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Question Number 162338 by HongKing last updated on 28/Dec/21

$$\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}-3\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)-4\mathrm{y}=\tan \left(\mathrm{x}\right)\log \left(\cos \left(\mathrm{x}\right)\right)$$

Answered by Ar Brandon last updated on 29/Dec/21

$${\mathrm{y}}^{″}-{3\mathrm{y}}^{\prime }-4\mathrm{y}=\tan x\cdot \ln \left(\cos x\right)$$$$\mathrm{For}{\mathrm{y}}_{\mathrm{gh}}:r^2-3r-4=0\Rightarrow r=\frac{3\pm 5}{2}$$$${\mathrm{y}}_{gh}={ae}^{4x}+{be}^{-x}$$$$\mathrm{By}\mathrm{varying}\mathrm{parameters},\mathrm{let}$$$${\mathrm{y}}_{\mathrm{p}}=a\left(x\right)e^{4x}+b\left(x\right)e^{-x}$$$$\{\begin{array}{l}{a}^{\prime }\left(x\right)e^{4x}+b^{\prime }\left(x\right)e^{-x}=0\\ 4a^{\prime }\left(x\right)e^{4x}-b^{\prime }\left(x\right)e^{-x}=\tan x\ln \left(\cos x\right)\end{array}$$$$\mathrm{W}=\left|\begin{array}{cc}{e}^{4x}& e^{-x}\\ 4e^{4x}& -e^{-x}\end{array}\right|=-e^{3x}-4e^{-3x}=-3e^{-3x}\ne 0$$$${\mathrm{W}}_a=\left|\begin{array}{cc}0& e^{-x}\\ \tan x\ln \left(\cos x\right)& -e^{-x}\end{array}\right|=e^{-x}\tan x\ln \left(\cos x\right)$$$${\mathrm{W}}_b=\left|\begin{array}{cc}{e}^{4x}& 0\\ 4e^x& \tan x\ln \left(\cos x\right)\end{array}\right|=e^{4x}\tan x\ln \left(\cos x\right)$$$$a\left(x\right)=\int \frac{{\mathrm{W}}_a}{\mathrm{W}}dx=-\frac{1}{3}\int e^{2x}\tan x\ln \left(\cos x\right)dx$$$$b\left(x\right)=\int \frac{{\mathrm{W}}_b}{\mathrm{W}}dx=-\frac{1}{3}\int e^{7x}\tan x\ln \left(\cos x\right)dx$$

Commented by HongKing last updated on 29/Dec/21

$$\mathrm{cool}\mathrm{my}\mathrm{dear}\mathrm{Sir}\mathrm{thank}\mathrm{you}\mathrm{so}\mathrm{much}$$

Commented by HongKing last updated on 29/Dec/21

$$\mathrm{but}\mathrm{what}\mathrm{is}\mathrm{the}\mathrm{general}\mathrm{solution}\mathrm{dear}\mathrm{Sir}$$

Commented by Ar Brandon last updated on 29/Dec/21

$$\mathrm{We}\mathrm{need}\mathrm{to}\mathrm{find}\mathrm{the}\mathrm{primitives},\mathrm{which}{\mathrm{isn}}^{\prime }\mathrm{t}\mathrm{evident}\mathrm{enough}.$$

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