Solve the Differential equations y'' + 4y = 4 tan^{2} x

Question
Differential equations
asked 2021-02-21
Solve the Differential equations \(y'' + 4y = 4 \tan^{2} x\)

Answers (1)

2021-02-22
We need to solve following differential equation
\(y" + 4y = 4 \tan^{2} x\)
Let us first solve the homogeneous part.
\(y" + 4y = 0\)
Let \(y = e^{m}x\) be the trial solution. Then we get the auxiliary equation
\(m^{2} + 4 = 0\)
This gives us
\(m = \pm 2i\)
Thus the solution of the homogeneous part of the differential equation be
\(y_{h}(x) = c_{1} \cos(2x) + c_{2} \sin(2x)\)
We use variation of parameter to calculate a
particular solution y_h. Let us take
\(y_{1} = \cos 2x, y_{2} = \sin 2x\)
Then we get
\(y`_{1} = -2 \sin 2x, y`_{2} = 2 \cos 2x\)
The wro
ian of \(y_{1}\ and\ y_{2}\) is
\(W = y_{1}y`_{2} - y`_{1}y_{2}\)
\(= (\cos 2x)(2 \cos 2x) - (-2 \sin 2x)(\sin 2x) = 2 \neq 0\)
In this case it is given that
\(f(x) = 4 \tan(2x)\)
Then by the formula we obtain
\(u_{1} = \int \frac{-y_{2}f}{W}dx\)
\(=\int \frac{-(\sin 2x)(4 \tan 2x)}{2}dx\)
\(=-2 \int \frac{\sin^{2}(2x)}{\cos(2x)}dx\)
\(=-2 \int (\frac{1- \cos^{2}2x}{\cos 2x})dx\)
\(=2 \int (\cos 2x- \sec 2x)dx\)
\(= 2 \int (\cos 2x - \sec 2x)dx\)
\(= \sin 2x - \ln |\sec 2x + \tan 2x|\)
Similarly
\(u_{2} = \int \frac{-y_{1}f}{W}dx\)
\(=\int \frac{(\cos 2x)(4 \tan 2x)}{2}dx\)
\(= 2 \int \sin(2x) dx\)
\(= -\cos (2x)\)
Thus by variation of parameters the patricular solution is
\(y_{p} = u_{1} y_{1} + u_{2} y_{2}\)
\(= (\sin 2x - \ln |\sec 2x + \tan 2x|)(\cos 2x) + (-\cos 2x)(\sin 2x)\)
\(y_{p} = -\cos 2x \ln |\sec 2x + \tan 2x|\)
Hence the general solution is
\(y = y_{c} + y_{p} = c_{1} \cos 2x + c_{2} \sin 2x - \cos 2x \ln |\sec 2x + \tan 2x|\)
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