solved particular integral of linear ode dy/dt+cos(t)y=3cos(t)

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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

\[y\left(t\right)=e^{-\sin{\left(t\right)}}\left(\frac{3t}{4}+\frac{3\sin{\left(2t\right)}}{4}\right)+5\]


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