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STD 12 - 9. differential equations — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 9. differential equations1 Mark
MCQ
If the general solution of the differential equation $y' = \frac{y}{x} + \phi \left( {\frac{x}{y}} \right)$ , for some function $\phi $, is given by $y \ln \,\left| {cx} \right| = x$, where $c$ is an arbitrary constant, then $\phi \,(2)$ is equal to:
  • A
    $4$
  • B
    $\frac{1}{4}$
  • C
    $-4$
  • ✓
    $-\frac{1}{4}$
✓

Answer

Correct option: D.
$-\frac{1}{4}$
d
$\text { Given } \frac{d y}{d x}=\frac{y}{x}+\phi\left(\frac{y}{x}\right)$      ....$(1)$

Let $\left(\frac{y}{x}\right)=v$ so that $y=x v$

or $\frac{d y}{d x}=x \frac{d h}{d x}+y$      ....$(2)$

from $(1)$ and $(2), x \frac{d v}{d x}+v=v+\phi\left(\frac{1}{v}\right)$

or. $\frac{d v}{d\left(\begin{array}{l}{1} \\ {v}\end{array}\right)}=\frac{d x}{x}$

Integrating both sides, we get

$\int \frac{d x}{x}=\int \frac{d r}{\phi\left(\begin{array}{l}{1} \\ {v}\end{array}\right)} \Rightarrow \ln x+c=\int \frac{d t}{\phi\left(\frac{1}{v}\right)}$

(where $c$ being constant of integration)

But, given $y=\frac{x}{\ln |c x|}$ is the general solution

so that $\frac{x}{y}=\frac{1}{v}=\ln |x|=\int \frac{d v}{\phi\left(\frac{1}{v}\right)}$

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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