Question

Show that if \[y = {\left( {{{\sin }^{ - 1}}x} \right)^2}\] then \[\left( {1 - {x^2}} \right)\dfrac{{{d^2}y}}{{d{x^2}}} - x\dfrac{{dy}}{{dx}} - 2 = 0\] \[?\]

Answer 1

Hint: First we have to find the first derivative of the given function \[y\] with respect to \[x\] and rewrite it so that the second order derivative of the function exists as a function of the first derivative. After that simplify or rewrite it to prove the differential equation.

Complete step-by-step answer:
Given \[y = {\left( {{{\sin }^{ - 1}}x} \right)^2}\] -----(1)
Differentiating with respect to \[x\] both sides of the equation (1), we get
\[\dfrac{{dy}}{{dx}} = 2{\sin ^{ - 1}}x\left( {\dfrac{1}{{\sqrt {1 - {x^2}} }}} \right)\] ------(2)
Squaring both sides of the equation (2) and rewriting it, we get
 \[\left( {1 - {x^2}} \right){\left( {\dfrac{{dy}}{{dx}}} \right)^2} = 4y\] -------(3)
Differentiating with respect to \[x\] both sides of the equation (3), we get
\[2\left( {1 - {x^2}} \right)\left( {\dfrac{{dy}}{{dx}}} \right)\left( {\dfrac{{{d^2}y}}{{d{x^2}}}} \right) - 2x{\left( {\dfrac{{dy}}{{dx}}} \right)^2} = 4\dfrac{{dy}}{{dx}}\]
\[ \Rightarrow \left( {1 - {x^2}} \right)\dfrac{{{d^2}y}}{{d{x^2}}} - x{\left( {\dfrac{{dy}}{{dx}}} \right)} = 2\]
\[ \Rightarrow \left( {1 - {x^2}} \right)\dfrac{{{d^2}y}}{{d{x^2}}} - x{\left( {\dfrac{{dy}}{{dx}}} \right)} - 2 = 0\] .

Note: A differential equation is the equation which contains dependent variables, independent variables and derivatives of the dependent variables with respect to the independent variables. Since differential equations are classifying into two types, Ordinary differential equations where dependent variables depend on one independent variable and Partial differential equations where dependent variables depend on two or more independent variables.