First order and first degree differential equations and their geometrical interpretations

A first order and first degree differential equation involves the independent variable x (say), dependent variable y (say) so, it can be put in any one of the following forms:

dy/ dx = f(x, y) or f (x, y) = 0, or f(x, y) dx + g(x, y)dy = 0

Where f(x, y) and g(x, y) are functions of x and y.

Geometrical interpretation

The general from of a first order and first degree differential equation is f(x, y, dy/dx) = 0 … (i)

We know that the tangent of the direction of a curve in Cartesian rectangular coordinates at any point is given by dy/dx, so the equation in (i) can be known as an equation which establishes the relationship between the coordinates of a point and the slope of the tangent i.e., dy/dx to the integral curve at that point. Solving the differential equation given by (i) means finding those curves for which the direction of tangent at each point coincides with the direction of the field. All the curves represented by the general solution when taken together will give the locus of the differential equation. Since there is one arbitrary constant in the general solution of the equation of first order, the locus of the equation can be said to be made up of single infinity of curves.

Formation of differential equations

Consider a family of exponential curves (y = Ae^x), where A is an arbitrary constant for different values of A, we get different members of the family. Differentiating the relation (y = Ae^x) w.r.t.x, we get dy/dx = Ae^x

Eliminating the arbitrary constant between y = Ae^x and dy/dx = Ae^x, we get dy/dx = y. This is the differential equation of the family of curves represented by y = Ae^x

Thus, by eliminating one arbitrary constant, a differential equation of first order is obtained.

Now consider the family of curves given by y = A cos 2x + B sin 2x ... (1)

Where A and B are arbitrary consists.

Differentiating (1) w.r.t, x we get dy/ dx = - 2Asin 2x + 2Bcos 2x ... (2)

Differentiating (2) w.r.t. x we get d²y/ dx² = - 4Acos ax - 4Bsin 2x ... (3)

Eliminating A and B from equations (1) and (2) (3), we get

d²y/ dx² = - 4y ⇒ d²y/ dx² + 4y = 0

Here we note that by eliminating two arbitrary consists, a differential equation of second order is obtained.

Step I: write the given equation involving independent variable x (say), dependent variable y (say) and the arbitrary constants.

Step II: obtained the numbers of a arbitrary constants in step in step I. let there be n arbitrary consists.

Step III: differentiate the relation in step in times with respect to x.

Step IV: eliminate arbitrary constants with the help of n equations involving differential coefficient obtained in step III and an equation in step I.

The equation so obtained is the desired differential equation.

Example: Show that the differential equation that represented all parabolas having their axis symmetry coincident with the axis of x is yy₂ + y₁² = 0.

Solution: The equation that represents a family of parabolas having their axis of symmetry coincident with the axis of x is y² = 4a(x - h) ... (1)

This equation contains two arbitrary constants, so we shall differentiate twice to obtain second order differential equation.

Differentiating (1) w.r.t x we get

2y dy/dx = 4a ⇒ y dy/ dx = 2a ... (2)

Differentiating (2) w.r.t, x we get

y d²y/ dx² + (dy/ dx)² = 0 ⇒ yy₂ + y₁² = 0

Which is the required differential equation.

Example: Find the differential equation of all non-horizontal lines in a plane.

Solution: The equation of the family of all non-horizontal line in a plane is given by

Ax + by = 1 ... (1)

Where a, b are arbitrary constants such that (a ≠ 0)

 Differentiating (1) w.r.t, x we get

a dx/dy + b = 0

Differentiating this w.r.t y we get 

ad²x/ dy² = 0

⇒ d²x/ dy² = 0

Hence, the differential equation of all non-horizontal lines in a plane is d²x/ dy² = 0.

Example: Find the differential equation of all non- vertical lines in a plane.

Solution: The general equation of all non-vertical lines in a plane is (ax + by = 1) where (b ≠ 0).

Now,

ax + by = 1

a + b dy/dx = 0            [differentiating w.r.t.x]

b d²y/ dx = 0                [differentiating w.r.t.x]

d²y/ dx² = 0                  [∵ b ≠ 0]

Hence, the differential equation is d²y/ dx² = 0

Solution of a differential equation: The solution of a differential equation is a relation between the variable involved which satisfies the differential equation. Such a relation and the derivates obtained therefore when substituted in the differential equation, makes left hand right hand sides identically equal.

General solution: The solution which contains as many as arbitrary constants as the order of the differential equations is called the general solution of the differential equation.

For example, y = Acos x + Bsin x is the general solutions one arbitrary constant.

Particular solution: Solution obtained by giving particular values to the arbitrary constant in the general solution of a differential equation is called a particular solution.

Example: Show that xy = ae^x + be^{- x} + x² is a solution of the differential equation  x d²y/ dx² + 2 dy/dx - xy + x² - 2 = 0

Solution: We are given that xy = ae^x + be^{- x} + x² ... (1)

Differentiating w.r.t.x, we get x dy/ dx + y = ae^x - be^{- x} + 2x

Differentiating again w.r.t.x, we get xd²y/ dx² + dy/dx + dy/dx = ae^x + be^{- x} + 2

xd²y/ dx² + 2dy/dx = ae^x + be^{- x} + 2 ... (2)

Now x d²y/ dx² + 2dy/dx - xy + x2 - 2

= [ae^x + be^{- x} + 2] - [ae^x + be^{- x} + x²] + x² - 2

= 0 [using (1) and (2)]

Thus, (xy = ae^x + be^{- x} + x²) is a solution of the given differential equation.

Ordinary D.F equations, their order and degree

Differential equation: An equation containing an independent variable, dependent variable and differential coefficient of dependent variable with respect to independent variable is called a differential equation.

Order of a differential equation: The order of a differential Equation is the order of the highest order derivative appearing in the equation.

For example, in the equation , the order of highest order derivative is 2. So. It is a differential equation of order 2.

The equation  is of the order 32, It is a differential equation of order 2.

Degree of a differential equation: The degree of a differential equation is the degree of the highest order derivative, when differential coefficient are made free from radicals and fractions.

Example I: Consider the differential equation

In this equation the number of highest order derivative is 2. So it is a differential equation of degree 1.

Example II: Consider the differential equation

In this equation, the order of the highest order derivative is 3. And its power its power is 2. So it is a differential equation of order 3 and degree 2.

Integral Function and its Properties

Let f(x) be a continuous function defined on [a, b], then a function φ(x) defined by  for all x ϵ [a, b] is called the integral function of the function f(x).

Property I: The integral function of an integrable function is always continuous.

Property II:  if φ(x) is differentiable on (a, b) and φ’(x) = f(x) for all x ϵ (a, b).

Property III: The integral function of an odd function is an even function.

If f(x) is an odd function, then  is an function.

Example: Find the greatest value of  in the interval [5π/3, 7π/4]

Solution:  We have,

⇒ F’(x) = 6 cosx - 2sinx

For all x ϵ [5π/3, 7π/4], we have

⇒ cos x > 0 and sin x < 0

⇒ F’(x) = 6 cosx - 2sinx > 0

⇒ F(x) is an increasing function on [5π/3, 7π/4]

⇒ F(x) attains greatest value at x = 7π/4.

Hence,

Greatest value = F (7π/4)

= 3√3 - 2√2 - 1.

Methods of integration

We have the following methods of integration:

 (i) Integration by substitution or change of independent of independent variable.

 (ii) Integration by parts.

 (iii) Integration of rotational algebraic function by using partial fractions.

Integration by substitution: If φ (x) is a continuous differentiable function, then to evaluate integrals of the form ∫f (φ(x)) φ’ (x) dx,

We substitute φ (x) = t and φ’ (x) dx = dt

We substitution reduces the integral ∫f(φ(x)) φ’ (x) dx

This substitution reduces the integral ∫f(φ(x)) φ’ (x) dx to the form ∫f (t) dt

∫f (φ(x)) φ’ (x) dx = ∫f (φ(x)) d (φ (x)) [∵ dφ (x) = φ’ (x) dx]

= ∫f (t) dt, where t = φ(x)

Example: e^x(x + 1) cos² (x.e^x) dx

Solution: we observe that e^x (x + 1) occurs in the derivative x.e^x. So let us substitute x.e^x = t

⇒ d (x.e^x).dx = dt

⇒ (x + 1) e^x dx = dt

⇒ 

∴ ∫ e^x(x + 1) cos² (x.e^x) dx

= ∫cos²t dt

= ½ ∫ (1 + cos2t) dt

Algorithm to evaluate integrals of the form: ∫sin^mx cosⁿ x dx, ∫sin^m x dx and ∫cosⁿ x dx, where m, n ϵ N

Step I: find m and n.

Step II: if m is odd i.e., power or index of sin x is odd, put cosx = t and reduce the integral in terms of t.

If n is odd i.e., power of cosx is odd, then put sinx=t, and reduce the integral in terms of t.

If m and n both are odd, then either of the above substitutions can be used.

Step III: evaluate the integral obtained in step II and replace t by its value.

Example: ∫x² sin³ x³ cos⁵³ dx

Solution: let I = ∫x² sin³ x³ cos⁵³ dx

Here, indices of both sine and cosine are odd. So, we may substitute sinx³ = t or cos³ = t.

Let sinx³ = t then,

d (sinx³) = dt

⇒ cosx³. 3x² dx = dt

⇒ 

∴ 

= ⅓ ∫t³ cos⁴ x³ dt

= ⅓ ∫t³ (1 - sin² x³)² dt

= ⅓ ∫t³ (1 - t²)² dt

= ⅓ ∫(t³ - 2t⁵ + t⁷) dt 

Integral as Anti Derivative

Primitive or anti-derivative of a function:

A function ø is called a primitive or an anti derivative of a function f(x) if ø’(x) = f(x).

Let ø (x) be a primitive of a function f(x) and let C be any constant then,

d/dx [ø (x) + C] = ø’ (x) = f (x) [∵ø’ (x) = f (x)]

⇒ ø (x) + C is also a primitive of f(x)

Thus, if a function f(x) posses a primitive, then it possess infinity many primitives which are contained in the expression ø (x) + C, where C is a constant.

Indefinite integral or indefinite integration:

Let f(x) be a function. Then the collection of all its primitives is called the indefinite integral of f(x) and is denoted by ∫f(x) dx.

Thus, d/dx [ø’ (x) + C] = f (x) ⇔ ∫f(x) dx = ø (x) + C ... (i)

Where ø (x) is primitive of f(x) and C is an arbitrary constant known as the constant of integration.

Here ∫ is the integral sign f(x) is the integral, x is the variable of integration and dx is the element of integration or differential of x.

The process of finding an indefinite integral of a given function is called integration of the following.

It follows from the above discussion that of a given function is called integration of the function.

It follows from the above discussion that integrating a function f(x) means finding a function ø (x) such that d/dx (ø(x)) = f (x).

Fundamental formulas on integration

• 
• (log x) = 1/x ⇒ ∫1/x dx = log |x| + C
• (ex) = ex ⇒ ex dx = ex + C
• (-cos x) = sin x ⇒ ∫ sin x dx = - cos x + C
• (sin x) = cos x ⇒ ∫cos x dx = sinx + C
• (tanx) = sec²x ⇒ ∫sec²x dx = tanx + C
• (-cotx) = cosec²x ⇒ ∫cosec²x dx = - cot x + C
• (sec x) = sec x tan x ⇒ ∫sec x tan x dx = sec x + C
• (- cosec x) = cosec x cot x ⇒ ∫cosec x cot x dx = - cosec x + C
• (log sin x) = cot x ⇒ cot x dx = log |sinx| + C
• (-log cos x) = tan x ⇒ tan x dx = - log |cos x| + C
• (log (sec x + tan x)) = sec x ⇒ ∫sec x dx = log |sec x + tan x| + C
•  (log (cosec x – cot x)) = cosec x ⇒ ∫cosec x dx = log |cosec x – cot x| + C
• 
• 
• 
• 
• 

Ellipse

Terms Related to an Ellipse:

	Fundamental Terms	Ellipse (Horizontal ellipse)	Conjugate Ellipse(Vertical ellipse)
(a)	Equation	x²/a² + y²/b² = 1 (a > b)	x²/a² + y²/b² = -1 (a < b)
(b)	Graph
(c)	Centre	C(0, 0)	C(0, 0)
(d)	Vertices	(±a, 0)	(0, ±b)
(e)	Length of major axis	2a	2b
(f)	Length of minor axis	2b	2a
(g)	Foci	(±ae, 0)	(0, ±be)
(h)	Equation of directrices	X = ±(a/e)	Y = ±(b/e)
(i)	Eccentricity	e = √(1 - [b²/a²])	e = √(1 - [a²/b²])
(j)	Length of latusrectum	2b²/a	2a²/b
(k)	Ends of latusrectum	(±ae, ±b²/a)	(±a²/b, ±be)
(l)	Parametric equations
(m)	Parametric coordinates	acosα, bsinα	acosα, bsinα
(n)	Focal distance or radii	|SP| = (a - ex₁) and |S’P| = (a + ex₁)	|SP|=(b - ey₁) and |S’P| = (b + ey₁)
(o)	Sum of local radii |SP| + |S’P|	2a	2b
(p)	Distance between foci	2ae	2be