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

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

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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