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Differential Equations Lecture Work Solutions 252

Differential Equations Lecture Work Solutions 252 - c The...

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Unformatted text preview: c. The set of ODEs is du =u dt dx = −t2 dt The solution of the ﬁrst is x = x0 − 13 t 3 Now integrate the second ODE ln u(x, t) = −t + ln K or u(x, t) = K e−t At t = 0 the solution is u(x0 , 0) = f (x0 ) = K plug t = 0 in the solution u Thus when substituting for x0 in the solution u(x, t) = e−t f x + d. The set of ODEs is 13 t 3 du =t dt dx =x dt The solution of the ﬁrst is ln x = ln x0 + t or x = x0 et Now integrate the second ODE u(x, t) = 12 t +K 2 At t = 0 the solution is u(x0 , 0) = f (x0 ) = K plug t = 0 in the solution u Thus when substituting for x0 in the solution u(x, t) = 12 t + f x e−t 2 252 ...
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