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FIRST ORDER DIFFERENTIAL EQUATION AND ITS ENGINEERING APPLICATIONS By GARLAPATI SUNODH KUMAR ECE 1 ST YEAR SONA COLLEGE OF TECHNOLOGY,SALEM
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Page 1: Ppt of first order differenatiol equation

FIRST ORDER DIFFERENTIAL EQUATION AND ITS ENGINEERING APPLICATIONS

ByGARLAPATI SUNODH KUMARECE 1ST YEARSONA COLLEGE OF TECHNOLOGY,SALEM

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WHAT IS DIFFERENTIAL EQUATION?

An equation contain the derivatives of one or more dependent variables, with respect to one of more independent variables, said to be differential equation

WHAT IS FIRST ORDER DIFFERETIAL EQUATION?

An equation of 1st degree only in respect to the dependent variables and its derivatives

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WE CAN SOLVING FIRST ORDER DIFFERENTIAL EQUATIONS BY

Separable variables

Homogenous

equation

Exact equation

Integrating factor

Linear equation

Bernoulli equation

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APPLICATION IN ENIGNEERING FIELD

1.BASIC GEOMETRIC APPLICATION

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ORTHOGONAL TRAJECTORIES

AN CURVE CUTS ANOTHER CURVE AT RIGHT ANGLE THEN TWO CURVES SAID TO BE ORTHONOGLY TO EACH OTHER

EX: ORTHOGONAL TRJECTION OF Y2 =KX IS X2 + Y2 = R2

X2 + Y2 = R2

Y2 = KX

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STEPS TO SLOVE ORTHOGONAL TRAJECTORIES OF THE FAMILY OF CURVES F(x, y, c) = 0 or F(r, ø, c) = 0

SOLVE DIFFERENTIAL EQUATION OF THE ORTHOGONAL TRAJECTORIES F(x, y, dy/dx) or F(r,

ø, dr/dø)

REPLAYES THIS DIFFERENTIAL EQUATION dy/dx BY –dx/dy or dr/dø = -r2 dø/dr ( so product of slope at

each point is -1)

FORM DIFFERENTIAL EQUATION IN FORM F(x, y, dy/dx) = 0 or F(r, ø, dr/dø) = 0

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Example : If stream lines of a flow around a corner are x y = constant find their orthogonal trajectories

As by steps

Differential equation of curve is x . dy/dx + y = 0

Replays dy/dx by -dx/dy

Finally we get x . dx – y dy = 0

Therefore X2 - Y2 = c2 is required orthogonal trajectories

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Physical applications

dx/dt

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NEWTON’S SECOND LAW

THE RATE OF CHANGE IN MOMENTUM ENCOUNTERED BY A MOVING OBJECT IS EQUAL TO THE NET FORCE APPLIED TO IT. IN MATHEMATICAL TERMS,

Since m is constant then dm/dt is

F = m a

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ELECTRIC R-L SERIES CIRCUIT

Kirchhoff's law , sum of voltage drop across R and L = E

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THIS CIRCUIT CAN SOLVE BY INTERGRATION FACTOR

WHEN COMPARE TO

I.F IS

MULITIPLING I.F BOTH SIDES

When ‘t=0’ then ‘i=0’ we get c = - E/R

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THANK YOU