Q120 / 160 · TS EAMCET 2022 (19 Jul Shift 2) · Mathematics

The general solution of $\left(\left(1+x^2\right) y \sin x-2 x y\right) d x-\log y^{1+x^2} d y=0$ is

  • A
    $\sin x-\log \left(1+x^2\right)=\log y+c$
  • Correct answer
    B
    $(\log y)^2+2 \cos x+\log \left(1+x^2\right)^2=c$
  • C
    $\log y=2 \cos x+\log \left(1+x^2\right)+c$
  • D
    $\frac{\log y}{y}=2 \sin x+\cos x \log \left(1+x^2\right)+c$

Solution

We have a differential equation $$ \begin{aligned} & \left(\left(1+x^2\right) y \sin x-2 x y\right) d x-\log y^{1+x^2} d y=0 \\ & \Rightarrow\left(\left(1+x^2\right) \sin -2 x\right) d x-\frac{\left(1+x^2\right) \log y}{y} d y=0 \\ & \Rightarrow\left(\sin (x)-2 \frac{x}{1+x^2}\right) d x=\frac{\log y}{y} d y \end{aligned} $$ Integrating both sides we get. $$ (\log y)^2+2 \cos x+\log \left(1+x^2\right)^2=c $$

Hence, the correct answer is option: 'B'.

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