Question

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

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

Answer 1

The correct option is D $-\frac{1}{4}$

To find $\varphi \left(2\right)$, the values of $x$ and $y$ should be such that $\frac{x}{y}=2$
Using the differential equation, we get $y^{\prime }=\frac{1}{2}+\varphi \left(2\right)$
$\varphi \left(2\right)=y^{\prime }-\frac{1}{2}$
The solution of the differential equation is
$y\ln |cx|=x\Rightarrow \ln |cx|=\frac{x}{y}$
Differentiating with respect to $x$ we get
$y\frac{1}{x}+y^{\prime }\ln |cx|=1$
$\Rightarrow y^{\prime }=\frac{1-y/x}{\ln |cx|}$
$\Rightarrow y^{\prime }=\frac{1-y/x}{x/y}$
Using $x/y=2$, we get $y^{\prime }=1/4$
Hence, $\varphi \left(2\right)=\frac{1}{4}-\frac{1}{2}=-\frac{1}{4}$