practice(exam1)MA262fall2011

7 solve dy dx 4 y tan x sin x y y 0 3 14 2 3 8 solve

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will it take so that the concentration of Nesquik is 106g/L? 7. Solve dy dx - 4 y tan x = - sin x y , y (0) = 3 14 2 / 3 . 8. Solve ( x + 4 y ) dy + ( y - x ) dx = 0 . 9. Solve ( x - 4 y ) dy - ( y + x ) dx = 0 . 1
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2 INSTRUCTOR: RAPHAEL HORA 10. Solve y x + 6 x dx + (ln x - 2) dy = 0 , x > 0. 11. Solve y 00 + 2 x - 1 y 0 = 4 x, y 0 (1) = 2 , y (1) = 2 . 12. Solve d 2 y dx 2 = 6 y 4 dy dx 3 - 2 y - 1 dy dx 2 Let v = dy dx d 2 y dx 2 = v dv dy v dv dy = 6 y 4 v 3 - 2 y - 1 v 2 dv dy = 6 y 4 v 2 - 2 y - 1 v dv dy + 2 y - 1 v = 6 y 4 v 2 . This is a Bernoulli equation, so rewrite it as v - 2 dv dy + 2 y - 1 v - 1 = 6 y 4 and make u = v - 1 . So du dy = - v - 2 dv dy ⇒ - du dy + 2 y - 1 u = 6 y 4 du dy - 2 y - 1 u = - 6 y 4 Computing the integrating factor I ( x ) = e R - 2 y - 1 dy = e - 2 ln | y | = y - 2 , so d dy ( y - 2 u ) = - 6 y 2 y - 2 u = - 2 y 3 + C 1 u = - 2 y 5 + C 1 y 2 v - 1 = - 2 y 5 + C 1 y 2 v = 1 C 1 y 2 - 2 y 5 Hence dy dx = 1 C 1 y 2 - 2 y 5 ( C 1 y 2 - 2 y 5 ) dy = dx C 1 3 y 3 - 1 3 y 6 = x + C 0 13. Let A = 1 2 3 - 4 - 4 - 4 5 6 7 and B = 2 - 5 1 0 3 - 2 1 2 - 4 . Verify that AB 6 = BA . 14. A real matrix A is orthogonal if A T = A - 1 , that is, if AA T = A T A = I . Thus A must necessarily be square and invertible. Verify that A = 1 / 9 8 / 9 - 4 / 9 4 / 9 - 4 / 9 - 7 / 9 8 / 9 1 / 9 4
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