Search any question & find its solution
Question:
Answered & Verified by Expert
If $y=2 \cos (2 \log x)+3 \sin (2 \log x)$, then $x^2 y^{\prime \prime}+x y^{\prime}+2 y=$
Options:
Solution:
1299 Upvotes
Verified Answer
The correct answer is:
$-2 y$
We have,
$$
\begin{aligned}
& y=2 \cos (2 \log x)+3 \sin (2 \log x) \\
\therefore \quad & y^{\prime}=-2 \sin (2 \log x) \cdot \frac{2}{x}+3 \cos (2 \log x) \cdot \frac{2}{x} \\
\Rightarrow \quad & x y^{\prime}=2[-2 \sin (2 \log x)+3 \cos (2 \log x)] \\
\Rightarrow & x y^{\prime \prime}+y^{\prime \prime}=2\left[-2 \cos (2 \log x) \cdot \frac{2}{x}-3 \sin (2 \log x) \cdot \frac{2}{x}\right] \\
\Rightarrow & x^2 y^{\prime \prime}+x y^{\prime}=-4[2 \cos (2 \log x)+3 \sin (2 \log x)] \\
\Rightarrow & x^2 y^{\prime \prime}+x y^{\prime}=-4 y \\
\Rightarrow & x^2 y^{\prime \prime}+x y^{\prime}+2 y=-2 y
\end{aligned}
$$
$$
\begin{aligned}
& y=2 \cos (2 \log x)+3 \sin (2 \log x) \\
\therefore \quad & y^{\prime}=-2 \sin (2 \log x) \cdot \frac{2}{x}+3 \cos (2 \log x) \cdot \frac{2}{x} \\
\Rightarrow \quad & x y^{\prime}=2[-2 \sin (2 \log x)+3 \cos (2 \log x)] \\
\Rightarrow & x y^{\prime \prime}+y^{\prime \prime}=2\left[-2 \cos (2 \log x) \cdot \frac{2}{x}-3 \sin (2 \log x) \cdot \frac{2}{x}\right] \\
\Rightarrow & x^2 y^{\prime \prime}+x y^{\prime}=-4[2 \cos (2 \log x)+3 \sin (2 \log x)] \\
\Rightarrow & x^2 y^{\prime \prime}+x y^{\prime}=-4 y \\
\Rightarrow & x^2 y^{\prime \prime}+x y^{\prime}+2 y=-2 y
\end{aligned}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.