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  • If the solution curve of the differential equation <img alt="\mathrm{\frac{d y}{d x}=\frac{x+y-2}{x-y}}" src="/latex-image/?%5Cmathrm%7B%5Cfrac%7Bd%20y%7D%7Bd%20x%7D%3D%5Cfrac%7Bx&plus;y-2%7D%7

If the solution curve of the differential equation \mathrm{\frac{d y}{d x}=\frac{x+y-2}{x-y}} passes through the points \mathrm{(2,1) \: and \: (k+1,2), k>0}, then

 

Option: 1

\mathrm{2 \tan ^{-1}\left(\frac{1}{k}\right)=\log _{e}\left(k^{2}+1\right)}


 


Option: 2

\mathrm{\tan ^{-1}\left(\frac{1}{k}\right)=\log _{e}\left(k^{2}+1\right)}



 


Option: 3

\mathrm{2 \tan ^{-1}\left(\frac{1}{k+1}\right)=\log _{e}\left(k^{2}+2 k+2\right)}



 


Option: 4

\mathrm{2 \tan ^{-1}\left(\frac{1}{k}\right)=\log _{e}\left(\frac{k^{2}+1}{k^{2}}\right) }


Answers (1)

best_answer

\begin{aligned} &\text{Let }\mathrm{ u=x+h, v=y+k.}\\ &\Rightarrow \mathrm{ d u=d x }\text{ and }\mathrm{ d r=d y}\\ & \therefore \mathrm{\frac{d v}{d u}=\frac{u-h+v-k-2}{u-h-v+k}} \\ & \mathrm{\frac{d v}{d u}=\frac{u+v+(-k-k-2)}{(u-v)+(-h+k)}} \\ & \mathrm{\Rightarrow-h-k-2=0} \text { and }\mathrm{-h+k=0} \\ & \Rightarrow \mathrm{h=k=-1} \\ & \mathrm{\frac{d v}{d u}=\frac{u+v}{u-v}}\\ &\text{Let }\mathrm{ \frac{v}{u}=w}\\ &\mathrm{v=u w }\\ &\mathrm{\quad \frac{d v}{d u}=w+u \frac{d w}{d u} }\\ &\mathrm{\therefore \quad w+\frac{u d w}{d u}=\frac{1+w}{1-w}} \\ &\mathrm{u \frac{d w}{d u}=\frac{1+w^{2}}{1-w}} \end{aligned}\begin{aligned} & \mathrm{\frac{(1-w) d w}{1+u^{2}}=\frac{d u}{u}} \\ & \mathrm{\int \frac{d u}{1+w^{2}}-\frac{1}{2} \int \frac{2 u}{1+w^{2}} d u^{2}=\int \frac{d u}{u}} \\ &\mathrm{ \tan ^{-1}(w)-\frac{1}{2} \log \left(1+w^{2}\right)=\log u+c }\\ &\mathrm{ \tan ^{-1}\left(\frac{u}{u}\right)-\frac{1}{2} \log \left(1+\frac{v^{2}}{u^{2}}\right)=\log (u)+c} \\ & \mathrm{\tan ^{-1}\left(\frac{y-1}{x-4}\right)-\frac{1}{2} \log \left(1+\frac{(y-1)^{2}}{(x-i)^{2}}\right)=\log (x-1)} \\ & \text { Using }\mathrm{(2,1)} \\ & \mathrm{\tan ^{-1}(\theta)-\frac{1}{2} \log (1)=c} \\ & \begin{aligned}&\mathrm{\Rightarrow c=0} \\&\text { Using }\mathrm{(k+1,2)}\end{aligned} \\ & \mathrm{\tan ^{-1}\left(\frac{1}{k}\right)-\frac{1}{2} \log \left(1+\frac{1}{k^{2}}\right)=\log (k)} \\ & \mathrm{2 \tan ^{-1}\left(\frac{1}{k}\right)-\log \left(\frac{k^{2}+1}{k^{2}}\right)-2 \log k=0} \\ & \mathrm{2 \tan ^{-1}\left(\frac{1}{k}\right)=\log \left(k^{2}+1\right)} \\ & \therefore \text { option(1) } \end{aligned}

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