ppt of first order differenatiol equation
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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