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 05 | Integrating Factors Found by Inspection

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

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

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

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

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

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

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

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.
 

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

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

 

$F = \text{ Sum}$

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

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

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.
 

Problem 04 | Integrating Factors Found by Inspection

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

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

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

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

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

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

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

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

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

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

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

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 02 | Integrating Factors Found by Inspection

Problem 02
$y(y^3 - x) \, dx + x(y^3 + x) \, dy = 0$
 

Solution 02
$y(y^3 - x) \, dx + x(y^3 + x) \, dy = 0$

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

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

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

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

Divide by y3
$d(xy) - \dfrac{x}{y} \left( \dfrac{y \, dx - x \, dy}{y^2} \right) = 0$

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

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