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  • If the solution curve  of the differential equation <img alt="y^{2} d x+\left(x^{2}-x y+y^

If the solution curve y=y(x) of the differential equation y^{2} d x+\left(x^{2}-x y+y^{2}\right) d y=0, which passes through the point \mathrm{\left ( 1,1 \right )} and intersects the line y=\sqrt{3}x at the point \mathrm{\left ( \alpha ,\sqrt{3} \alpha \right )}, then value of \log _{e}(\sqrt{3} \alpha) is equal to:

Option: 1

\frac{\pi }{3}


Option: 2

\frac{\pi }{2}


Option: 3

\frac{\pi }{12}


Option: 4

\frac{\pi }{6}


Answers (1)

best_answer

\mathrm{given \; y^{2} d x+\left(x^{2}-x y+y^{2}\right) d y=0}

\mathrm{\Rightarrow \frac{d x}{d y}=\frac{-\left(x^{2}-x y+y^{2}\right)}{y^{2}}---(1)}
\mathrm{Now\; let \; x=v y \Rightarrow v+y \frac{d v}{d y}=\frac{d x}{d y}}
Now putting in eqn (i) we get
\mathrm{\Rightarrow v+y \frac{d v}{d y}=-\left(v^{2}-v+1\right.)}
\mathrm{\Rightarrow y \frac{d v}{d y}=-v^{2}-1}
\mathrm{\Rightarrow-\frac{d v}{1+v^{2}}=\frac{d y}{y} }
Now Integrating both side
\mathrm{\Rightarrow-\int \frac{d v}{1+v^{2}}=\int \frac{d y}{y} }
\mathrm{\Rightarrow-\tan^{-1} v=\log |y|+c \Rightarrow-\tan^{-1}\left(\frac{x}{y}\right)=\log |y|+c }
Also this curve passes through \mathrm{(1,1) }
\mathrm{\Rightarrow -\tan ^{-1} 1=\log| 1|+c \Rightarrow c=-\frac{\pi}{4} }
\mathrm{\Rightarrow -\tan^{-1}\frac{x}{y}=\log |y|-\frac{\pi}{4} }
\mathrm{Now \: putting\: \: y=\sqrt{3} x \; and \; \; x=\alpha\; \; we \; get. }
\mathrm{=-\tan ^{-1}\frac{\alpha}{\sqrt{3} \alpha}=\log |\sqrt{3} \alpha|-\frac{\pi}{4}}
\mathrm{\Rightarrow \log |\sqrt{3} \alpha|=\frac{\pi}{12} }

Posted by

Pankaj Sanodiya

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