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Thank you for taking my question. I am studying for a test; I can figure out my other 49 practice problems, but I just can't figure this one out (just practice,not homework). I was hoping you could provide your answer as well as a basic idea of how you got your answer; this one keeps tripping me up. Thank you for your time, Clark

Solve the initial value problem:

dy/dx=-(1/x^2)-(3/x^4)+12 and y=3 when x=1

The problem is a first order differential equation, and can be reduced directly by integration.

Integrating both sides wrt x:

y = - x^(-2+1)/ (-2+1) - 3 x^ (-4+1)/ (-4+1) +12x +C

= 1/x + 1/(x^3) + 12x + C ------(1)

When x=1 and y=3, substituting these into (1) gives

3 = 1 + 1 + 12(1) +C

C= 3-14 =-11

As such, the particular solution to the differential equation dy/dx=-(1/x^2)-(3/x^4)+12

is given by

Hope this helps. Good luck for your test. Peace.

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