Pre-Calc Homework Solutions 246

Pre-Calc Homework Solutions 246 - 246 Section 6.2 1 3 1/2 d...

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Unformatted text preview: 246 Section 6.2 1 3 1/2 d 2 1/2 3/2 35. Let u du 2 du 3 1 0 39. dy (y 5)(x 2) dx dy (x 2)dx y 5 Integrate both sides. d d 2 (10) 3 2 dy y 5 (x 2) dx 10 (1 3/2 2 ) u 1 2 2 1 du On the left, let u y du 5 dy 1 2 x 2 1 2 x 2 1 2 x 2 20 1 u 3 20 1 3 2 20 3 1 2 1 10 3 1 du u 2x 2x 2x C C C ln u ln y y 5 5 5 36. Let u du 1 du 3 4 4 3 sin x e (1/2)x 2 2x C 2 2x 3 cos x dx y cos x dx cos x 3 sin x 1 3 4 e Ce(1/2)x We now let C u 1/2 eC or C e C, depending on whether dx du 0 4 (y y 5) is positive or negative. Then 5 y C e (1/2)x C e (1/2)x 2 2x 2 2x 37. Let u du 1 0 t5 (5t 4 2t 2) dt 5 Since C represents an arbitrary constant (note that even the 3 t5 2t (5t 4 2) dt 0 u1/2 du 3 0 value C 0 gives a solution to the original differential 2 3/2 u 3 equation), we may write the solution as y 40. dy dx 2 3/2 (3) 3 2 3 Ce (1/2)x x dy y cos2 dy y cos2 y 2 2x 5. y 27 2 3 y cos2 x dx 38. Let u du 1 du 2 /6 cos 2 2 sin 2 d sin 2 d 3 Integrate both sides. x dx y On the left, let u 1 2 1 2 1/2 y cos 0 2 sin 2 d u 1 3 du 1/2 du 2 du 1 1 u 2 2 2 1 1/2 y dy 2 y 1/2 dy 1 1 4 2 1 (3) 4 1 3 4 1 2 x C 2 1 2 2 sec2 u du x C 2 1 2 2 tan u x C 2 1 2 2 tan y x C 2 1 2 tan y x C 4 2 du cos2 u (Note: technically, C is now C y y tan tan 2 1 x C . But C's are generic.) 2 4 2 C 2 1 x 4 C ...
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