Question

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

• A. $-x^2/y^2$
• B. $y^2/x^2$
• C. $x^2/y^2$
• D. $-y^2/x^2$

Answer 1

Answer: (D)

$-y^2/x^2$

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