Engineering Math
posted by Thomas .
solve the 1st differential equation: y1+4y=cos(x) with intial condition y(pie/3)=1

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.