MAP
de-t1-a

# B 4 e 2 x y 2 dx 3 y 2 2 xy dy this varmint is

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(b) 4 e 2 x y 2 dx 3 y 2 2 xy dy 0 This varmint is plainly exact. A one-parameter family of solutions is given by , where C is an arbitrary constant. (c) dy dx 1 x y 2 x 2 1 ; y (1) ln(4). The ODE of problem (c) is linear as written with an obvious integrating factor of µ ( t ) e ln( x ) x for x > 0. An explicit solution to the IVP is given by y ( x ) ln( x 2 1) ln(2) x for x > 0.

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TEST1/MAP2302 Page 2 of 3 ______________________________________________________________________ (d) ( x sec y x y ) dx ( x ) dy 0 ; y (1) π 6 The ODE is homogeneous. The degree of homogeneity is 1. Look at the " y / x " as a hint. Then write the equation in the form of dy / dx = g ( y / x ) by doing suitable algebra carefully. After setting y = vx , substituting, and doing a bit more algebra, you will end up looking at the separable equation sec( v ) dx - ( x ) dv = 0. Separating variables and integrating leads you to . cos( v ) dv 1 x dx C After doing that and integrating, and applying the initial condition you’ll obtain , sin( y x ) ln( x ) 1 2 for x > 0 or an equivalent explicit solution like y ( x ) x sin 1 (ln( x ) 1 2 ) for x > 0. ______________________________________________________________________ 2. (5 pts.) For certain values of the constant m the function f ( x ) = x m is a solution to the differential equation . x 2 y ( x ) 2 xy ( x ) 0 Determine all such values of m.
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