Posted by Thomas on Friday, June 24, 2011 at 2:26pm.
solve the 1st differential equation: y1+4y=cos(x) with intial condition y(pie/3)=1

Engineering Math  Count Iblis, Friday, June 24, 2011 at 2:53pm
cos(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) = (4i)/(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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