265_pdfsam_math 54 differential equation solutions odd

265_pdfsam_math 54 differential equation solutions odd -...

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Unformatted text preview: Exercises 5.2 7. In order to eliminate u, we multiply the first equation by (D − 1), the second equation – by (D + 1), and subtract the results. (D − 1) {(D + 1)[u] − (D + 1)[v ]} = (D − 1) [et ] = (et ) − et = 0, (D + 1) {(D − 1)[u] + (2D + 1)[v ]} = (D + 1) [5] = (5) + 5 = 5 ⇒ ⇒ (D 2 − 1) [u] − (D 2 − 1) [v ] = 0, (D 2 − 1) [u] + {(D + 1)(2D + 1)} [v ] = 5 (D + 1)(2D + 1) + D2 − 1 [v ] = 5 ⇒ {D (D + 1)} [v ] = 5 . 3 (5.2) The corresponding homogeneous equation, {D (D + 1)} [v ] = 0, has the characteristic equation r (r + 1) = 0 and so its general solution is vh (t) = c1 + c2 e−t . Applying the method of undetermined coefficients, we look for a particular solution to (5.2) of the form vp (t) = cts , where we choose s = 1 (because the homogeneous equation has constant solutions and does not have solutions of the form ct). Substitution v = ct into (5.2) yields {D (D + 1)} [ct] = (D + 1)[c] = c = Therefore, a general solution to (5.2) is v (t) = vh (t) + vp (t) = c1 + c2 e−t + 5 t. 3 5 3 ⇒ vp (t) = 5 t. 3 ⇒ r = 0, −1, We now go back to the original system and subtract the second equation from the first one. 2u − (3D + 2)[v ] = et − 5 1 3 5 D + 1 [v ] + et − ⇒ u= 2 2 2 1 5 5 5 3 ⇒ u= c1 + c2 e−t + t + c1 + c2 e−t + t + et − 2 3 3 2 2 1 1 5 ⇒ u(t) = c1 − c2 e−t + et + t. 2 2 3 261 ...
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