Question

If the general solution of the differential equation $y^{\prime }=\frac{y}{x}+\Phi \left(\frac{x}{y}\right),$ for some function $\Phi ,$ is given by $yln\left|cx\right|=x,$ 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)

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