solutions2bd - f x,y = C From ∂f ∂x = 2 x y 3 we see...

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Math 33b, Quiz 2bd, April 19, 2007 Name: UCLA ID: 1. Find the general solution to the differential equation (2 x 2 + xy 3 ) dx + (3 x 2 y 2 ) dy = 0 . (An integrating factor of the form μ ( x ) will be required to make the given equation exact.) Solution. Set M ( x, y ) = (2 x 2 + xy 3 ) and N ( x, y ) = 3 x 2 y 2 . We compute ∂M ∂y = 3 xy 2 and ∂N ∂x = 6 xy 2 , so the differential equation is not exact. However, note that ∂M ∂y - ∂N ∂x = - 3 xy 2 , and that ∂M ∂y - ∂N ∂x N = - 3 xy 2 3 x 2 y 2 = - 1 x , which is a function only of x . Therefore, we can use the function μ ( x ) = e R - 1 x = e - log | x | = 1 x as an integrating factor: (2 x + y 3 ) dx + 3 xy 2 dy = 0 As we have seen in class/in the text, this equation is guaranteed to be exact (you can check this for yourself, and it is probably a good idea to do so, but it is not necessary for the purposes of scoring on this quiz). We seek a function f ( x, y ) such that ∂f ∂x = 2 x + y 3 and ∂f ∂y = 3 xy 2
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Unformatted text preview: f ( x,y ) = C . From ∂f ∂x = 2 x + y 3 , we see that f ( x,y ) = x 2 + xy 3 + φ ( y ). From ∂f ∂y = 3 xy 2 and f ( x,y ) = x 2 + xy 3 , we see that 3 xy 2 = 3 xy 2 + φ ( y ), so φ ( y ) = 0. We can choose φ ( y ) to be 0 as well. (In general, φ ( y ) could be some constant K , but this will turn out not to matter.) The solution to the differential equation is given by f ( x,y ) = C , or x 2 + xy 3 = C. (Note that replacing x 2 + xy 3 with x 2 + xy 3 + K on the left-hand side doesn’t change anything, as both constants can be moved to the right-hand side.) The solution can further be rewritten as xy 3 = C-x 2 y 3 = C x-x y = ( C x-x ) 1 3 . 1...
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