# Y 2 D X D Y + X − 1 Y = 0 - Mathematics

Sum
$y^2 \frac{dx}{dy} + x - \frac{1}{y} = 0$

#### Solution

We have,
$y^2 \frac{dx}{dy} + x - \frac{1}{y} = 0$
$\Rightarrow y^2 \frac{dx}{dy} + x = \frac{1}{y}$
$\Rightarrow \frac{dx}{dy} + \frac{1}{y^2}x = \frac{1}{y^3} . . . . . . . . \left( 1 \right)$
Clearly, it is a linear differential equation of the form
$\frac{dx}{dy} + Px = Q$
where
$P = \frac{1}{y^2}$
$Q = \frac{1}{y^3}$
$\therefore I . F . = e^{\int P\ dy}$
$= e^{\int\frac{1}{y^2}dy}$
$= e^\frac{- 1}{y}$
$\text{ Multiplying both sides of }\left( 1 \right)\text{ by }e^\frac{- 1}{y} ,\text{ we get }$
$e^\frac{- 1}{y} \left( \frac{dx}{dy} + x\frac{1}{y^2} \right) = e^\frac{- 1}{y} \frac{1}{y^3}$
$\Rightarrow e^\frac{- 1}{y} \frac{dx}{dy} + x\frac{1}{y^2} e^\frac{- 1}{y} = e^\frac{- 1}{y} \frac{1}{y^3}$
Integrating both sides with respect to y, we get
$x\ e^\frac{- 1}{y} = \int e^\frac{- 1}{y} \frac{1}{y^3}dy + C$
$\Rightarrow x\ e^\frac{- 1}{y} = I + C . . . . . . . . \left( 2 \right)$
where
$I = \int e^\frac{- 1}{y} \frac{1}{y^3}dy$
$\text{Putting }t = \frac{1}{y}, \text{ we get }$
$dt = - \frac{1}{y^2}dy$

$= - t\int e^{- t} dt + \int\left[ \frac{d}{dt}\left( t \right)\int e^{- t} dt \right]dt$
$= t e^{- t} + e^{- t}$
$= \left( t + 1 \right) e^{- t}$
$= \left( \frac{1}{y} + 1 \right) e^{- \frac{1}{y}}$
$\text{Putting the value of I in }\left( 2 \right),\text{ we get }$
$x\ e^\frac{- 1}{y} = \left( \frac{1}{y} + 1 \right) e^{- \frac{1}{y}} + C$
$\Rightarrow x = \left( \frac{y + 1}{y} \right) + C e^\frac{1}{y}$
$\text{Hence, }x = \left( \frac{y + 1}{y} \right) + C e^\frac{1}{y} \text{ is the required solution.}$

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Chapter 22: Differential Equations - Exercise 22.10 [Page 106]

#### APPEARS IN

RD Sharma Class 12 Maths
Chapter 22 Differential Equations
Exercise 22.10 | Q 23 | Page 106

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