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

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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yotaniwc
$$\displaystyle{\cos{{2}}}{t}={2}{{\cos}^{{2}}{t}}-{1}$$ 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
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
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 $$\displaystyle{C}_{{2}}$$
$$\displaystyle=-{\frac{{{1}}}{{{4}}}}{\cos{{2}}}{t}+{\frac{{12}}{{\log{{\left({\cos{{t}}}\right)}}}}}$$
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
$$\displaystyle{y}{\left({t}\right)}={A}{\sin{{2}}}{t}+{B}{\cos{{2}}}{t}-{\frac{{12}}{{t}}}{\cos{{2}}}{t}+{\frac{{12}}{{\sin{{2}}}}}{t}{\log{{\left({\cos{{t}}}\right)}}}$$