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The solution of the differential equation $\log x \frac{d y}{d x}+\frac{y}{x}=\sin 2 x$ is
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Verified Answer
The correct answer is:
$y \log |x|=C-\frac{1}{2} \cos 2 x$
$\frac{d y}{d x}+\frac{y}{x \log x}=\frac{\sin 2 x}{\log x}$
$$
\begin{array}{l}
\text { I.F. }=e^{\int \frac{d x}{x \log x}} \\
\therefore \text { I.F. }=e^{\int \frac{1}{t} d t}=e^{\log t}=t=\log |x|
\end{array}
$$
solution is given by
$$
\begin{array}{l}
\mathrm{y}(\text { I.F. })=\int \mathrm{Q} .(\mathrm{I} . \mathrm{F}) \mathrm{d} \mathrm{x}+\mathrm{C} \\
\mathrm{y} \log |\mathrm{x}|=\int \frac{\sin 2 \mathrm{x}}{\log |\mathrm{x}|}(\log |\mathrm{x}|) \mathrm{d} \mathrm{x}+\mathrm{C} \\
=-\frac{\cos 2 \mathrm{x}}{2}+\mathrm{C}
\end{array}
$$
$$
\begin{array}{l}
\text { I.F. }=e^{\int \frac{d x}{x \log x}} \\
\therefore \text { I.F. }=e^{\int \frac{1}{t} d t}=e^{\log t}=t=\log |x|
\end{array}
$$
solution is given by
$$
\begin{array}{l}
\mathrm{y}(\text { I.F. })=\int \mathrm{Q} .(\mathrm{I} . \mathrm{F}) \mathrm{d} \mathrm{x}+\mathrm{C} \\
\mathrm{y} \log |\mathrm{x}|=\int \frac{\sin 2 \mathrm{x}}{\log |\mathrm{x}|}(\log |\mathrm{x}|) \mathrm{d} \mathrm{x}+\mathrm{C} \\
=-\frac{\cos 2 \mathrm{x}}{2}+\mathrm{C}
\end{array}
$$
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