mathcal{L} \left\{ f'''(t) \right\} = s^3 \mathcal{L} \left\{ f(t) \right\} - s^2 f(0) - s \, f'(0) - f''(0)$
 

For nth order derivative:

$\mathcal{L} \left\{ f^n(t) \right\} = s^n \mathcal{L} \left\{ f(t) \right\} - s^{n - 1} f(0) - s^{n - 2} \, f'(0) - \dots - f^{n - 1}(0)$

 

Problem 01 | Laplace Transform of Derivatives

Problem 01
Find the Laplace transform of   $f(t) = t^3$   using the transform of derivatives.
 

Solution 01
$f(t) = t^3$       ..........       $f(0) = 0$

$f'(t) = 3t^2$       ..........       $f'(0) = 0$

$f''(t) = 6t$       ..........       $f''(0) = 0$

$f'''(t) = 6$
 

$\mathcal{L} \left\{ f'''(t) \right\} = s^3 \mathcal{L} \left\{ f(t) \right\} - s^2 f(0) - s \, f'(0) - f''(0)$

Problem 02 | Laplace Transform of Derivatives

Problem 02
Find the Laplace transform of   $f(t) = \sin^2 t$   using the transform of derivatives.
 

Solution 02
$f(t) = \sin^2 t$       ..........       $f(0) = 0$

$f'(t) = 2\sin t \, \cos t = \sin 2t$
 

$\mathcal{L} \left\{ f'(t) \right\} = s \, \mathcal{L} \left\{ f(t) \right\} - f(0)$

$\mathcal{L} (\sin 2t) = s \, \mathcal{L} (\sin^2 t) - 0$

$\dfrac{2}{s^2 + 2^2} = s \, \mathcal{L} (\sin^2 t)$

$s \, \mathcal{L} (\sin^2 t) = \dfrac{2}{s^2 + 4}$

Problem 03 | Laplace Transform of Derivatives

Problem 03
Find the Laplace transform of   $f(t) = e^{5t}$   using the transform of derivatives.
 

Solution 03
$f(t) = e^{5t}$       ..........       $f(0) = 1$

$f'(t) = 5e^{5t}$
 

$\mathcal{L} \left\{ f'(t) \right\} = s \, \mathcal{L} \left\{ f(t) \right\} - f(0)$

$\mathcal{L} (5e^{5t}) = s \, \mathcal{L} (e^{5t}) - 1$

$1 = s \, \mathcal{L} (e^{5t}) - \mathcal{L} (5e^{5t})$

$s \, \mathcal{L} (e^{5t}) - 5 \, \mathcal{L} (e^{5t}) = 1$

Problem 04 | Laplace Transform of Derivatives

Problem 04
Find the Laplace transform of   $f(t) = t \, \sin t$   using the transform of derivatives.
 

Solution 04
$f(t) = t \, \sin t$       ..........       $f(0) = 0$

$f'(t) = t \, \cos t + \sin t$       ..........       $f'(0) = 0$

$f''(t) = (-t \, \sin t + \cos t) + \cos t = -t \, \sin t + 2\cos t$
 

$\mathcal{L} \left\{ f''(t) \right\} = s^2 \mathcal{L} \left\{ f(t) \right\} - s \, f(0) - f'(0)$

Laplace Transform of Intergrals

Theorem
If   $\mathcal{L} \left\{ f(t) \right\} = F(s)$,   then
 

$\displaystyle \mathcal{L} \left[ \int_0^t f(u) \, du \right] = \dfrac{F(s)}{s}$

 

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