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May 13th, 2013, 05:27 AM   #1
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differential equation

Given is the differential equation where b is constant and real.

How do I show that for any the solution to the equation is equal to , where and A and B are constant and real?

And I also wonder how to determine for b=2, y(0)=2 and y'(0)=8 the solution to the differential equation?
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May 13th, 2013, 05:36 AM   #2
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Re: differential equation

I would multiply the ODE by , then write the characteristic equation and use the quadratic formula to determine the roots. What do you find?
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May 13th, 2013, 06:44 AM   #3
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Re: differential equation

I find . So ? Almost there, only I have a instead of in the sinus and cosinus...
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May 13th, 2013, 07:26 AM   #4
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Re: differential equation

That is a second order, linear homogeneous differential equation. The theory says that the set of all solutions is a vector space of dimension 2 which means that every solution can be written as a linear combination of two functions that satisfy the equation. Clearly is a linear combination of and . So it is sufficient to show that and satisfy the differential equation and are independent.
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May 13th, 2013, 08:47 AM   #5
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Re: differential equation

I forgot to take the square root of . When I do that, the answer is correct.

And the second question is just substituting b by 2 and equate y and y' to the given values? It gives me A=2 and B=5/4
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