# hw1 - Section 1.2 5. y + 3 y = e- t , y (0) =- 1 2 y = 1 2...

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Unformatted text preview: Section 1.2 5. y + 3 y = e- t , y (0) =- 1 2 y = 1 2 e- t- e- 3 t y =- 1 2 e- t + 3 e- 3 t ⇒ y + 3 y =- 1 2 e- t + 3 e- 3 t + 3 1 2 e- t- e- 3 t ¶ = e- t y (0) = 1 2- 1 =- 1 2 6. y = 2 y + 1- 2 t 2 , y (0) = 2 y = t + t 2 + 2 e 2 t y = 1 + 2 t + 4 e 2 t ⇒ y = 1 + 2 t + 4 e 2 t = 2 ( t + t 2 + 2 e 2 t ) + 1- 2 t 2 = 1 + 2 t + 4 e 2 t y (0) = 0 + 0 + 2 = 2 8. y- y =- e t sin( t ) , y (0) =- 1 y = e t cos( t ) + ce t y = e t cos( t )- e t sin( t ) + ce t ⇒ y- y = e t cos( t )- e t sin( t ) + c e t- ( e t cos( t ) + c e t ) =- e t sin( t ) y (0) =- 1 ⇒ y (0) = e cos(0) + c e = 1 + c =- 1 ⇒ c =- 2 13. y = 1- y-2-1.5-1-0.5 0.5 1 1.5 2-2-1.5-1-0.5 0.5 1 1.5 2 1 1- y = 0 ⇒ y = 1, which is a stable equilibrium solution. When y < 1, slopes are positive, and when y > 1, slopes are negative. 14. y = y ( y + 1)-2-1.5-1-0.5 0.5 1 1.5 2-2-1.5-1-0.5 0.5 1 1.5 2 y ( y + 1) = 0 ⇒ y = 0, which is an unstable equilibrium solution; y =- 1, which is stable....
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## This note was uploaded on 03/23/2010 for the course APPM 2360 taught by Professor Williamheuett during the Spring '07 term at Colorado.

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hw1 - Section 1.2 5. y + 3 y = e- t , y (0) =- 1 2 y = 1 2...

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