Introduction : The differential equation dy/dt+cos(t)y=3cos(t) is a linear ode. We shall apply integrating factor method to solve this. Finally using initial condition we shall get it’s particular integral.
Question : Find particular integral of the ode
\[\frac{dy}{dt}+ycos\left(t\right)=6\cos{\left(t\right)}\]
satisfying
\[y\left(0\right)=5\]
Solution :
\[\frac{dy}{dt}+ycos\left(t\right)=6\cos{\left(t\right)}\]
\[y\left(0\right)=8\]
Compare with Linear ODE
\[\frac{dy}{dt}+P\left(t\right)y=Q\left(t\right)\]
\[P(t)=\cos{\left(t\right)}, Q(t)=6\cos{\left(t\right)}\]
integrating factor
\[IF(t) =e^{\int{P\left(t\right)}dt} = e^{\int{cos\left(t\right)}dt}\]
Hence general solution of the ODE is
\[y\left(t\right)e^{\int{cos\left(t\right)}dt}=3\int{cos\left(t\right)}\cos{\left(t\right)}dt\]
\[y\left(t\right)e^{\int{cos\left(t\right)}dt}=\int{Q\left(t\right)}IF(t)dt\]
\[y\left(t\right)e^{\sin{\left(t\right)}}=\frac{3}{2}\left(t+\frac{\sin{\left(2t\right)}}{2}\right)+A\]
to determine value of A applying
\[y\left(0\right)=5\]
\[5=0+A\]
\[A=5\]
Hence particular integral is
\[y\left(t\right)=e^{-\sin{\left(t\right)}}\left(\frac{3t}{4}+\frac{3\sin{\left(2t\right)}}{4}\right)+5\]