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Unformatted text preview: Exercises 6.1
Integration yields v1 (x) = v2 (x) = w1 (x) dx = w2 (x) dx = −3e−3x dx = e−3x , 2e2x dx = e2x , where we have chosen zero integration constants. (d) With functions v1 (x) and v2 (x) obtained in (c), we have y1 (x) = v1 (x)ex = e−3x ex = e−2x , y2 (x) = v2 (x)ex = e2x ex = e3x . To show that the functions ex , e−2x , and e3x are linearly independent on (−∞, ∞), we can use the approach similar to that in Problem 7. Alternatively, since these functions are solutions to the diﬀerential equation (32), one can apply Theorem 3, as we did in Problem 15. To this end, ex W e ,e
x −2x e−2x −2e
−2x e3x 3e
3x 1 =e e
x −2x 3x 11 = −30e2x = 0 49 ,e 3x (x) = e x e 1 −2 3 1 ex 4e−2x 9e3x on (−∞, ∞) and so the functions ex , e−2x , and e3x are linearly independent on (−∞, ∞). 33. Let y (x) = v (x)e2x . Diﬀerentiating y (x), we obtain y (x) = [v (x) + 2v (x)] e2x , y (x) = [v (x) + 4v (x) + 4v (x)] e2x , y (x) = [v (x) + 6v (x) + 12v (x) + 8v (x)] e2x . Substituting these expressions into the given diﬀerential equation yields [(v + 6v + 12v + 8v ) − 2 (v + 4v + 4v ) + (v + 2v ) − (2v )] e2x = 0 ⇒ [v + 4v + 5v ] e2x = 0 ⇒ v + 4v + 5v = 0. With w (x) := v (x), the above equation becomes w (x) + 4w (x) + 5w (x) = 0. 349 ...
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