Differential Equation:

An equation containing the derivatives of one or more dependent variables (unknown functions), with respect to one or more independent variables, is said to be a differential equation (DE).

CLASSIFICATION BY TYPEi) Ordinary Differential Equation (ODE) ii) Partial Differential Equation (PDE).

If an equation contains only ordinary derivatives of one or more dependent variables with respect to a single independent variable it is said to be an ordinary differential equation (ODE).

Example: dx / dt + dy / dt = 2x + yAn equation involving partial derivatives of one or more dependent variables of two or more independent variables is called a partial differential equation (PDE).

Example: ∂2u / ∂x2 + ∂2u / ∂y2 = 0

Notationsi) Leibniz notation dy / dx, d2y / dx2, d3y / dx3, . . . ii) Prime notation y`, y``, y```, . . .iii) Dot notation (flyspeck notation) denotes derivatives with respect to time t.

So d2s / dt2 = - 32 becomes s¨ = -32.iv) Partial derivatives are often denoted by a subscript notation indicating the

independent variables.

For example ∂2u / ∂x2 + ∂2u / ∂y2 = 0 uxx + uyy = 0.

Classification by OrderThe order of a differential equation (either ODE or PDE) is the largest derivative present in the differential equation.

It is second-order ordinary differential equation.

CLASSIFICATION BY LINEARITY

1. Linear 2. Nonlinear

A linear differential equation is any differential equation that can be written in the following form.

Two properties of a linear ODE are as follows: The dependent variable y and all its derivatives y`, y``, . . . , y(n) are of the first

degree, that is, the power of each term involving y is 1. The coefficients a0, a1, . . . , an of y, y`, . . . , y(n) depend at most on the independent

variable x.

Nonlinear ordinary differential equation is simply one that is not linear.Nonlinear functions of the dependent variable or its derivatives, such as sin y or ey,

cannot appear in a linear equation.

first-order ODE second-order ODE fourth-order ODE

See EXERCISES 1.1 Page#10 (A First Course in DE 9th Edition)

Solution of an ODE

A solution to a differential equation on an interval   is any function   which satisfies

the differential equation in question on the interval  . 

Separable Equation

See EXERCISES 2.2 Page No. 50

2.3 LINEAR EQUATIONSHomogeneous vs non-homogeneous

When g(x) = 0, the linear equation (1) is said to be homogeneous;

otherwise, it is nonhomogeneous.

EXERCISES 2.3 Page No. 61

Degree of a Differential Equation:

The Degree of a differential equation is the power of the highest order derivative in the equation.

Order: 3

Degree: 1

Order: 2

Degree: 3

Exact Equation

2nd MethodThis method is so easy for Exact DE.

⌠ M dx + ⌠ term N (not containing x) dy = C`

∂ M∂ y

=∂ N∂ x Exact Equation

Q: (2x – y + 1) dx + (2y – x - 1) dy = 0∂ M∂ y

=¿-1 ; ∂ N∂ x

=−1

Hence ∂ M∂ y

=∂ N∂ x and the equation is Exact equation

⌠ (2x – y + 1) dx + ⌠ (2y - 1) dy = 0

x2 – xy + x + y2 – y = 0

So, required solution is:

x2 + y2 – xy + x – y = 0

SOLUTIONS BY SUBSTITUTIONSWhen solution is impossible with Separable Method.We use SUBSTITUTIONS Method.Example:

dydx

=( y+x )2

Put u = y + x

And dudx

=dydx

+1 ; dydx

=dudx

−1

Put in question

dudx

−1=u2

dudx

=u2 + 1 now this is separable eq.

duu2+1

=dx

Integrating both sides:

Tan-1 u = x + c u = tan (x + c)

we know that u = y + x

so, y + x = tan (x + c)

Ans: y = tan (x + c) – x

Another Example:dydx

= y2−x2

yx 2nd degree equation

multiplying∧dividingboth sides by 1x2

We get dydx

=( y

x )2

−1

( yx)

Substitute v = yx

y=vx dydx

=v+x dvdx

Put in question v+x dvdx

= v2−1v

v+x dvdx

=v−1v

x dvdx

=−1v

Now it is separable eq.

vdv=−1x

dx

By integrating, we get

v2

2=−ln|x|+c

( yx )

2

=−2 ln|x|+c y2=−2 x2(ln|x|+c )

2.6 A NUMERICAL METHOD

Not important

Another type of problem consists of solving a linear differential equation of order two or greater in which the dependent variable y or its derivatives are specified at different points. A problem such as

is called a boundary-value problem (BVP). where y0 and y1 denote arbitrary constants.

The prescribed values y(a) = y0 and y(b) = y1 are called boundary conditions.

A BVP Can Have Many, One, or No Solutions

has infinitely many solutions.

Def: In calculus differentiation is often denoted by the capital letter D —that is, dydx

=Dy. The symbol D is called

a Differential Operator, because it transforms a differentiable function into another function.

Higher-order derivatives can be expressed in terms of D in a natural manner:

General Form:In general, we define an nth-order differential operator or polynomial operator to be:

L is a linear operator.

It means that there should be one or more than one constant i.e ≠ 0.

A Second Solution by Reduction of Orderhttps://www.youtube.com/watch?v=oQSFW8BIrY0

Q: y1 = ex; y`` - y = 0

Solution:

y`` - y = 0 (1)

Let y = y1 v

y = ex. v (2)

y` = ex v` + v ex

Y``= (ex v`` + v` ex) + (vex + ex v`)

Y``= ex v`` + v` ex + vex + ex v`

Y``= ex v`` + v` ex + vex + ex v`

Y``= ex v`` + 2v` ex + vex

Put these in eq. (1)

y`` - y = 0

(ex v`` + 2v` ex + vex) - vex = 0

ex v`` + 2v` ex + vex - vex =0

ex v`` + 2v` ex =0 (3)

Now let w = v`

w` = v``

put in eq. (2)

ex w` + 2w ex =0

ex dwdx + 2w ex =0

ex dwdx = - 2w ex

dww = - 2ex dx

Taking Integration on both sides:

ln |w| = -2 ex + c

Taking antilog:

w = e(-2 ex + c)

w = e-2 ex. e C0

w = e-2 ex.C1

w = C1e-2 ex

we know that w = v`

v` = C1e-2 ex

Taking integration on both sides:

v = C 1e−2ex

−2 + C

C 1−2

=new constant=¿C

v = C e-2ex + C

put this in eq. (2)

y = ex. v = ex. (C e-2ex + C)

y = (C ex e-2ex + C ex)

y = C ex (e-2ex + 1) General Solution

y2 = (C ex e-2ex + C ex) Second Solution

we find two solutions

It follows that the general solution of on this interval is

When m1 = m2, we necessarily obtain only one exponential solution,

Higher-Order EquationsIf all the roots of (12) are real and distinct, then the general solution of (1) is

Case ll:

EXERCISES 4.3 Q NO. 1 to 14