Question

If $y$ is a function of $x$, then $\dfrac{{{d^2}y}}{{d{x^2}}} + y\dfrac{{dy}}{{dx}} = 0$ . If $x$ is a function of $y,$ then the equation becomes:
A.$\dfrac{{{d^2}x}}{{d{y^2}}} + x\dfrac{{dx}}{{dy}} = 0$
B.$\dfrac{{{d^2}x}}{{d{y^2}}} + y{\left( {\dfrac{{dx}}{{dy}}} \right)^3} = 0$
C.$\dfrac{{{d^2}x}}{{d{y^2}}} + y{\left( {\dfrac{{dx}}{{dy}}} \right)^2} = 0$
D.$\dfrac{{{d^2}x}}{{d{y^2}}} + x{\left( {\dfrac{{dx}}{{dy}}} \right)^2} = 0$

Answer 1

Hint: In this question we have to find the differential equation if $x$ is a function of $y,$.Here we will first separate the parts into simpler terms and assume that $\dfrac{{dy}}{{dx}} = p$ . We will put this term in the equation and then we use the linear differential equation to solve this question.
WE know the formula of linear differential equation is:
$\dfrac{{dy}}{{dx}} + P(x)y = Q(x)$ .

Complete answer:
Here we have been given in the question : $\dfrac{{{d^2}y}}{{d{x^2}}} + y\dfrac{{dy}}{{dx}} = 0$
We can write the given equation also as
$\dfrac{d}{{dx}}\left( {\dfrac{{dy}}{{dx}}} \right) + y\dfrac{{dy}}{{dx}} = 0$
Now, let us assume that $\dfrac{{dy}}{{dx}} = p$ .
By putting this value in the equation we can write:
$\dfrac{d}{{dx}}p + yp = 0$
Now by comparing the above equation with the linear differential equation we have:
$p(x) = y$ and $Q(x) = 0$ .
Now we will find the integrating factor. We can write the integrating factor as:
$ = {e^{\int {ydx} }}$
Now we will write the solution for this i.e.
$ \Rightarrow P \times I.F = \int {Q(IF)} dx$
We will simplify this now:
$ \Rightarrow P\int {{e^{ydx}} = \int {0(IF)dx} } $
It gives us value here:
$ \Rightarrow P\int {{e^{ydx}} = 0} $
Therefore it gives us value $p = 0$ .
By putting the value we have :
$ \Rightarrow \dfrac{{dy}}{{dx}} = 0$
So it means that we have the value of $y$ as some constant.
Now if the value of $y$ is constant then the value of $x$ is also constant because it is given that $y$ is a function of $x$ .
Then the equation will get converted in the form of :
$\dfrac{{{d^2}x}}{{d{y^2}}} + x\dfrac{{dx}}{{dy}} = 0$

Hence the correct option is (A) .

Note:
We should note that we have the expression in the form of
$ \Rightarrow P\int {{e^{ydx}} = 0} $ .
We should know the condition that the value of Integrating factor cannot be equal to zero i.e. $I.F \ne 0$. Hence it gives us value as $P = 0$ and we have assumed $p = \dfrac{{dy}}{{dx}}$ .