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Unformatted text preview: y = 2e x 3 + 2 y , y (0) = 0 and determine where the solution attains its maximum value. ANSWER: This is a separable equation: (3 + 2 y ) dy = (2e x ) dx, so (integrate) 3 y + y 2 = 2 xe x + C. Since y (0) = 0 , we have 0 =1 + C, so that C = 1 . We get 2 xe x + 1 = 3 y + y 2 = ± y + 3 2 ¶ 29 4 , so that ± y + 3 2 ¶ 2 = 2 xe x + 13 4 . We take the positive square root to get: y =3 2 + r 2 xe x + 13 4 . The x value where y attains its maximum is where 2 xe x + 13 4 attains its maximum. Settin the derivative to be zero, we get 0 = d dx ± 2 xe x + 13 4 ¶ = 2e x . So it is where e x = 2 , that is, x = ln2 ....
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 Spring '06
 Mohanty
 Math, Boundary value problem, e−t sin tdt

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