# solutions2ac - Math 33b Quiz 2ac Name UCLA ID 1 Find the...

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Math 33b, Quiz 2ac, January 22, 2008 Name: UCLA ID: 1. Find the general solution to the initial-value problem ( x 2 + xy 3 ) dx + ( - 3 x 2 y 2 ) dy = 0 , y (1) = 8 . (An integrating factor of the form μ ( x ) will be required to make the given equation exact.) Solution. Set M ( x, y ) = ( x 2 + xy 3 ) and N ( x, y ) = ( - 3 x 2 y 2 ). Note that M y = 3 xy 2 and N x = - 6 xy 2 , so the equation is not exact. However, M y - N x N = 9 xy 2 - 3 x 2 y 2 = - 3 x , which is a function only of x . Hence we can use the integrating factor μ ( x ) = e R - 3 x dx = e - 3 log x = x - 3 : ( x - 1 + x - 2 y 3 ) dx + ( - 3 x - 1 y 2 ) dy = 0 . As we have seen in class, this equation should be exact. Hence we want to find a function f ( x, y ) such that f x = ( x - 1 + x - 2 y
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Unformatted text preview: f x = ( x-1 + x-2 y 3 ) and integrating by x , we get f ( x,y ) = log x-x-1 y 3 + φ ( y ) . Diﬀerentiating by y , we have f y =-3 x-1 y 2 + φ ( y ), which must equal-3 x-1 y 2 . Hence we must have φ ( y ) = 0, so we can take φ ( y ) = 0. The solution to the diﬀerential equation is simply f ( x,y ) = C , so log x-x-1 y 3 = C. Using the initial condition y (1) = 8, we have-8 3 = C , so the solution to the initial value problem is log x-x-1 y 3 =-8 3 . 1...
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• Spring '07
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• Math, Boundary value problem, x-1 + x-2

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