2302-practice-mid1-soln

Now we see that the relevant second partials are in

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Now we see that the relevant second partials are in fact equal: ∂M ∂y = 2 x = ∂N ∂x , so the equation is exact. 1 If we wanted to, we could verify this by considering the four points ( x,y ) = (1 , 1) , (1 , 2) , (2 , 1) , (2 , 2), but that won’t be required on the midterm.
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MAC 2313, Fall 2010 — Midterm 1 Review Solutions 3 We now proceed to solve the equation; remember that the solution will be an implicit solution of the form F ( x, y ) = C where d dx F ( x, y ) = M ( x, y ) + N ( x, y ) dy dx . First we integrate M ( x, y ) with respect to x : F ( x, y ) = integraldisplay 2 xy dx = x 2 y + g ( y ) . Note that, as usual, our constant of integration is a function of y . Now we want to have ∂F ∂y = N ( x, y ) , so we get x 2 + dg dy = x 2 + y 2 . Therefore dg/dy = y 2 , so g ( y ) = y 3 / 3 + C . Thus our solution can be written as x 2 y + y 3 3 + C = D, where D is another arbitrary constant. Since we don’t need both of these constants, we can also write this solution as x 2 y + y 3 3 = C. In this particular case, we could solve to get y as an explicit function of x , but there is no need, so we might as well leave it in the implicit form above. b. dy dx = xy sin x . Solution: This equation is separable, because we can write it as 1 y dy dx = x sin x. Therefore, by our previous problem, this equation is also exact. Furthermore, the equation is also linear, because we can express it as dy dx ( x sin x ) y = 0 . Therefore we have three methods available to solve the equation. Viewing it as a separable equation, we separate to get 1 y dy dx = x sin x.
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MAC 2313, Fall 2010 — Midterm 1 Review Solutions 4 We then integrate both sides with respect to x , integraldisplay 1 y dy dx dx = integraldisplay x sin x dx, integraldisplay 1 y dy = sin x x cos x + C (obtained by integration by parts), ln | y | = sin x x cos x + C, y ( x ) = ± e C e sin x - x cos x , y ( x ) = Ce sin x - x cos x . Note that in going from the second-to-last line to the last line above, we have done our standard abuse of constants, replacing ± e C by C . It is worth noting that C cannot be 0. c. dy dx = sin( x + y 2 ). Solution: The equation is not separable 2 . The equation is also not linear, because we cannot express it in the form dy dx + P ( x ) y = Q ( x ). Finally, the equation is not exact; to see this, we express it in the form M ( x, y ) + N ( x, y ) dy dx = 0, where we see that M ( x, y ) = sin( x + y 2 ) and N ( x, y ) = 1 , and the partials do not match: ∂M ∂y = 2 y cos( x + y 2 ) negationslash = 0 = ∂N ∂x . Since none of our methods apply to this equation, we cannot solve it (yet). d. dy dx = x - y 2 x . 2 Again, if we wanted to, we could verify this by considering the three points ( x,y )=(0 , 0) , ( π / 2 , 0) , (0 , radicalbig π / 2 ), but again, that won’t be required on the midterm.
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MAC 2313, Fall 2010 — Midterm 1 Review Solutions 5 Solution: The equation is not separable (again this could be verified with a chart). The
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