Question

If the general solution of the differential equation $y' = \frac{y}{x} + \Phi\left(\frac{x}{y}\right),$ for some function $\Phi,$ is given by $y \,ln\,\left|cx\right| \, = x,$ where $c$ is an arbitrary constant, then $\Phi \left(2\right)$ is equal to :

• A. 4
• B. $\frac{1}{4}$
• C. -4
• D. $-\frac{1}{4}$

Answer 1

Answer: (D)

$y' = \frac{y}{x} + \phi\left(\frac{x}{y}\right)\quad...\left(1\right)$ is solution of
$y \,In |cx| = x\quad\quad...\left(2\right)$
d.w.r. to x
$\frac{y}{ \left|cx|\right|}\cdot\frac{\left|cx\right|}{cx|}\cdot c+In \left|cx\right| y' = 1$
$\frac{y}{x} + \frac{x}{y}y' = 1\quad\quad\left(use In c\left|cx\right| = \frac{x}{y}\right)$
$y' = \left(1-\frac{y}{x}\right)\frac{y}{x}$
use y' in equation $\left(1\right)$
$\frac{y}{x}\left(1-\frac{y}{x}\right) = \frac{y}{x}+\phi\left( \frac{x}{y}\right)$
put $\left( \frac{x}{y}\right) = 2 \Rightarrow \left(1- \frac{1}{2}\right)\left( \frac{1}{2}\right) = \frac{1}{2}+\phi\left(2\right)$
$= \frac{1}{4} = \frac{1}{2} +\phi \left(2\right)$
$\phi \left(2\right) = -\frac{1}{4}$