For the differential equation $\left(x \log _e x\right) d y=\left(\log _e x-y\right) d x$
(A) Degree of the given differential equation is 1.
(B) It is a homogeneous differential equation.
(C) Solution is $2 y \log _e x+A=\left(\log _e x\right)^2$, where A is an arbitrary constant.
(D) Solution is $2 y \log _e x+A=\log _e\left(\log _e x\right)$, where A is an arbitrary constant.
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A) and (C) only
$\left(x \log x\right) d y=\left(\log x-y\right) d x$
degree → I
Not homogenous as $\frac{dy}{dx}≠λ^nf(x/y)$
$\left(x \log x\right) d y=\left(\log x-y\right) d x$
$\frac{dy}{dx}=\frac{1}{x}-\frac{y}{x\log x}$
$\int\frac{y}{x\log x}+\frac{dy}{dx}=\int\frac{1}{x}dx$
$I.F.=e^{\int\frac{y}{x\log x}dx}=e^{\log\log x}=\log x$
multiplying eq. with I.F.
$\int\frac{dy}{dx}\log x+\frac{y}{x}dx=\int\frac{\log x}{x}dx$
$=y\log x=\frac{(\log x)^2}{2}+C$
$≡2y\log x+A=(\log x)^2$
only A, C correct