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Active learning Assignment
Topic : CLASSIFICATION OF SECOND ORDER PARTIAL
DIFFERENTIAL EQUATION
BRANCH : ELECTRICAL ENGINE...
β€’ The general form of a non Homogeneous second order
P.D.E is
β€’ 𝐴 π‘₯, 𝑦
πœ•2 𝑒
πœ•π‘₯2+B π‘₯, 𝑦
πœ•2 𝑒
πœ•π‘₯πœ•π‘¦
+C π‘₯, 𝑦
πœ•2 𝑒
πœ•π‘¦2+f
π‘₯, 𝑦, ...
β€’ EXAMPLE:-1
Classify the Following P.D.E
πœ•π‘’
πœ•π‘‘
=
πœ•2 𝑒
πœ•π‘₯2
Ans:- Comparing this equation with (1) we get
A=1 , B=0 , C=0
S...
β€’ EXAMPLE:-2
Classify the following P.D.E
πœ•2 𝑒
πœ•π‘₯2 +
πœ•2 𝑒
πœ•π‘¦2=0
Ans:- Comparing this given P.D.E with (1) we get
A=C=1 , B...
β€’ EXAMPLE:- 3
Classify the Following P.D.E
πœ•2 𝑒
πœ•π‘₯2 + 3
πœ•2 𝑒
πœ•π‘₯πœ•π‘‘
+
πœ•2 𝑒
πœ•π‘‘2 =0
Ans:- Comparing this given P.D.E with (1) ...
classification of second order partial differential equation
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classification of second order partial differential equation

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classification of second order partial differential equation

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classification of second order partial differential equation

  1. 1. Active learning Assignment Topic : CLASSIFICATION OF SECOND ORDER PARTIAL DIFFERENTIAL EQUATION BRANCH : ELECTRICAL ENGINEERING BATCH : B1 SUBJECT : ADVANCED ENGINEERING MATHEMATICS Prepared By : JIGAR METHANIYA(150120109021) Guided By : Prof. MIHIR SUTHAR 1
  2. 2. β€’ The general form of a non Homogeneous second order P.D.E is β€’ 𝐴 π‘₯, 𝑦 πœ•2 𝑒 πœ•π‘₯2+B π‘₯, 𝑦 πœ•2 𝑒 πœ•π‘₯πœ•π‘¦ +C π‘₯, 𝑦 πœ•2 𝑒 πœ•π‘¦2+f π‘₯, 𝑦, 𝑒, πœ•π‘’ πœ•π‘₯ , πœ•π‘’ πœ•π‘¦ =F π‘₯, 𝑦 ………..(1) β€’ Equation (1) is said to be β€’ Elliptic , if 𝐡2 -4AC < 0 β€’ Parabolic , if 𝐡2 -4AC = 0 β€’ Hyperbolic , if 𝐡2-4AC > 0 β€’ CLASSIFICATION OF SECOND-ORDER PARTIAL DIFFERENTIAL EQUATION
  3. 3. β€’ EXAMPLE:-1 Classify the Following P.D.E πœ•π‘’ πœ•π‘‘ = πœ•2 𝑒 πœ•π‘₯2 Ans:- Comparing this equation with (1) we get A=1 , B=0 , C=0 So , 𝐡2 -4AC = 0 Hence given P.D.E. is parabolic.
  4. 4. β€’ EXAMPLE:-2 Classify the following P.D.E πœ•2 𝑒 πœ•π‘₯2 + πœ•2 𝑒 πœ•π‘¦2=0 Ans:- Comparing this given P.D.E with (1) we get A=C=1 , B=0 So , 𝐡2-4AC = -4<0 Hence , given P.D.E is elliptic.
  5. 5. β€’ EXAMPLE:- 3 Classify the Following P.D.E πœ•2 𝑒 πœ•π‘₯2 + 3 πœ•2 𝑒 πœ•π‘₯πœ•π‘‘ + πœ•2 𝑒 πœ•π‘‘2 =0 Ans:- Comparing this given P.D.E with (1) we get A=1 , B=3 , C=1 So , 𝐡2 -4AC = 9-4 = 5>0 Hence the given P.D.E is hyperbolic.

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