Prog 25 Second-order differential equations

Prog 25 Second-order differential equations - PROGRAMME 25...

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Worked examples and exercises are in the text STROUD PROGRAMME 25 SECOND-ORDER DIFFERENTIAL EQUATIONS
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Worked examples and exercises are in the text STROUD Programme 25: Second-order differential equations Introduction Homogeneous equations The auxiliary equation Summary Inhomogeneous equations
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Worked examples and exercises are in the text STROUD Programme 25: Second-order differential equations Introduction Homogeneous equations The auxiliary equation Summary Inhomogeneous equations
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Worked examples and exercises are in the text STROUD Programme 25: Second-order differential equations Introduction For any three numbers a , b and c , the two numbers: are solutions to the quadratic equation: with the properties: 2 2 1 2 4 4 and 2 2 b b ac b b ac m m a a - + - - - - = = 2 0 am bm c + + = 1 2 1 2 and b c m m m m a a + = - =
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Worked examples and exercises are in the text STROUD Programme 25: Second-order differential equations Introduction The differential equation: can be re-written to read: that is: 2 2 0 d y dy a b cy dx dx + + = 2 2 0 provided 0 d y b dy c y a dx a dx a + + = 2 2 0 d y b dy c a y dx a dx a + + =
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Worked examples and exercises are in the text STROUD Programme 25: Second-order differential equations Introduction The differential equation can again be re-written as: where: ( 29 2 2 1 2 1 2 2 2 1 2 1 2 0 d y b dy c d y dy y m m m m y dx a dx a dx dx d dy dy m y m m y dx dx dx dz m z dx + + = - + + = - - - = - = 2 2 1 2 1 4 4 , and 2 2 b b ac b b ac dy m m z m y a a dx - + - - - - = = = -
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Worked examples and exercises are in the text STROUD Programme 25: Second-order differential equations Introduction The differential equation: has solution: This means that: That is: 2 0 dz m z dx - = 2 1 m x dy z m y dx Ce = - = 2 : being the integration constant m x z Ce C = 2 1 m x dy m y Ce dx - =
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Worked examples and exercises are in the text STROUD Programme 25: Second-order differential equations Introduction The differential equation: has solution: where: and are constants A B 1 2 1 1 2 1 2 : if ( ) : if m x m x m x y Ae Be m m A Bx e m m = + = + = 2 1 m x dy m y Ce dx - =
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