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WrittenHW01-Solutions

# WrittenHW01-Solutions - Math 216(Section 50 1(1.14 Written...

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Math 216 (Section 50) Written Homework #1 – Solutions Fall 2007 1 ( 1.1–4 ) y 1 y 1 = 3 e 3 x = y 1 = 3 2 e 3 x = 9 e 3 x y 1 = 9 y 1 . Therefore y 1 is a solution of the differential equation y = 9 y . y 2 y 2 = - 3 e - 3 x = y 2 = ( - 3) 2 e - 3 x = 9 e - 3 x y 2 = 9 y 2 . Therefore y 2 is a solution of the differential equation y = 9 y . 2 ( 1.2–10 ) dy dx = x e - x = dy = x e - x dx = dy = x e - x dx Now, integration by parts yields x e - x dx = x ( - e - x ) - ( - e - x ) dx = - x e - x + e - x dx = - x e - x - e - x + C = - ( x + 1) e - x + C. Thus dy = x e - x dx = y ( x ) = - ( x + 1) e - x + C. However, since y (0) = 1, plugging x = 0 into the above equation, y (0) = - 1 + C = 1 = C = 2 . Therefore the solution is given by y ( x ) = - ( x + 1) e - x + 2 . 3 ( 1.3–16 ) Let us write f ( x, y ) = x - y. Then our differential equation can be written as dy dx = f ( x, y ) . 1

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Now f ( x, y ) = x - y is defined and also continuous for x - y > 0 ⇐⇒ y < x . Let us look at D y f ( x, y ): D y f ( x, y ) = ∂f ( x, y ) ∂y = ∂y x - y = - 1 2 1 x - y , which is defined and also continuous for, again x - y > 0 ⇐⇒ y < x .
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WrittenHW01-Solutions - Math 216(Section 50 1(1.14 Written...

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