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# Cos v dv 1 x dx c after doing that and integrating

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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. It turns out that the function f ( x )= x m is a solution to the differential equation precisely when 0 2 x ( mx m 1 ) x 2 ( m ( m 1) x m 2 ( m ( m 3)) x m for all x. This is equivalent to m =0o r m =3 . ______________________________________________________________________ 3. (5 pts.) It is known that every solution to the differential equation y - y = 0 is of the form . Which of these functions satisfies the initial conditions y (0) = 2 and y ( 0

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TEST1/MAP2302 Page 3 of 3 ______________________________________________________________________ ______________________________________________________________________ 4. (10 points) The following differential equation may be solved by either performing a substitution to reduce it
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cos v dv 1 x dx C After doing that and integrating and...

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