Question

If for the differential equation $y^1=\frac{y}{x}+\varphi \left(\frac{x}{y}\right)$ the general solution is $y=\frac{x}{log|Cx|}$ then $\varphi \left(\frac{x}{y}\right)$is given by

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

Answer 1

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

Putting so that $\frac{xdv}{dx}+v=\frac{dy}{dx}$, we have
$\frac{xdv}{dx}+v=v+\varphi \left(\frac{1}{v}\right)\Rightarrow \frac{dv}{\varphi \left(\frac{1}{v}\right)}=\frac{dx}{x}$
$\Rightarrow log|Cx|=\int \frac{dv}{\varphi \left(\frac{1}{v}\right)}$ (being constant of integration.)
But $y=\frac{x}{log|Cx|}$ is the general solution so
$\frac{x}{y}=\frac{1}{v}=\int \frac{dv}{\varphi \left(\frac{1}{v}\right)}\Rightarrow \varphi \left(\frac{1}{v}\right)=-{\frac{1}{v}}^2\Rightarrow \varphi \left(\frac{x}{y}\right)=-\frac{y^2}{x^2}$