Question

Consider the differential equation, y² dx +(x - 1/y) dy = 0. If value of y is 1 when x = 1, then the value of x for which y = 2, is : (A) (3/2) - √e
(B) (1/2) + (1/√e)
(C) (3/2) - (1/√e)
(D) (5/2) + (1/√e)

Answer 1

The correct option is option (C)

Therefore the value of x is $\frac32-\frac1 {\sqrt e}$.

Step-by-step explanation:

Given differential equation is

$y^2dy +(x-\frac{1}{y})dy= 0$

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

$\Rightarrow \frac{dx}{dy} +\frac{x}{y^2}-\frac{1}{y^3}=0$

$\Rightarrow \frac{dx}{dy} +\frac{x}{y^2}=\frac{1}{y^3}$......(i)

Here $P=\frac{1}{y^2}$  and $Q=\frac{1}{y^3}$

The integrating factor $=e^{\int P \ dy}$

                                    = $e^{\int \frac{1}{y^2} \ dy$

                                   $=e^{-\frac{1}{y}}$

Multiplying the integrating factor both sides of (i)

$e^{-\frac{1}{y}} \frac{dx}{dy} +\frac{x}{y^2}e^{-\frac{1}{y}}=\frac{1}{y^3}e^{-\frac{1}{y}}$

Integrating both sides

$\int e^{-\frac{1}{y}} dx +\int \frac{x}{y^2}e^{-\frac{1}{y}}dy=\int\frac{1}{y^3}e^{-\frac{1}{y}}dy$ ........(ii)

Let

$I_1=\int\frac{1}{y^3}e^{-\frac{1}{y}}dy$

   $=\int (\frac1 y)(\frac1{y^2}) e^{-\frac{1}{y} dy$

Let $-\frac{1}{y} =z$, Differentiating $\frac1{y^2} dy=dz$

Then,

$I_1=\int (-z)e^zdz$

   $=-[z\int e^z -\int(\frac{dz}{dz}\int e^zdz)dz]$

   $=-ze^z +e^z+c$

  Putting the value of z

  $=(\frac{1}y}+1)e^{-\frac{1}{y}}+c$

From (ii) we get

$\int e^{-\frac{1}{y}} dx +\int \frac{x}{y^2}e^{-\frac{1}{y}}dy=\int\frac{1}{y^3}e^{-\frac{1}{y}}dy$

$\Rightarrow xe^{-\frac1y}=(\frac{1}y}+1)e^{-\frac{1}{y}}+c$

$\Rightarrow x=(\frac{1}y}+1)+c e^{\frac1 y$

Given that y=1 when x=1

Putting the value of x and y

$\Rightarrow 1=(\frac{1}1}+1)+c e^{\frac 11$

$\Rightarrow ce=1-2$

$\Rightarrow c=-e^{-1}$

Therefore the required solution  is

$x=(\frac{1}y}+1)-e^{-1} e^{\frac1 y$

 $x=(\frac{1}y}+1)- e^{\frac1 y-1}$

Now putting x= 2

$\therefore x=(\frac12+1)-e^{\frac12-1}$

       $=\frac32-e^{-\frac12}$

       $=\frac32-\frac1 {\sqrt e}$

Therefore the value of x is $\frac32-\frac1 {\sqrt e}$.