Differential Equations Lecture Work Solutions 258

Differential Equations Lecture Work Solutions 258 - 8 a Use...

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8. a. Use the method of characteristics to solve u t + uu x + cu =0 with u ( x, 0) = f ( x )where c is a positive constant. Let’s rewrite the equation as u t + uu x = cu Now the characteristic equation is dx dt = u x (0) = ξ In order to solve this, we have to Frst Fnd u from the second ODE, which is du dt = cu u ( ξ, 0) = f ( ξ ) This equation is separable du u = cdt ln u = ct +ln f ( ξ ) or u ( x ( t ) ,t )= f ( ξ ) e ct Taking this to the right hand side of the characteristic equation we have dx dt = f ( ξ
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This note was uploaded on 12/22/2011 for the course MAP 2302 taught by Professor Bell,d during the Fall '08 term at UNF.

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