Unformatted text preview: Exercises 4.2
To ﬁnd the constants c1 and c2 , we substitute the initial conditions into the general solution and its derivative, y (t) = −c1 e−t + c2 (e−t − te−t ), and obtain y (0) = 1 = c1 e0 + c2 · 0 = c1 , y (0) = −3 = −c1 e0 + c2 (e0 − 0) = −c1 + c2 . So, c1 = 1 and c2 = −2. Therefore, the solution that satisﬁes the initial conditions is given by y (t) = e−t − 2te−t . 17. The auxiliary equation for this problem, r 2 − 2r − 2 = 0, has roots r = 1 ±
√ √ √ 3. Thus, a general solution is given by z (t) = c1 e(1+ 3)t + c2 e(1− 3)t . Diﬀerentiating, we ﬁnd that √ √ √ √ z (t) = c1 (1 + 3)e(1+ 3)t + c2 (1 − 3)e(1− 3)t . Substitution of z (t) and z (t) into the initial conditions yields the system z (0) = c1 + c2 = 0, √ √ √ z (0) = c1 (1 + 3) + c2 (1 − 3) = 3(c1 − c2 ) = 3 ⇒ 3/2, √ c2 = − 3 / 2 . c1 = √ Thus, the solution satisfying the given initial conditions is √ √ √ √ 3 (1+√3)t 3 (1−√3)t 3 (1+√3)t e e e − = − e(1− 3)t . z (t) = 2 2 2 19. Here, the auxiliary equation is r 2 − 4r − 5 = (r − 5)(r + 1) = 0, which has roots r = 5, −1. Consequently, a general solution to the diﬀerential equation is y (t) = c1 e5t + c2 e−t , where c1 and c2 are arbitrary constants. To ﬁnd the solution that satisﬁes the initial conditions, y (−1) = 3 and y (−1) = 9, we ﬁrst diﬀerentiate the solution found above, then plug in y and y into the initial conditions. This gives y (−1) = 3 = c1 e−5 + c2 e y (−1) = 9 = 5c1 e−5 − c2 e. Solving this system yields c1 = 2e5 , c2 = e−1 . Thus y (t) = 2e5(t+1) + e−(t+1) is the desired solution. 171 ...
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 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations, Derivative, Substitute good, initial conditions

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