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January 30, 2015

January 30, 2015

Posted by **Thomas** on Friday, June 24, 2011 at 2:26pm.

- Engineering Math -
**Count Iblis**, Friday, June 24, 2011 at 2:53pmcos(x) is the real part of exp(i x). You can then solve:

y' + 4 y = exp(i x)

and take the real part of the general solution (and then impose the boundary conditions). This is because the differential equation is linear and taking the ral part of both sides gives:

Re[y' + 4 y] = Re[exp(i x)] = cos(x)

And

Re[y' + 4 y] = Re(y)' + 4 Re(y)

The homogeneous part:

yh' + 4 yh = 0

as the solution:

yh = A exp(-4 x)

A solution to

y' + 4 y = exp(i x)

can be found by putting y = B exp(i x):

i B + 4 B = 1 ------>

B = 1/(i + 4) = (4-i)/(17)

So, we have:

y = (4/17 - i/17) exp(ix)

The real part of this is:

Re(y) = 4/17 cos(x)+ 1/17 sin(x)

The general solution is thus:

y = A exp(-4 x) +

4/17 cos(x)+ 1/17 sin(x)

You can then insert x = pi/3 in here and equate that to 1 to find A.

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