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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 $y \ln |c x|=x$, where $\mathrm{c}$ is an arbitrary constant, then $\Phi(2)$ is equal to:
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$-\frac{1}{4}$
$-\frac{1}{4}$
Given $\frac{d y}{d x}=\frac{y}{x}+\phi\left(\frac{y}{x}\right)$
Let $\left(\frac{y}{x}\right)=v$ so that $y=x v$ or $\frac{d y}{d x}=x \frac{d v}{d x}+v$
from (1) \& (2), $x \frac{d v}{d x}+v=v+\phi\left(\frac{1}{v}\right)$ or, $\frac{d v}{\phi\left(\frac{1}{v}\right)}=\frac{d x}{x}$
Integrating both sides, we get $\int \frac{d x}{x}=\int \frac{d v}{\phi\left(\frac{1}{v}\right)} \Rightarrow \ln x+c=\int \frac{d v}{\phi\left(\frac{1}{v}\right)}$
(where $\mathrm{c}$ being constant of integration)
But, given $\mathrm{y}=\frac{x}{\ln |c x|}$ is the general solution
so that $\frac{x}{y}=\frac{1}{v}=\ln |c x|=\int \frac{d v}{\phi\left(\frac{1}{v}\right)}$
Differentiating w.r.t $v$ both sides, we get $\phi\left(\frac{1}{v}\right)=\frac{-1}{v^2} \Rightarrow \phi\left(\frac{x}{y}\right)=-\frac{y^2}{x^2}$
when $\frac{x}{y}=2$ i.e. $\phi(2)$
$$
=-\left(\frac{y}{x}\right)^2=-\left(\frac{1}{2}\right)^2=\left(\frac{-1}{4}\right)
$$
Let $\left(\frac{y}{x}\right)=v$ so that $y=x v$ or $\frac{d y}{d x}=x \frac{d v}{d x}+v$
from (1) \& (2), $x \frac{d v}{d x}+v=v+\phi\left(\frac{1}{v}\right)$ or, $\frac{d v}{\phi\left(\frac{1}{v}\right)}=\frac{d x}{x}$
Integrating both sides, we get $\int \frac{d x}{x}=\int \frac{d v}{\phi\left(\frac{1}{v}\right)} \Rightarrow \ln x+c=\int \frac{d v}{\phi\left(\frac{1}{v}\right)}$
(where $\mathrm{c}$ being constant of integration)
But, given $\mathrm{y}=\frac{x}{\ln |c x|}$ is the general solution
so that $\frac{x}{y}=\frac{1}{v}=\ln |c x|=\int \frac{d v}{\phi\left(\frac{1}{v}\right)}$
Differentiating w.r.t $v$ both sides, we get $\phi\left(\frac{1}{v}\right)=\frac{-1}{v^2} \Rightarrow \phi\left(\frac{x}{y}\right)=-\frac{y^2}{x^2}$
when $\frac{x}{y}=2$ i.e. $\phi(2)$
$$
=-\left(\frac{y}{x}\right)^2=-\left(\frac{1}{2}\right)^2=\left(\frac{-1}{4}\right)
$$
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