Differential Equations Lecture Work Solutions 240

# Differential Equations Lecture Work Solutions 240 - + c ) (...

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6. a. Solve the initial value problem for the inhomogeneous equation v t + cv x = f ( x, t ) v ( x, 0) = F ( x ) where f ( x, t )and F ( x ) are speciFed functions. The ODEs are dx dt = c, dv dt = f ( x ( t ) ,t ) The initial condition for each: x (0) = ξ, v ( ξ, 0) = F ( ξ ) Solve the characteristic equation to get x = ct + ξ. Now solve the other ODE to get v ( x ( t ) ,t )= Z t 0 f ( x ( τ ) ) + F ( ξ ) Substitute the value of x ( τ )toget v ( x ( t ) ,t )= Z t 0 f ( + ξ,τ ) + F ( ξ ) Now get ξ from the characteristics and substitute here to get ξ = x ct v ( x, t )= Z t 0 f ( x ct + cτ, τ ) + F ( x ct ) To check the answer, we can di±erentiate the solution. v t =( c ) · F 0 ( x ct )+ f ( x, t )+ Z t 0 ∂f ( x ct + cτ, τ ) ( x ct
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Unformatted text preview: + c ) ( x ct + c ) t | {z } c d v x = F ( x ct ) + Z t f ( x ct + c, ) ( x ct + c ) ( x ct + c ) x | {z } 1 d Substitute these two derivatives into the left side of the equation and Fnd that the only term left is f ( x, t ) which is on the right. We can also check the initial condition by substituting t = 0 in the solution. In this case the integral is zero and we get F ( x ). 240...
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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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