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,

& {{\left( \sqrt{y} \right)}^{2}}={{\left( {{\cos }^{-1}}x \right)}^{2}} \\\ & \Rightarrow y={{\left( {{\cos }^{-1}}x \right)}^{2}} \\\ \end{aligned}$$ Let us differentiate both sides of the expression with respect to ‘x’. $$\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{aligned} & \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{aligned}$$ [$$\therefore $$While differentiation, $$\dfrac{d}{dx}\sqrt{1-{{x}^{2}}}$$ $$\begin{aligned} & =\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{aligned}$$ $$\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{aligned} & \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{aligned}$$ We can compare the differential equation given in the question to equation (1). $$\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.