Search any question & find its solution
Question:
Answered & Verified by Expert
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
Options:
Solution:
1334 Upvotes
Verified Answer
The correct answer is:
$(\log y)^2+2 \cos x+\log \left(1+x^2\right)^2=c$
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.
$$
(\log y)^2+2 \cos x+\log \left(1+x^2\right)^2=c
$$
$$
\begin{aligned}
& \left(\left(1+x^2\right) y \sin x-2 x y\right) d x-\log y^{1+x^2} d y=0 \\
& \Rightarrow\left(\left(1+x^2\right) \sin -2 x\right) d x-\frac{\left(1+x^2\right) \log y}{y} d y=0 \\
& \Rightarrow\left(\sin (x)-2 \frac{x}{1+x^2}\right) d x=\frac{\log y}{y} d y
\end{aligned}
$$
Integrating both sides we get.
$$
(\log y)^2+2 \cos x+\log \left(1+x^2\right)^2=c
$$
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.