MATH111-200630-PS08-Solutions

4 first we verify that the function is a solution to

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Unformatted text preview: have 1 2 sec3 θ dθ ¦ ¥ ¡ ¤ 1 dy 1 ¨ ¦ © ¡ ¥ ¡ dx dy 2 y2 dy 4 ¦ ¨ ¥ ¨ ¨  ¡ ¦ ¥ © ¡ ¡ ¨ © ¡ ¦ ¥ © © © so 1 1 1 which agrees with the right hand side. 4. First we verify that the function is a solution to the differential equation. The left hand side is which agrees with the right hand side, so the function is a solution to the differential equation. Second, we verify that the given function satisfies the initial condition: 5. (a) If y (b) If y (c) If y et then y e te y et , y e et and y e t, 2y e y 4et so te (d) If y t is a solution. so t 2 e 6. We have y y t is not a solution. rert , y 6y r2 ert so 7. We did this one in class. 8. (a) The left hand side of the equation is by the chain rule. The right hand side of the equation is The two sides agree, so the given function is a solution to the differential equation. £ £ xy3 x c x2 1 2 xc x2 ¤  ¦ ¥ ¡ 3 £  ¦ £ ¥ £ y x2 3 2 2x xc x2 3 2  ¦ ¥ ¡ £  ¦ 1 c 2 3 2 3 £ £ The above expression is zero if and only if r2 r 6 0, i.e., if r 3 or r ¡ ¡ ¤ ¡ £ ¨ £ ¥ ¡ ¡ £ y r2 ert rert 6ert r2 r 6 ert £ ¦ ¨ ¥ ¡ £ y 2y y 2e t 4te t t 2e t 4te t 2t 2e t t 2e t 2e t 0 2. £ $¡  ¡  ¨  £  £ ¨  ¨  £ t 2e t then y 2te t t 2e t , y 2e t 2te t 2te t t 2e t 2e t 4te t t 2e  ¨   ¡  ¨    £ ¡    £ 2y y 2e t te t 2e t 2te t te t £ ¡  ¨  £  £ ¨  ¨  £ t then y t te y t e t te t £ ¡  ¨   ¡  ¨...
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