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Question Number 97794 by abdomathmax last updated on 09/Jun/20

$$\mathrm{solve}{\mathrm{y}}^{{}^{″}}+\mathrm{y}=\frac{1}{\mathrm{cosx}}$$

Commented by john santu last updated on 10/Jun/20

$${\mathrm{y}}^{″}+{\mathrm{y}}^{\prime }=\sec \mathrm{x}$$$$\mathrm{homogenous}\mathrm{solution}$$$${\mathrm{y}}_{\mathrm{h}}={\mathrm{C}}_1\cos \mathrm{x}+{\mathrm{C}}_2\sin \mathrm{x}$$$$\mathrm{particular}\mathrm{solution}$$$${\mathrm{y}}_{\mathrm{p}}={\mathrm{v}}_1\left(\mathrm{x}\right)\cos \mathrm{x}+{\mathrm{v}}_2\left(\mathrm{x}\right)\sin \mathrm{x}$$$$\mathrm{where}{\mathrm{v}}_1^{\prime }\cos \mathrm{x}+{\mathrm{v}}_2^{\prime }\sin \mathrm{x}=0$$$$-{\mathrm{v}}_1^{\prime }\sin \mathrm{x}+{\mathrm{v}}_2^{\prime }\cos \mathrm{x}=\sec \mathrm{x}$$$$\mathrm{we}\mathrm{get}\{\begin{array}{l}{\mathrm{v}}_1\left(\mathrm{x}\right)=\int -\tan \mathrm{x}\mathrm{dx}=\ln \mid \cos \mathrm{x}\mid \\ {\mathrm{v}}_2\left(\mathrm{x}\right)=\int \mathrm{dx}=\mathrm{x}\end{array}$$$$\mathrm{so}{\mathrm{y}}_{\mathrm{p}}=(\ln \mid \cos \mathrm{x}\mid )\cos \mathrm{x}+\sin \mathrm{x}$$$$\mathrm{generall}\mathrm{solution}$$$$\mathrm{y}=(\ln \mid \cos \mathrm{x}\mid )\cos \mathrm{x}+\sin \mathrm{x}+{\mathrm{C}}_1\cos \mathrm{x}+{\mathrm{C}}_2\sin \mathrm{x}◼$$

Commented by bobhans last updated on 10/Jun/20

nice.....very...nice

Commented by bemath last updated on 10/Jun/20

very....afdolll

Answered by niroj last updated on 09/Jun/20

$${\mathrm{y}}^{{}^{″}}+\mathrm{y}=\frac{1}{\cos \mathrm{x}}$$$$\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}+\mathrm{y}=\frac{1}{\cos \mathrm{x}}$$$$({\mathrm{D}}^2+1)\mathrm{y}=\sec \mathrm{x}$$$$\mathrm{A}.\mathrm{E}.,{\mathrm{m}}^2+1=0$$$$\mathrm{m}=0\stackrel{-}{+}1\mathrm{i}$$$$\mathrm{CF}={\mathrm{C}}_1\cos \mathrm{x}+{\mathrm{C}}_2\sin \mathrm{x}$$$$\mathrm{Here},\mathrm{By}\mathrm{the}\mathrm{help}\mathrm{of}\mathrm{variation}\mathrm{parameters}:$$$${\mathrm{y}}_1=\cos \mathrm{x},{\mathrm{y}}_2=\sin \mathrm{x},\mathrm{w}=\sec \mathrm{x}$$$$\mathrm{W}=\left|\begin{array}{c}{\mathrm{y}}_1{\mathrm{y}}_2\\ {\mathrm{y}}_1^{{}^{\prime }}{\mathrm{y}}_2^{{}^{\prime }}\end{array}\right|$$$$\mathrm{W}=\left|\begin{array}{c}\cos \mathrm{x}\sin \mathrm{x}\\ -\sin \mathrm{x}\cos \mathrm{x}\end{array}\right|$$$$\mathrm{W}={\cos }^2\mathrm{x}-(-{\sin }^2\mathrm{x})=1$$$$\mathrm{W}=1$$$$\mathrm{PI}=-{\mathrm{y}}_1\int \frac{{\mathrm{y}}_2\mathrm{x}}{\mathrm{w}}\mathrm{dx}+{\mathrm{y}}_2\int \frac{{\mathrm{y}}_1\mathrm{x}}{\mathrm{w}}\mathrm{dx}$$$$=-\mathrm{cosx}\int \frac{\sin \mathrm{x}\sec \mathrm{x}}{1}\mathrm{dx}+\sin \mathrm{x}\int \frac{\cos \mathrm{x}\sec \mathrm{x}}{1}\mathrm{dx}$$$$=-\cos \mathrm{x}\int \tan \mathrm{x}\mathrm{dx}+\mathrm{sinx}\int \mathrm{dx}$$$$=-\cos \mathrm{x}\log \left(\sec \mathrm{x}\right)+\sin \mathrm{x}.\mathrm{x}$$$$\mathrm{y}=\mathrm{CF}+\mathrm{PI}$$$$\therefore \mathrm{y}={\mathrm{C}}_1\cos \mathrm{x}+{\mathrm{C}}_2\sin \mathrm{x}-\cos \mathrm{x}\log \left(\sec \mathrm{x}\right)+\mathrm{x}\sin \mathrm{xo}$$

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