Question

If \[\sqrt{y}={{\cos }^{-1}}x\], then it satisfies the differential equation \[\left( 1-{{x}^{2}} \right)\dfrac{{{d}^{2}}y}{d{{x}^{2}}}-x\dfrac{dy}{dx}=c\], where c is equal to,
(a)0
(b)3
(c)1
(d)2

Answer 1

Hint: Differentiate the expression to get \[\dfrac{dy}{dx}\]. Differentiate the expression again to get \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}\]. Now simplify the expression and it changes into a differential equation in question and finds the value of c.

Complete step-by-step answer:
Given to us is the expression, \[\sqrt{y}={{\cos }^{-1}}x\].
Let us square both sides of the expression,
\[\begin{align}
  & {{\left( \sqrt{y} \right)}^{2}}={{\left( {{\cos }^{-1}}x \right)}^{2}} \\
 & \Rightarrow y={{\left( {{\cos }^{-1}}x \right)}^{2}} \\
\end{align}\]
Let us differentiate both sides of the expression with respect to ‘x’.
\[\dfrac{dy}{dx}=2\left( {{\cos }^{-1}}x \right)\times \dfrac{-1}{\sqrt{1-{{x}^{2}}}}\]
The derivative of \[{{\cos }^{-1}}x\]with respect to \[x=-\dfrac{1}{\sqrt{1-{{x}^{2}}}}\].
Thus we got, \[{{y}^{'}}=\dfrac{dy}{dx}=\dfrac{-2\left( {{\cos }^{-1}}x \right)}{\sqrt{1-{{x}^{2}}}}\].
Now let us differentiate it again, w.r.t x. thus \[\dfrac{dy}{dx}\]becomes \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}\].
The division rule for differentiation says that, if
\[u=f\left( x \right)\]and \[v=g\left( x \right)\]then,
\[\dfrac{d}{dx}\left[ \dfrac{u}{v} \right]=\dfrac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{{{v}^{2}}}\]
Here, \[u=-2{{\cos }^{-1}}x\]and \[v=\sqrt{1-{{x}^{2}}}\].
\[\begin{align}
  & \therefore \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-2\left[ \dfrac{\sqrt{1-{{x}^{2}}}\dfrac{d}{dx}\left( {{\cos }^{-1}}x \right)-\left( {{\cos }^{-1}}x \right)\dfrac{d}{dx}\sqrt{1-{{x}^{2}}}}{{{\left( \sqrt{1-{{x}^{2}}} \right)}^{2}}} \right] \\
 & =-2\left[ \dfrac{\sqrt{1-{{x}^{2}}}\left( \dfrac{-1}{\sqrt{1-{{x}^{2}}}} \right)-\left( {{\cos }^{-1}}x\times \dfrac{1}{2}\times \dfrac{-2x}{\sqrt{1-{{x}^{2}}}} \right)}{1-{{x}^{2}}} \right] \\
\end{align}\]
[\[\therefore \]While differentiation, \[\dfrac{d}{dx}\sqrt{1-{{x}^{2}}}\]
\[\begin{align}
  & =\dfrac{d}{dx}{{\left( 1-{{x}^{2}} \right)}^{\dfrac{1}{2}}}=\dfrac{1}{2}\dfrac{d}{dx}{{\left( 1-{{x}^{2}} \right)}^{1-\dfrac{1}{2}}}\times \left( -2x \right) \\
 & =\dfrac{1}{2}\times \dfrac{-2x}{\sqrt{1-{{x}^{2}}}}=\dfrac{-x}{\sqrt{1-{{x}^{2}}}} ]\\

 & =\dfrac{-2\left[ -1-\left[ {{\cos }^{-1}}x\times \dfrac{-x}{\sqrt{1-{{x}^{2}}}} \right] \right]}{1-{{x}^{2}}} \\
 & =\dfrac{-2\left[ -1+\dfrac{x{{\cos }^{-1}}x}{\sqrt{1-{{x}^{2}}}} \right]}{1-{{x}^{2}}}=\left[ \dfrac{2-\dfrac{2x{{\cos }^{-1}}x}{\sqrt{1-{{x}^{2}}}}}{1-{{x}^{2}}} \right] \\
\end{align}\]
\[\therefore \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\left[ \dfrac{2-\dfrac{2x{{\cos }^{-1}}x}{\sqrt{1-{{x}^{2}}}}}{1-{{x}^{2}}} \right]\]and \[\dfrac{dy}{dx}=\dfrac{-2{{\cos }^{-1}}x}{\sqrt{1-{{x}^{2}}}}\].
By substituting \[\dfrac{dy}{dx}\]in \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}\].
\[\begin{align}
  & \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{2+x.\dfrac{dy}{dx}}{1-{{x}^{2}}} \\
 & \Rightarrow \left( 1-{{x}^{2}} \right)\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=2+x.\dfrac{dy}{dx} \\
 & \left( 1-{{x}^{2}} \right)\dfrac{{{d}^{2}}y}{d{{x}^{2}}}-x.\dfrac{dy}{dx}=2-(1) \\
\end{align}\]
We can compare the differential equation given in the question to equation (1).
\[\left( 1-{{x}^{2}} \right)\dfrac{{{d}^{2}}y}{d{{x}^{2}}}-x.\dfrac{dy}{dx}=c\], (by comparing to equation (1))
We get, c = 2.
Thus, \[\sqrt{y}={{\cos }^{-1}}x\], satisfies the differential equation \[\left( 1-{{x}^{2}} \right)\dfrac{{{d}^{2}}y}{d{{x}^{2}}}-x.\dfrac{dy}{dx}=2\], where value of c = 2.
Hence, option (d) is correct.

Note: We have used lots of trigonometric functions and their differentiations. Remember these formulae as they will be useful while solving problems like these.
Square the expression \[\sqrt{y}={{\cos }^{-1}}x\]to make the differentiation and simplification easy or else it becomes very complex to solve.