I am trying to solve the following: NSK PSKy''+4y'=\tan(t)ZSK NSK I have used the

Roger Smith 2021-12-10 Answered
I am trying to solve the following:
\(\displaystyle{y}{''}+{4}{y}'={\tan{{\left({t}\right)}}}\)
I have used the method of variation of parameters. Currently I am at a point in the equation where I have this:
\(\displaystyle{u}_{{1}}=\int{\frac{{{\tan{{t}}}{\cos{{2}}}{t}}}{{{2}}}}\)
I am stuck here

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Expert Answer

yotaniwc
Answered 2021-12-11 Author has 4975 answers
\(\displaystyle{\cos{{2}}}{t}={2}{{\cos}^{{2}}{t}}-{1}\) hence
\(\displaystyle\int{\frac{{{\tan{{t}}}{\cos{{2}}}{t}}}{{{2}}}}\)dt=\int\sin t\cos tdt-\frac{1}{2}\tan tdtask
\(\displaystyle={\frac{{{1}}}{{{2}}}}{{\sin}^{{2}}{t}}-{\frac{{{1}}}{{{2}}}}{\ln{{\left({\sec{{t}}}\right)}}}+{C}\)
\(\displaystyle={\frac{{{1}}}{{{2}}}}{\left({{\sin}^{{2}}{t}}+{\ln{{\left({\cos{{t}}}\right)}}}\right)}+{C}\)
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SlabydouluS62
Answered 2021-12-12 Author has 1854 answers
Start with the ODE
The homogeneous are \(\displaystyle{y}_{{1}}{\left({t}\right)}={\sin{{2}}}{t}{a}{s}{k}{\quad\text{and}\quad}{P}{S}{K}{\cos{{2}}}{t}{a}{s}{k}.{I}{n}{t}{h}{e}{m}{e}{t}{h}{o}{d}{o}{f}{v}{a}{r}{i}{a}{t}{i}{o}{n}{s},{w}{e}{f}{\quad\text{or}\quad}{m}{t}{h}{e}{p}{a}{r}{t}{i}{c}\underline{{a}}{r}{s}{o}{l}{u}{t}{i}{o}{n}{P}{S}{K}{y}_{{p}}{\left({t}\right)}\) as
\(\displaystyle{y}_{{p}}{\left({t}\right)}={C}_{{1}}{\left({t}\right)}{y}_{{1}}{\left({t}\right)}+{C}_{{2}}{\left({t}\right)}{y}_{{2}}{\left({t}\right)}\)
where the functions \(\displaystyle{C}_{{1}}\) and \(\displaystyle{C}_{{2}}\) are given by
PSKC_1=-\int\frac{1}{W(t)}y_2(t)\tan tdtask
PSKC_2=+\int\frac{1}{W(t)}y_1(t)\tan tdtask
where \(\displaystyle{W}{\left({t}\right)}\) is the Wronskian for \(\displaystyle{y}_{{1}}\) and \(\displaystyle{y}_{{2}}\)
First, the Wronskian is trivially evaluated to be \(\displaystyle{W}=-{2}\)
Second, we evaluate \(\displaystyle{C}_{{1}}\) and \(\displaystyle{C}_{{2}}\)
PSKC_1=\frac{1}{2}\int\cos2t\tan tdtask
PSK=\frac{1}{2}\int(\sin2t-\tan t)dtask
\(\displaystyle=-{\frac{{{1}}}{{{4}}}}{\cos{{2}}}{t}+{\frac{{12}}{{\log{{\left({\cos{{t}}}\right)}}}}}\)
PSKC_2=-\frac12\int\sin2t\tan tdtask
PSK=-\int\sin^2 tdtask
PSK=-\frac{1}{2}t+\frac{1}{4}\sin2task
Third, we determine \(\displaystyle{y}_{{p}}\) as
\(\displaystyle{y}_{{p}}{\left({t}\right)}={\left(-{\frac{{14}}{{\cos{{2}}}}}{t}+{\frac{{12}}{{\log{{\left({\cos{{t}}}\right)}}}}}\right)}{\sin{{t}}}+{\left(-{\frac{{12}}{{t}}}+{\frac{{14}}{{\sin{{2}}}}}{t}\right)}{\cos{}}\)
\(\displaystyle=-{\frac{{12}}{{t}}}{\cos{{2}}}{t}+{\frac{{12}}{{\sin{{2}}}}}{t}{\log{{\left({\cos{{t}}}\right)}}}\)
Finally, the total solution to the ODE is
\(\displaystyle{y}{\left({t}\right)}={A}{\sin{{2}}}{t}+{B}{\cos{{2}}}{t}-{\frac{{12}}{{t}}}{\cos{{2}}}{t}+{\frac{{12}}{{\sin{{2}}}}}{t}{\log{{\left({\cos{{t}}}\right)}}}\)
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