Thus b is a solution c if y x 3 x 2 then y 3 x 2 2 x

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Thus (b) is a solution. c. If y = x 3 x 2 then y = 3 x 2 2 x and y ′′ = 6 x 2. Then xy ′′ + y = x (6 x 2) + ( 3 x 2 2 x ) = 9 x 2 4 x negationslash = 0 . Thus (c) is not a solution. The only solution is (b). 6. Find and classify the critical points for the function f ( x, y ) = x 3 3 xy + y 3 6. We must solve the system f x = 3 x 2 3 y = 0 (1) f y = 3 x + 3 y 2 = 0 . (2) Solving (1) for y yields y = x 2 . (3) Plugging (3) into (2) yields 3 x + 3 x 4 = 0 3 x (1 x 3 ) = 0 x = 0 or x = 1 . If x = 0 then (3) says that y = 0. Thus (0 , 0) is a critical point. If x = 1 then (3) says y = 1. Thus (1 , 1) is a critical point. These are the only two critical points. 4
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To test what kind of points these are, we form the function D ( x, y ) = f xx f yy ( f xy ) 2 = (6 x )(6 y ) ( 3) 2 = 36 xy 9 We see that D (0 , 0) = 9 < 0 so (0 , 0) is a saddle point. Next, D (1 , 1) = 36 9 > 0 and f xx (1 , 1) = 6 > 0 . Thus (1 , 1) is a relative minimum. 7. a. Find the general solution to the differential equation 3 y = cos x y 2 sin x . We observe that this differential equation is separable since dx dy = parenleftbigg 1 3 y 2 parenrightbigg parenleftBig cos x sin x parenrightBig . Separating variables, we get 3 y 2 dy = cos x sin x dx. Taking the integral of both sides, integraldisplay 3 y 2 dy = integraldisplay cos x sin x dx y 3 = ln | sin x | + C ( u -substitution with u = sin x ) y = 3 radicalbig ln | sin x | + C. b. Find the particular solution of the above differential equation with the initial condition y ( π/ 2) = 1. The general solution from above is y = 3 radicalbig ln | sin x | + C . The initial condition says 1 = 3 radicalbig ln | sin( π/ 2) | + C 1 3 = ln | sin( π/ 2) | + C 1 = ln | 1 | + C ( sin( π/ 2) = 1) 1 = C. ( ln(1) = 0) Thus the particular solution is y = 3 radicalbig ln | sin x | + 1. 5
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8. Use Lagrange multipliers to find the maximum and minimum of the function f ( x, y ) = xy subject to the constraint 2 x 2 + 3 y 2 = 9. We write our constraint in the form g ( x, y ) = 2 x 2 + 3 y 2 9 = 0, then form the new function F ( x, y, λ ) = xy + λ (2 x 2 + 3 y 2 9) or F ( x, y, λ ) = xy + 2 λx 2 + 3 λy 2 9 λ.
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