The solution of dy/dx=x+y/x-y is

Correct Answer is : tan−1(yx)=logx2+y2+C{tan}^{-1}≤ft(y/x)={log}{x²+y²}+Ctan−1(xy)=logx2+y2+C

{tan}^{-1}≤ft(y/x)={log}{x²+y²}+C

{tan}^{-1}≤ft(y/x)={log}{x²-y²}+C

{sin}^{-1}≤ft(y/x)={log}{x²+y²}+C

{cos}^{-1}≤ft(y/x)={log}{x²-y²}+C

Correct Answer is : tan−1(yx)=logx2+y2+C{tan}^{-1}≤ft(y/x)={log}{x²+y²}+Ctan−1(xy)=logx2+y2+C

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TS EAMCET 2017 Code A Solved Paper

Section: Mathematics

Question : 28 of 160

Marks: +1, -0

The solution of

\frac{dy}{dx}=\frac{x+y}{x-y}

Solution:

The given function is,

\;\;\frac{dy}{dx}=\;\;\frac{x+y}{x-y}

Substitute

y=v x

\;\;\frac{dy}{dx}=v + x \;\;\frac{dv}{dx}

v + x \;\;\frac{dv}{dx}=\;\;\frac{x+vx}{x-vx}

v + x \;\;\frac{dv}{dx}=\;\;\frac{1+v}{1-v}

x \;\;\frac{dv}{dx}=\;\;\frac{1+v^{2}}{1-v}

Integrate the above equation

\int\;\;\frac{1-v}{1+v^{2}}\,dv=\int\;\;\frac{dx}{x}

\int\;\;\frac{1}{1+v^{2}}\,dv-\;\;\frac{1}{2}\int\;\;\frac{2v}{1+v^{2}}\,dv=\int\;\;\frac{dx}{x}

\tan^{-1}v-\;\;\frac{1}{2}\log|1+v^{2}|=\log x+C

\tan^{-1}\;\;\frac{y}{x}=\log x+\;\;\frac{1}{2}\log\left(1+\;\;\frac{y^{2}}{x^{2}}\right)+C

\tan^{-1}\;\;\frac{y}{x}=\log\sqrt{x^{2}+y^{2}}+C

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