MATH201 Midterm 2015 F Solution

MATH201 Midterm 2015 F Solution - MATH 201Midterm Exam...

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MATH 201—Midterm Exam Solutions 1. Consider the differential equation cos x + 4 x 2 y + 2 3 y 3 dx + x 3 + xy 2 + y x dy = 0, x > 0. (a) Determine whether the equation is exact. If it is not, find an integrating factor of the form μ = μ ( x ) to make it exact. [4] Solution: Let M = cos x + 4 x 2 y + 2 3 y 3 , N = x 3 + xy 2 + y x . Then ∂M ∂y = 4 x 2 + 2 y 2 but ∂N ∂x = 3 x 2 + y 2 - y x 2 so ∂M ∂y 6 = ∂N ∂x . The required integrating factor is simply μ ( x ) = x . Letting ˜ M = xM , ˜ N = xN , we get ˜ M ∂y = 4 x 3 + 2 xy 2 = ˜ N ∂x , so now the differential equation is exact. (b) Find the general solution of the differential equation from part (a). (You may express the solution in implicit form.) [4] The solution is given by integrating: F ( x, y ) = ˜ Mdx = Z x cos x + 4 x 3 y + 2 3 xy 3 dx = x sin x +cos x + x 4 y + 1 3 x 2 y 3 + C ( y ) . Differentiate this with respect to y and compare to ˜ N to see that C 0 ( y ) = y , so C ( y ) = 1 2 y 2 , up to a constant. Thus, the solution of the DE is given in implicit form as x sin x + cos x + x 4 y + 1 3 x 2 y 3 + 1 2 y 2 = const.
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2. Find the general solution of 4 y 00 ( t ) + y ( t ) = 2 cos( t/ 2) + te - t/ 2 . [10] Solution: It is easy to see that the corresponding homogeneous equation 4 y 00 ( t ) + y ( t ) = 0 has general solution y H = C 1 cos( t/ 2) + C 2 sin( t/ 2) .
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  • Fall '16
  • dt, Recurrence relation, general solution

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