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 $k$.
A.-3
B.3
C.1
D.-1

Answer 1

Hint: Look at the equation, which is satisfied by $xy - \log y = 1$ and notice that it has double derivative of y. Thus, differentiate the equation, $xy - \log y = 1$ twice to get this equation in terms of the equation having a double derivative of y. Once you have determined the equation obtained by differentiating twice the equation, $xy - \log y = 1$; then all you need is to compare it with $x\left( {yy'' + {{y'}^2}} \right) - y'' + kyy' = 0$ to find the value of k.

Complete step-by-step answer:
Let us start by considering the given equation which is;
$xy - \log y = 1$
It can be clearly seen from the other given equation that it has double derivative of y. Also, it is given that $x\left( {yy'' + {{y'}^2}} \right) - y'' + kyy' = 0$ is satisfied by $xy - \log y = 1$.
Thus, we will be differentiating the equation, $xy - \log y = 1$ twice so that we get the equation in terms of double derivative of y.
Let us first differentiate the equation, $xy - \log y = 1$, with respect to $x$ , as follows;
$
  xy' + y.1 - \dfrac{1}{y} \cdot y' = 0 \\
   \Rightarrow xyy' + {y^2} - y' = 0 \\
 $
In order to get the double derivative of y, we will again differentiate the obtained equation with respect to $x$.
\[
  xy'y' + xyy'' + yy' + 2yy' - y'' = 0 \\
   \Rightarrow x\left( {yy'' + {{y'}^2}} \right) - y'' + 3yy' = 0 \\
 \]
Now we will compare the obtained equation with the given equation;
$x\left( {yy'' + {{y'}^2}} \right) - y'' + kyy' = 0$
Thus, it can be easily seen that the value of k is 3.
Hence, option (b) is the correct option.

Note: While differentiating the given equation, with respect to $x$, make sure that product rule is applied as both x and y are variables. Also, in order to compare the obtained equation with, $x\left( {yy'' + {{y'}^2}} \right) - y'' + kyy' = 0$, make sure that the obtained equation should be completely in this form, with not even a difference of a term or even a sign. Any change in the form of the obtained equation and the equation $x\left( {yy'' + {{y'}^2}} \right) - y'' + kyy' = 0$ will lead to a wrong answer.