Question and Answers Forum

All Questions      Topic List

Differential Equation Questions

Previous in All Question      Next in All Question      

Previous in Differential Equation      Next in Differential Equation      

Question Number 104893 by bramlex last updated on 24/Jul/20

$$\frac{d^{2}y}{{dx}^2}+\tan x\frac{dy}{dx}=\sec x+\cot x$$

Answered by mathmax by abdo last updated on 24/Jul/20

$${\mathrm{y}}^{{}^{″}}+\mathrm{tanx}{\mathrm{y}}^{{}^{\prime }}=\frac{1+\mathrm{sinx}}{\mathrm{cosx}}\mathrm{let}{\mathrm{y}}^{{}^{\prime }}=\mathrm{z}\mathrm{so}\mathrm{e}\Rightarrow {\mathrm{z}}^{{}^{\prime }}+\mathrm{tanx}\mathrm{z}=\frac{1+\mathrm{sinx}}{\mathrm{cosx}}$$$$\mathrm{h}\to {\mathrm{z}}^{{}^{\prime }}=-\mathrm{tanx}\mathrm{z}\Rightarrow \frac{{\mathrm{z}}^{{}^{\prime }}}{\mathrm{z}}=-\mathrm{tanx}\Rightarrow \ln \mid \mathrm{z}\mid =-\int \mathrm{tanx}\mathrm{dx}+\mathrm{c}$$$$=-\int \frac{\mathrm{sinx}}{\mathrm{cosx}}\mathrm{dx}+\mathrm{c}=\ln \mid \mathrm{cosx}\mid +\mathrm{c}\Rightarrow \mathrm{z}=\mathrm{k}\mid \mathrm{cosx}\mid \mathrm{let}\mathrm{determine}\mathrm{solution}\mathrm{on}$$$$\{\mathrm{x}/\mathrm{cosx}>0\}\mathrm{mvc}\mathrm{method}\Rightarrow {\mathrm{z}}^{{}^{\prime }}={\mathrm{k}}^{{}^{\prime }}\mathrm{cosx}-\mathrm{ksinx}$$$$\mathrm{e}\Rightarrow {\mathrm{k}}^{{}^{\prime }}\mathrm{cosx}-\mathrm{ksinx}+\mathrm{tamx}\times \mathrm{k}\mathrm{cosx}=\frac{1+\mathrm{sinx}}{\mathrm{cosx}}\Rightarrow {\mathrm{k}}^{{}^{\prime }}=\frac{1+\mathrm{sinx}}{{\cos }^2\mathrm{x}}\Rightarrow$$$$\mathrm{k}=\int \frac{1+\mathrm{sinx}}{{\cos }^2\mathrm{x}}\mathrm{dx}=\int \frac{\mathrm{dx}}{{\cos }^2\mathrm{x}}+\int \frac{\mathrm{sinx}}{{\cos }^2\mathrm{x}}\mathrm{dx}=\frac{1}{\mathrm{cosx}}+\int \frac{\mathrm{dx}}{{\cos }^2\mathrm{x}}\mathrm{we}\mathrm{have}$$$$\int \frac{\mathrm{dx}}{{\cos }^2\mathrm{x}}=\int (1+{\tan }^2\mathrm{x})\mathrm{dx}=\mathrm{tanx}+\mathrm{c}\Rightarrow \mathrm{k}\left(\mathrm{x}\right)=\frac{1}{\mathrm{cosx}}+\mathrm{tanx}+\lambda \Rightarrow$$$$\mathrm{z}\left(\mathrm{x}\right)=(\frac{1}{\mathrm{cosx}}+\mathrm{tanx}+\lambda )\mathrm{cosx}=1+\mathrm{sinx}+\lambda \mathrm{cosx}$$$${\mathrm{y}}^{{}^{\prime }}=\mathrm{z}\Rightarrow {\mathrm{y}}^{{}^{\prime }}=1+\mathrm{sinx}+\lambda \mathrm{cosx}\Rightarrow \mathrm{y}\left(\mathrm{x}\right)=\int (1+\mathrm{sinx}+\lambda \mathrm{cosx})\mathrm{dx}$$$$\mathrm{y}\left(\mathrm{x}\right)=\mathrm{x}-\mathrm{cosx}+\lambda \mathrm{sinx}$$

Commented by mathmax by abdo last updated on 24/Jul/20

$$\mathrm{sorry}\mathrm{i}\mathrm{have}\mathrm{solved}{\mathrm{y}}^{{}^{″}}+\mathrm{tanx}{\mathrm{y}}^{{}^{\prime }}=\mathrm{secx}+\mathrm{tanx}\mathrm{but}\mathrm{the}\mathrm{way}\mathrm{is}\mathrm{the}\mathrm{same}...$$

Commented by bramlex last updated on 24/Jul/20

$$oksir.thankyou$$

Commented by mathmax by abdo last updated on 24/Jul/20

$$\mathrm{you}\mathrm{are}\mathrm{welcome}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com