Pre-Calc Homework Solutions 273

Pre-Calc Homework Solutions 273 - Section 6.6 Section 6.6...

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Unformatted text preview: Section 6.6 Section 6.6 Exercises 1. Check the differential equation: y x d (x dx 273 y y 1 2e x eC ex Ae x 1 1 1 (x x 2e x) 1 y. 1 2e ) x 2e x( 1) 1 2e x y x Therefore, y Initial condition: y(0) 1 1 Ae0 1 2 A Solution: y 2 e x 1 6. dy dx dy 1 y Check the initial condition: y(0) 0 1 2e (0) x(1 x dx 1 2 x 2 y) 1 2 1 ln 1 2. Check the differential equation: y x d (x dx y y y y C C 1 1 (x x e x) 1 y. 1 e ) x e (x2/2) e x( 1) 1 e x 1 e x 1 e e Ae (x2/2) C C y x e x2/2 1 Therefore, y y Check the initial condition: y(0) 0 1 e (0) x2/2 1 1 1 2 3. Check the differential equation: y 2y d e dx 2x Initial condition: y( 2) 0 2 0 Ae ( 2) /2 1 2 0 Ae 1 e2 A 2 Solution: y e2 e x /2 1 or y 2 cos x 5 sin x e (x2/2) 2 1 2 sin x 5 cos x 2e 2x 7. sin x 2 e2x 2e2x 2 e2x 2 sin x cos x 5 4 sin x 2 cos x 5 2 cos x sin x 5 sin x 5 sin x dy dx dy y dy y 2y(x 2(x (2x x2 1) 1) dx 2) dx 2x C ln y y y y 2 e x 2x C Therefore, y 2y sin x eC e x A ex 2 2 2x Check the initial condition: y(0) e2(0) 2 sin 0 5 cos 0 1 5 1 2x 0 4. Check the differential equation: y y d x (e dx Initial condition: y( 2) 2 2 Ae( 2) 2( 2) 2 A 2 Solution: y 2e x 2x 8. dy dx 2 e2x (e x 1) e e 2x ex 1) 1. 2e2x e 2x y 2(1 (1 x x2 2x) 2x) dx x2 1 x 1 ( 1) e 2x 1 1 e x 2e 2x y 2 dy 1 Therefore, y y 2x y C C Check the initial condition: y(0) e 0 y e 2(0) 1 1 Initial condition: y( 1) 1 1 ( 1)2 1 C C 1 5. Note that we are finding an exact solution to the initial value problem discussed in Examples 14. dy dx dy 1 y 1 dx y C 1 Solution: y x ex C C ln 1 1 1 y y y x2 1 x 1 ex C ...
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This note was uploaded on 10/05/2011 for the course MAC 1147 taught by Professor German during the Spring '08 term at University of Florida.

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