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Of the differential equation y y 4sec x obviously the

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, of the differential equation y y 4sec( x ). Obviously the driving function here is NOT a UC function. Thus, we must use variation of parameters to nab the culprit. Corresponding Homogeneous: F.S. = {sin( x ), cos( x )}. y y 0. If y p = v 1 cos( x ) + v 2 sin( x ) then v 1 and v 2 are solutions to the following system: Solving the system yields v 1 = -4tan( x) and v 2 = 4. Thus, by integrating, we obtain v 1 4ln sec( x ) c and v 2 4 x d . Thus, a particular integral of the ODE above is y p v 1 cos( x ) v 2 sin( x ) 4ln sec( x ) cos( x ) 4 x sin( x ). ______________________________________________________________________ 4. (10 pts.) Set up the correct linear combination of undetermined coefficient functions you would use to find a particular integral, y p , of the O.D.E. y 4 y 5 y x 2 e 2 x sin( x ) e 2 x . [ Warning: (a) If you skip a critical initial step, you will get no credit!! (b) Do not waste time attempting to find the numerical values of the coefficients!! ] First, the corresponding homogeneous equation is y 4 y 5 y 0. which has an auxiliary equation given by 0 = m 2 - 4 m + 5. Thus, m = 2 + i or m = 2 - i , and a fundamental set of solutions for the corresponding homogeneous equation is { exp(2 x )cos( x ), exp(2 x )sin( x ) }. Taking this into account, we may now write y p ( x ) Ax 2 e 2 x Bxe 2 x Ce 2 x Dx cos( x ) e 2 x Ex sin( x ) e 2 x or something equivalent. ______________________________________________________________________ Silly 10 Point Bonus: Let f ( x ) = x and g ( x ) = sin( x ). (a) It is trivial to obtain a 4th order homogeneous linear constant coefficient ordinary differential equation with f and g as solutions. Do so. (b) It’s only slightly messier to obtain a 2nd order homogeneous linear ordinary differential equation with { f , g } as a fundamental set of solutions. Do so.
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