Differential Equations Lecture Work Solutions 233

Differential Equations Lecture Work Solutions 233 - du u =...

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2. b. The ODEs in this case are dx dt = x du dt =1 Solve the characteristic equation ln x = t +ln x 0 or x = x 0 e t The solution of the second ODE is u = t + K and the constant is f ( x 0 ) u ( x, t )= t + f ( x 0 ) Substitute x 0 from the characteristic curve u ( x, t )= t + f xe t 2. c. The ODEs in this case are dx dt =3 t du dt = u Solve the characteristic equation x = 3 2 t 2 + x 0 The second ODE can be written as
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Unformatted text preview: du u = dt Thus the solution of the second ODE is ln u = t + ln K and the constant is f ( x ) u ( x, t ) = f ( x ) e t Substitute x from the characteristic curve u ( x, t ) = f x 3 2 t 2 e t 233...
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