Q53 / 120 · VITEEE 2017 · Mathematics

The solution of the differential equation $\log x \frac{d y}{d x}+\frac{y}{x}=\sin 2 x$ is

  • A
    $y \log |x|=C-\frac{1}{2} \cos x$
  • B
    $y \log |x|=C+\frac{1}{2} \cos 2 x$
  • Correct answer
    C
    $y \log |x|=C-\frac{1}{2} \cos 2 x$
  • D
    $\quad x y \log |x|=C-\frac{1}{2} \cos 2 x$

Solution

$\frac{d y}{d x}+\frac{y}{x \log x}=\frac{\sin 2 x}{\log x}$ $$ \begin{array}{l} \text { I.F. }=e^{\int \frac{d x}{x \log x}} \\ \therefore \text { I.F. }=e^{\int \frac{1}{t} d t}=e^{\log t}=t=\log |x| \end{array} $$ solution is given by $$ \begin{array}{l} \mathrm{y}(\text { I.F. })=\int \mathrm{Q} .(\mathrm{I} . \mathrm{F}) \mathrm{d} \mathrm{x}+\mathrm{C} \\ \mathrm{y} \log |\mathrm{x}|=\int \frac{\sin 2 \mathrm{x}}{\log |\mathrm{x}|}(\log |\mathrm{x}|) \mathrm{d} \mathrm{x}+\mathrm{C} \\ =-\frac{\cos 2 \mathrm{x}}{2}+\mathrm{C} \end{array} $$

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

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