Question

The function defined by the equation $xy - \log y = 1$ satisfies $x\left( {yy'' + y{'^2}} \right) - y'' + kyy' = 0$ . Find the value of k.

Answer 1

Hint:
It is given that the equation $xy - \log y = 1$ satisfies $x\left( {yy'' + y{'^2}} \right) - y'' + kyy' = 0$. Also, it can be seen that on doing the double derivative of the equation $xy - \log y = 1$, we get the type of equation $x\left( {yy'' + y{'^2}} \right) - y'' + kyy' = 0$.
Then, write the derived equation in the form of $x\left( {yy'' + y{'^2}} \right) - y'' + kyy' = 0$ and compare to get the value of k.

Complete step by step solution:
The given equation in the question is $xy - \log y = 1$ .
Also, it is given that the equation $xy - \log y = 1$ satisfies $x\left( {yy'' + y{'^2}} \right) - y'' + kyy' = 0$ . It can be seen that on doing the double derivative of the equation $xy - \log y = 1$ , we get the type of equation $x\left( {yy'' + y{'^2}} \right) - y'' + kyy' = 0$ .
So, firstly, we will differentiate the equation $xy - \log y = 1$ with respect to x.
 $
   \Rightarrow \dfrac{d}{{dx}}\left( {xy - \log y} \right) = \dfrac{d}{{dx}}\left( 1 \right) \\
   \Rightarrow \dfrac{d}{{dx}}\left( {xy} \right) - \dfrac{d}{{dx}}\left( {\log y} \right) = 0 \\
 $
Apply the formula for derivative of product of two functions u and v $\dfrac{d}{{dx}}\left( {uv} \right) = u\dfrac{{dv}}{{dx}} + v\dfrac{{du}}{{dx}}$ on the function xy.
 $
   \Rightarrow x\dfrac{{dy}}{{dx}} + y\dfrac{{dx}}{{dx}} - \dfrac{1}{y}\dfrac{{dy}}{{dx}} = 0 \\
   \Rightarrow xy' + y - \dfrac{1}{y}y' = 0\left( { \Rightarrow \dfrac{{dy}}{{dx}} = y'} \right) \\
   \Rightarrow \dfrac{{xyy' + {y^2} - y'}}{y} = 0 \\
   \Rightarrow xyy' + {y^2} - y' = 0 \\
 $
Now, in order to get the double derivative, we need to differentiate the equation $xyy' + {y^2} - y' = 0$ with respect to x.
 $
   \Rightarrow \dfrac{d}{{dx}}\left( {xyy' + {y^2} - y'} \right) = \dfrac{d}{{dx}}\left( 0 \right) \\
   \Rightarrow \dfrac{d}{{dx}}\left( {xyy'} \right) + \dfrac{d}{{dx}}\left( {{y^2}} \right) - \dfrac{d}{{dx}}\left( {y'} \right) = 0 \\
 $
Applying the formula for derivative of triple product $\dfrac{d}{{dx}}\left( {uvw} \right) = \dfrac{{du}}{{dx}}vw + u\dfrac{{dv}}{{dx}}w + uv\dfrac{{dw}}{{dx}}$ on function $xyy'$
 $
   \Rightarrow \dfrac{{dx}}{{dx}}\left( {yy'} \right) + x\left( {\dfrac{{dy}}{{dx}}} \right)y' + xy\left( {\dfrac{{dy'}}{{dx}}} \right) + 2y\dfrac{{dy}}{{dx}} - y'' = 0 \\
   \Rightarrow yy' + xy'y' + xyy'' + 2yy' - y'' = 0 \\
   \Rightarrow x\left( {y{'^2} + yy''} \right) + 3yy' - y'' = 0 \\
   \Rightarrow x\left( {yy'' + y{'^2}} \right) - y'' + 3yy' = 0 \\
 $
Now, to get the required value of k, we will compare the equation $x\left( {yy'' + y{'^2}} \right) - y'' + 3yy' = 0$ with $x\left( {yy'' + y{'^2}} \right) - y'' + kyy' = 0$ .

Thus, on comparing we get \[k = 3\].

Note:
While differentiating the given equation $xy - \log y = 1$ , we have to differentiate both x and y as they both are variables.
Also, when comparing the derived equation after derivation the equation $xy - \log y = 1$ twice, we have to write the derived equation in the form of $x\left( {yy'' + y{'^2}} \right) - y'' + kyy' = 0$ or else we may get the required answer wrong.