Question

If $y=\frac{x}{ln|cx|}$ (where c is an arbitrary constant) is the general solution of the differential equation $\frac{dy}{dx}=\frac{y}{x}+\varphi \left(\frac{x}{y}\right)$ then the function $\varphi \left(\frac{x}{y}\right)$

• A. $\frac{x^2}{y^2}$
• B. $-\frac{x^2}{y^2}$
• C. $\frac{y^2}{x^2}$
• D. $-\frac{y^2}{x^2}$

Answer 1

The correct option is D $-\frac{y^2}{x^2}$

$y=\frac{x}{ln|cx|}\Rightarrow \frac{y}{x}=\frac{1}{ln|cx|}$

$\frac{dy}{dx}=\frac{lncx-x\times \frac{1}{cx}\times c}{(ln|cx|{)}^2}$

$=\frac{lncx-1}{(ln|cx|{)}^2}$

$\therefore \frac{dy}{dx}=\frac{lncx-1}{(ln|cx|{)}^2}$

Putting values in the differential equation

$\frac{dy}{dx}=\frac{y}{x}+\varphi \left(x/y\right)$

$\Rightarrow \frac{lncx-1}{(lncx{)}^2}=\frac{y}{x}+\varphi \left(x/y\right)$

$\Rightarrow \frac{lncx-1}{(lncx{)}^2}-\frac{y}{x}=\varphi \left(x/y\right)$

$\Rightarrow \varphi \left(x/y\right)=\frac{lncx-1}{(lncx{)}^2}-\frac{1}{lncx}$

$=\frac{lncx-1-lncx}{(lncx{)}^2}=\frac{-1}{(lncx{)}^2}$

$\therefore \varphi \left(x/y\right)=-{\left(\frac{1}{lncx}\right)}^2=-\frac{y^2}{x^2}$

$\therefore \varphi \left(x/y\right)=\frac{-y^2}{x^2}$