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Question: If \(y\) is a function of \(x\), then \(\dfrac{{{d^2}y}}{{d{x^2}}} + y\dfrac{{dy}}{{dx}} = 0\) . If ...

If yy is a function of xx, then d2ydx2+ydydx=0\dfrac{{{d^2}y}}{{d{x^2}}} + y\dfrac{{dy}}{{dx}} = 0 . If xx is a function of y,y, then the equation becomes:
A.d2xdy2+xdxdy=0\dfrac{{{d^2}x}}{{d{y^2}}} + x\dfrac{{dx}}{{dy}} = 0
B.d2xdy2+y(dxdy)3=0\dfrac{{{d^2}x}}{{d{y^2}}} + y{\left( {\dfrac{{dx}}{{dy}}} \right)^3} = 0
C.d2xdy2+y(dxdy)2=0\dfrac{{{d^2}x}}{{d{y^2}}} + y{\left( {\dfrac{{dx}}{{dy}}} \right)^2} = 0
D.d2xdy2+x(dxdy)2=0\dfrac{{{d^2}x}}{{d{y^2}}} + x{\left( {\dfrac{{dx}}{{dy}}} \right)^2} = 0

Explanation

Solution

In this question we have to find the differential equation if xx is a function of y,y,.Here we will first separate the parts into simpler terms and assume that dydx=p\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:
dydx+P(x)y=Q(x)\dfrac{{dy}}{{dx}} + P(x)y = Q(x) .

Complete answer:
Here we have been given in the question : d2ydx2+ydydx=0\dfrac{{{d^2}y}}{{d{x^2}}} + y\dfrac{{dy}}{{dx}} = 0
We can write the given equation also as
ddx(dydx)+ydydx=0\dfrac{d}{{dx}}\left( {\dfrac{{dy}}{{dx}}} \right) + y\dfrac{{dy}}{{dx}} = 0
Now, let us assume that dydx=p\dfrac{{dy}}{{dx}} = p .
By putting this value in the equation we can write:
ddxp+yp=0\dfrac{d}{{dx}}p + yp = 0
Now by comparing the above equation with the linear differential equation we have:
p(x)=yp(x) = y and Q(x)=0Q(x) = 0 .
Now we will find the integrating factor. We can write the integrating factor as:
=e∫ydx= {e^{\int {ydx} }}
Now we will write the solution for this i.e.
⇒P×I.F=∫Q(IF)dx\Rightarrow P \times I.F = \int {Q(IF)} dx
We will simplify this now:
⇒P∫eydx=∫0(IF)dx\Rightarrow P\int {{e^{ydx}} = \int {0(IF)dx} }
It gives us value here:
⇒P∫eydx=0\Rightarrow P\int {{e^{ydx}} = 0}
Therefore it gives us value p=0p = 0 .
By putting the value we have :
⇒dydx=0\Rightarrow \dfrac{{dy}}{{dx}} = 0
So it means that we have the value of yy as some constant.
Now if the value of yy is constant then the value of xx is also constant because it is given that yy is a function of xx .
Then the equation will get converted in the form of :
d2xdy2+xdxdy=0\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
⇒P∫eydx=0\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≠0I.F \ne 0. Hence it gives us value as P=0P = 0 and we have assumed p=dydxp = \dfrac{{dy}}{{dx}} .