Q3B-230

# Q3B-230 - μ x,y = xe x Solve this new equation Comparing...

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Math 415 Name: Quiz 3B Determine whether the following equation is exact, and if it is, solve it. dy dx = x - 3 y 3 x + 2 y Writing this in standard form, we get (3 y - x ) dx + (3 x + 2 y ) dy = 0. The equation is exact as (3 y - x ) y = 3 = (3 x + 2 y ) x . So there’s some function Ψ( x,y ) with Ψ x = 3 y - x and Ψ y = 3 x + 2 y . Integrating Ψ x (with respect to x ) gets us Ψ( x,y ) = 3 xy - 1 2 x 2 + h ( y ). If we take the partial derivative of this with respect to y and compare to the Ψ y that we already know we see that 3 x +2 y = 3 x + h 0 ( y ). So h 0 ( y ) = 2 y and h ( y ) = y 2 (ignoring the constant for now). Since our ﬁnal solution should be Ψ( x,y ) = C , it is 3 xy - 1 2 x 2 + y 2 = C . Show that ( x + 2) sin ydx + x cos ydy = 0 , is not exact, but becomes exact when multiplied through by the integrating factor
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Unformatted text preview: μ ( x,y ) = xe x . Solve this new equation. Comparing partial derivatives to see whether the equation is exact gets us (( x + 2) sin y ) y = ( x + 2) cos y and ( x cos y ) x = cos y , so the equation is not exact. But if we multiply through by xe x we see that (( x + 2) xe x sin y ) y = ( x + 2) xe x cos y and ( x 2 e x cos y ) x = 2 xe x cos y + x 2 e x cos y , and these two are equal. So the equation is exact now. So Ψ( x,y ) = Z x 2 e x cos y dy = x 2 e x sin y + h ( x ) . And, by comparison h ( x ) = 0, leaving us with the solution x 2 e x sin y = C . 1...
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