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Differential Equations Lecture Work Solutions 305

# Differential Equations Lecture Work Solutions 305 - Now...

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Now collect terms to compute the terms on the right at time n + 1 u n +1 j +1 2 u n +1 j + u n +1 j 1 = 0 · u + 0 · u t + 0 · u x + 0 · u tt + 0 · u tx + (∆ x ) 2 u xx + 0 · u ttt + 0 · u ttx + t 2 (∆ x ) 2 u txx + 0 · u xxx + ( t 2 ) 2 (∆ x ) 2 2 u ttxx + (∆ x ) 4 12 u xxxx + · · · Divide by 2(∆ x ) 2 we get u n +1 j +1 2 u n +1 j + u n +1 j 1 2(∆ x ) 2 = 1 2 u xx + t 4 u txx + (∆ t ) 2 16 u ttxx + (∆ x ) 2 24 u xxxx + · · · Now to terms at time n u n j +1 = u t 2 u t + ∆ xu x + 1 2 ( t 2 ) 2 u tt t 2 xu tx + 1 2 (∆ x ) 2 u xx 1 6 ( t 2 ) 3 u ttt + 3( t 2 ) 2 x 6 u ttx 3 t 2 (∆ x ) 2 6 u txx + (∆ x ) 3 6 u xxx + 1 24 ( t 2 ) 4 u tttt 4( t 2 ) 3 x 24 u tttx + 6( t 2 ) 2 (∆ x ) 2 24 u ttxx 4 t 2 (∆ x ) 3 24 u txxx + (∆ x ) 4 24 u xxxx + · · · u n j 1 = u t 2 u t xu x + 1 2 ( t 2 ) 2 u tt + t 2 xu tx + 1 2 (∆ x ) 2 u xx 1 6 ( t 2 ) 3 u ttt 3( t 2 ) 2 x 6 u ttx 3 t 2 (∆ x ) 2 6 u txx (∆ x ) 3 6 u xxx + 1 24 ( t 2 ) 4 u tttt + 4( t 2 ) 3 x 24 u tttx + 6( t 2 ) 2 (∆ x ) 2 24
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