Question

Help me with this question, find the general solution to the equation dy/dx + y = 5cos2x, the answer should be y = cos2x + 2sin2x + Ae^-x, but I don't know how to get there so show some working

Answer 1

\[y \prime+y=5\cos 2x\]
\[I.F.=e ^{\int\limits 1 dx}=e^x\]
\[C.S.~is~y~e^x=\int\limits5 \cos 2x~e^x~dx+c=5 I _{1}+c\]
\[I _{1}=\int\limits \cos 2x~e^x~dx=\cos 2x~e^x-\int\limits\left( -2\sin 2x \right)e^x~dx\]
\[I _{1}=e^x \cos 2x+2\left[ \sin 2x~e^x-\int\limits \left(2 \cos 2x \right)e^xdx \right]\]
\[I _{1}=e^x \cos 2x+2e^x \sin 2x-4 ~I _{1}\]
complete it.

Answer 2

The following is known:
For a differential equation of the form
\[\frac{ dy }{ dx }+a(x)y(x)=b(x)\]
the general solution is given by
\[y(x)=e^{-\int\limits a(x)dx}\left( \int\limits b(x)\: e^{\int\limits a(x)dx}dx +c \right)\]
Now in your ODE we identify
\[a(x)=1\] and \[b(x)=5\cos(2x)\]
For simplicity you can integrate a(x) first
\[\int\limits a(x)\:dx=\int\limits 1\:dx=x\]
Now we have got that
\[y(x)=e^{-x}\left( \int\limits\limits 5\cos(2x)\: e^{x}dx +c \right)\]
And if you evaluate the last integral you get
\[y(x)=e^{-x}\left( e^x (\cos(2 x)+2\sin(2 x)) +c \right)\]
Which can be further simplified
\[y(x)=\cos(2 x)+2\sin(2 x)+c\:e^{-x}\]