# DIFFERENTIAL EQUATIONS

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### DIFFERENTIAL EQUATIONS :

DIFFERENTIAL EQUATIONS

### Slide 2:

DE is an equation containing derivatives. Partial DE : Has more than one independent variable First Order DE, First Degree Second Order DE, First Degree Third Order DE, First Degree Second Order DE, Second Degree

### ORDER & DEGREE OF A DE:- :

ORDER & DEGREE OF A DE:- Order of a DE : It is the order of the highest derivative. Degree of a DE: It is the degree of the highest ordered derivative. Linear DE: It is a DE in which, the dependant variable and its differential co-efficients occur in it in the first degree only, and are not multiplied together.

### Solve the following:- :

Solve the following:- 1. A curve is defined by the condition that at each of its points (x,y), its slope is equal to twice the sum of the co-ordinates of the point. Express the condition by a DE.

### Slide 5:

2. 100 gm of sugar in water is converted into dextrose at a rate which is proportional to the amount unconverted. Find the DE expressing the rate of conversion after t minutes.

### Slide 6:

3. Obtain the DE associated with the primitive y=Ax2+Bx+C Primitive equation is one which does not have derivatives. In the eqn given, arbitrary constants are 3 (A,B and C) To get the required DE we must eliminate the constants.

### Examples of DE of Physical systems:- :

Examples of DE of Physical systems:-

### SOLUTION OF FIRST ORDER LINEAR DIFFERENTIAL EQUATION (LDE):- :

SOLUTION OF FIRST ORDER LINEAR DIFFERENTIAL EQUATION (LDE):- Finding the relationship between x and y from the DE is known as the solution of a DE.

### Case 1: k is positive s=Ae+kt :

Case 1: k is positive s=Ae+kt This will be an increasing exponential & divergent If A is the amplitude at t=0, the time t2 taken for s to double its value is called “time to double amplitude” It is given by

### Case 2: k is Negative s=Ae-kt :

Case 2: k is Negative s=Ae-kt This will be an decreasing exponential & convergent A (½)A s = Ae - kt s t t1/2 If A is the amplitude at t=0, the time t1/2 taken for s to halve its value is called “time to half amplitude” It is given by

### General Solution of a First Order LDE:- :

General Solution of a First Order LDE:-

### Slide 14:

First Order LDE represents a family of curves s=Ae-kt+C A is known as the parameter of the family. C is known as the initial bias in the system. Eg. Initial bias in a strain gauge.

### SECOND ORDER LDE:- :

SECOND ORDER LDE:- General form of a 2nd Order LDE with constant co-efficients:-

### 3 Possible sets of Roots:- :

3 Possible sets of Roots:- Two distinct real roots. Two identical roots. Pair of Complex roots.

### DISTINCT REAL ROOTS:- :

DISTINCT REAL ROOTS:-

### IDENTICAL REAL ROOTS:- :

IDENTICAL REAL ROOTS:-

### COMPLEX PAIR OF ROOTS:- :

COMPLEX PAIR OF ROOTS:-

### Solve the following:- :

Solve the following:-

### General Solution of a First Order LDE:- :

General Solution of a First Order LDE:- Linear DE: It is a DE in which, the dependant variable and its differential co-efficients occur in it in the first degree only, and are not multiplied together. Lebnitz Linear Equation

### DIFFERENTIAL EQUATIONS ASSIGNMENT : DOS – 18 Aug 08 :

DIFFERENTIAL EQUATIONS ASSIGNMENT : DOS – 18 Aug 08

### DIFFERENTIAL EQUATIONS ASSIGNMENT : DOS – 18 Aug 08 :

DIFFERENTIAL EQUATIONS ASSIGNMENT : DOS – 18 Aug 08

### Bernoulli’s Equation:- :

Bernoulli’s Equation:-

### Determining values of Arbitrary Constants:- :

Determining values of Arbitrary Constants:- We have seen that the solution of a DE contains one or more arbitrary constants. We will now find these constants; given the initial conditions of the DE. For a DE with a solution having two arbitrary constants, we require two independent initial conditions to determine the values of the two constants.

### Solve the following LDE:- :

Solve the following LDE:-

### Significance of a LDE in a Physical System:- :

Significance of a LDE in a Physical System:- Mass (m) Dashpot Damping Co-eff (b) Spring Const (k) Aircraft Pitching Motion x

### Mass – Damper Analogy (2nd Order LDE):- :

Mass – Damper Analogy (2nd Order LDE):-

### Case 1 : Zero Damping (b=0) (Inference : Dashpot is removed) :

Case 1 : Zero Damping (b=0) (Inference : Dashpot is removed)

### Problem:- :

Problem:- A Sopwith Camel moved 0.5 m during its routine tyre check. On braking its oscillations were of the order of 6 sec. If the mass of the aircraft was to be assumed as 300 kg, (a) Determine the stiffness of the oleo spring. (b) Plot the response of the aircraft. Note: If nothing is given wrt oscillations, consider the case as one with free oscillations without damping.

### Case 2 : Low Damping (b≠0 & a small value) :

Case 2 : Low Damping (b≠0 & a small value)

### Case 3 : High Damping (b≠0 & a high value) :

Case 3 : High Damping (b≠0 & a high value)

### Case 4 : Critical Damping :

Case 4 : Critical Damping Critical Damping is defined as the condition wherein the damping is just sufficient to prevent oscillatory response. The necessary condition is that (b2-4mk)=0 Critical Damping co-efficient

### Critical Damping Response:- :

Critical Damping Response:- x0 x t

### Relative Damping Co-efficient:- :

Relative Damping Co-efficient:-

### 2nd order LDE Response wrt  & n:- :

2nd order LDE Response wrt  & n:-

### Zero Stiffness (k=0):- :

Zero Stiffness (k=0):-

### ROOT LOCUS PLOTS/ ARGAND DIAGRAM:- (STABILITY DOMAIN) :

ROOT LOCUS PLOTS/ ARGAND DIAGRAM:- (STABILITY DOMAIN) Study of stability characteristics with variabtion of b, m & k is undertaken by plotting the values of the roots of the 2nd order LDE =j

### ROOT LOCUS PLOTS/ ARGAND DIAGRAM:- :

ROOT LOCUS PLOTS/ ARGAND DIAGRAM:-