MATH111-200630-PS08-Solutions

y0 sin 0 cos 0 cos 0 01 1 1

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Unformatted text preview: t then y e t, y t and y 2y y e t 2e 2e 0 t e t 0 so e and t te t,   ¡  ¨  ¨   £ so the function also satisfies the initial condition y 0 problem. ¡ $¡ ¡ ¦ ¥ ¨ ¡ £ ¨ ¨ £ ¨   ¡  ¡ £ y0 sin 0 cos 0 cos 0 01 1 1 1, so the function is a solution to the initial value 0 so et is not a solution. t is a solution. £ £ £ ¡ ¦ ¥ ¡ ¦ ¥ ¦ ¥ ¡ ¡ ¡ ¡ ¡ £ ¡ ¡ ¨ ¨ ¦ ¥ ¡  ¨ ¨   ¡  ¡ ¦ ¥  ¡ ¡ ¡ ¡ ¨ £ y tan x y cos2 x sin2 x sin x tan x sin x cos x cosx cos2 x sin2 x sin x sin2 x sin x t and ¡ ¨ ¨ ¡ ¦ ¥ ¦ ¥ ¨ ¨ £ xy y x x ¡ ¨  3. We have y 1 1 x2 so the left hand side is 1 x 1 x 2x ¤ cos2 x £ £ £ sx x2 dx x2 ¨ ¡ ¡ ¨ © ¡ ¨ x 1 4x2 2 x 1 dx 4x2 x3 3 1 4x x x3 3 1 4x 1 3 1 4 ¨ ¨ ¡ ¦ ¡ © ¡ ¦ ¥ ¡ ¨ ¥ (b) We try to find a value of c so that the function y c x2 1 2 has the property y 0 2. In terms of 2 1 2 1 2 . If y 0 c we have y 0 c 0 c 2 we have c 1 2 2 which implies c1 2 1 2 which implies c 1 2 2 1 4. Checking, the function y 1 4 x2 1 2 is a solution to the given initial value problem. (c) You can graph members of the family of functions, or try letting c ∞, or try solving the separable differential equation dy dx xy3 , and you’ll see that there is an “envelope solution” y 0 which is not of the form y c x2 1 2 . 9. Differen...
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This note was uploaded on 01/12/2010 for the course MATH 111 taught by Professor Doolittle during the Fall '06 term at University of California, Berkeley.

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