Differential Equations Lecture Work Solutions 241

Differential Equations Lecture Work Solutions 241 - ct 3...

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6. b. Solve this problem when f ( x, t )= xt and F ( x )=s in x . In this case the integral becomes Z t 0 ( x ct + ) τdτ = x t 2 2 ct t 2 2 + c t 3 3 = 1 2 xt 2 1
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Unformatted text preview: ct 3 Thus the solution is v ( x, t ) = x 1 3 ct t 2 2 + sin( x ct ) 241...
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