de-t2-a

By using the two initial conditions now you can

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. By using the two initial conditions now, you can obtain an easy to solve linear system involving the two constants. Solving the system reveals that the solution to the initial value problem is y ( x ) 4 3 e x 2 xe x
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TEST2/MAP2302 Page 2 of 4 ______________________________________________________________________ 3. (15 pts.) Find a particular integral, y p , 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)
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By using the two initial conditions now you can obtain an...

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