"I am trying to solve the following:y″+4y′=tan⁡(t)I have..." was answered by students like you

Roger Smith

Answered question

2021-12-10

I am trying to solve the following:

y^{″} + 4 y^{'} = \tan (t)

I have used the method of variation of parameters. Currently I am at a point in the equation where I have this:

u_{1} = \int \frac{\tan t \cos 2 t}{2}

I am stuck here

Answer & Explanation

yotaniwc

Beginner2021-12-11Added 34 answers

$\cos 2 t = 2 \cos^{2} t - 1$ hence
$\int \frac{\tan t \cos 2 t}{2}$ $d t = \int \sin t \cos t d t - \frac{1}{2} \tan t d t$
$= \frac{1}{2} \sin^{2} t - \frac{1}{2} \ln (\sec t) + C$
$= \frac{1}{2} (\sin^{2} t + \ln (\cos t)) + C$

SlabydouluS62

Skilled2021-12-12Added 52 answers

Start with the ODE
The homogeneous are $y_{1} (t) = \sin 2 t$ and $\cos 2 t$ . In the method of variations, we form the particular solution $y_{p} (t)$ as
$y_{p} (t) = C_{1} (t) y_{1} (t) + C_{2} (t) y_{2} (t)$
where the functions $C_{1}$ and $C_{2}$ are given by
$C_{1} = - \int \frac{1}{W (t)} y_{2} (t) \tan t d t C_{2} = + \int \frac{1}{W (t)} y_{1} (t) \tan t d t$
where $W (t)$ is the Wronskian for $y_{1}$ and $y_{2}$
First, the Wronskian is trivially evaluated to be $W = - 2$
Second, we evaluate $C_{1}$ and $C_{2}$
$C_{1} = \frac{1}{2} \int \cos 2 t \tan t d t = \frac{1}{2} \int (\sin 2 t - \tan t) d t$
$= - \frac{1}{4} \cos 2 t + \frac{12}{\log (\cos t)}$
$C_{2} = - \frac{1}{2} \int \sin 2 t \tan t d t = - \int \sin^{2} t d t = - \frac{1}{2} t + \frac{1}{4} \sin 2 t$
Third, we determine $y_{p}$ as
$y_{p} (t) = (- \frac{14}{\cos 2} t + \frac{12}{\log (\cos t)}) \sin t + (- \frac{12}{t} + \frac{14}{\sin 2} t) \cos$
$= - \frac{12}{t} \cos 2 t + \frac{12}{\sin 2} t \log (\cos t)$
Finally, the total solution to the ODE is
$y (t) = A \sin 2 t + B \cos 2 t - \frac{12}{t} \cos 2 t + \frac{12}{\sin 2} t \log (\cos t)$

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