Question

If $2y = \sqrt {x + 1} + \sqrt {x - 1}$, show that $4\left( {{x^2} - 1} \right)\dfrac{{{d^2}y}}{{d{x^2}}} + 4x\dfrac{{dy}}{{dx}} - y = 0$.

Answer 1

We will find the value of $\dfrac{{dy}}{{dx}}$ by differentiating $2y = \sqrt {x + 1} + \sqrt {x - 1}$ with respect to $x$, using the differentiation of $\dfrac{{d\left( {\sqrt x } \right)}}{{dx}} = \dfrac{1}{{2\sqrt x }}$ and then we will differentiate it to find the second order derivative of $y$ i.e., $\dfrac{{{d^2}y}}{{d{x^2}}}$. We will simplify the obtained equation using $\dfrac{{dy}}{{dx}}$ and will reduce it in the required form.

Complete step-by-step answer:
We are given that $2y = \sqrt {x + 1} + \sqrt {x - 1}$.
We are required to prove that $4\left( {{x^2} - 1} \right)\dfrac{{{d^2}y}}{{d{x^2}}} + 4x\dfrac{{dy}}{{dx}} - y = 0$.
We will differentiate $2y = \sqrt {x + 1} + \sqrt {x - 1}$ with regard to $x$ for calculating $\dfrac{{dy}}{{dx}}$ and then, on differentiating $\dfrac{{dy}}{{dx}}$ again with respect to $x$, we will get $\dfrac{{{d^2}y}}{{d{x^2}}}$.
Differentiating $2y = \sqrt {x + 1} + \sqrt {x - 1}$ both sides with respect to $x$, we get
$\Rightarrow 2\dfrac{{dy}}{{dx}} = \dfrac{{d\left( {\sqrt {x + 1} + \sqrt {x - 1} } \right)}}{{dx}}$
Using the addition rule i.e., $\dfrac{{d\left( {m + n} \right)}}{{dx}} = \dfrac{{dm}}{{dx}} + \dfrac{{dn}}{{dx}}$, we get
$\Rightarrow 2\dfrac{{dy}}{{dx}} = \dfrac{{d\left( {\sqrt {x + 1} } \right)}}{{dx}} + \dfrac{{d\left( {\sqrt {x - 1} } \right)}}{{dx}}$
Using the differentiation of $\dfrac{{d\left( {\sqrt x } \right)}}{{dx}} = \dfrac{1}{{2\sqrt x }}$ in the above equation, we get
$\Rightarrow 2\dfrac{{dy}}{{dx}} = \dfrac{1}{{2\sqrt {x + 1} }} + \dfrac{1}{{2\sqrt {x - 1} }}$
Or, we can write this equation as:
$\Rightarrow 2\dfrac{{dy}}{{dx}} = \dfrac{{\sqrt {x - 1} + \sqrt {x + 1} }}{{2\left( {\sqrt {x + 1} \sqrt {x - 1} } \right)}}$
Using the formula: $\sqrt m \sqrt n = \sqrt {mn}$ in the denominator and putting $\sqrt {x - 1} + \sqrt {x + 1} = 2y$ in the numerator, we get
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{2y}}{{4\sqrt {\left( {x + 1} \right)\left( {x - 1} \right)} }}$
Using the algebraic identity $\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}$ in the denominator, we get
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{2y}}{{4\sqrt {{x^2} - 1} }}$
Now, differentiating $\dfrac{{dy}}{{dx}}$ with respect to $x$, we get
$\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{1}{4}\left( {\dfrac{{d\left( {\dfrac{{2y}}{{\sqrt {{x^2} - 1} }}} \right)}}{{dx}}} \right)$
Using the quotient rule: $\dfrac{{d\left( {\dfrac{p}{q}} \right)}}{{dx}} = \dfrac{{q\dfrac{{d\left( p \right)}}{{dx}} - p\dfrac{{d\left( q \right)}}{{dx}}}}{{{{\left( q \right)}^2}}},q \ne 0$ of the differentiation, we get
$\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{1}{4}\left( {\dfrac{{2\dfrac{{dy}}{{dx}}\sqrt {{x^2} - 1} - 2y\left( {\dfrac{1}{{2\sqrt {{x^2} - 1} }} \times 2x} \right)}}{{{{\left( {\sqrt {{x^2} - 1} } \right)}^2}}}} \right)$
Upon simplifying this equation, we get
$\Rightarrow 4\dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{{2\sqrt {{x^2} - 1} \dfrac{{dy}}{{dx}} - x\dfrac{{2y}}{{\sqrt {{x^2} - 1} }}}}{{{x^2} - 1}}$
Substituting the values of $\dfrac{{dy}}{{dx}}$ and putting $\dfrac{{2y}}{{\sqrt {{x^2} - 1} }} = 4\dfrac{{dy}}{{dx}}$, we get
$\Rightarrow 4\left( {{x^2} - 1} \right)\dfrac{{{d^2}y}}{{d{x^2}}} = 2\sqrt {{x^2} - 1} \left( {\dfrac{{2y}}{{4\sqrt {{x^2} - 1} }}} \right) - 4x\dfrac{{dy}}{{dx}}$
On further simplifying this equation, we get
$\Rightarrow 4\left( {{x^2} - 1} \right)\dfrac{{{d^2}y}}{{d{x^2}}} = y - 4x\dfrac{{dy}}{{dx}}$
Or, we can write this as:
$\Rightarrow 4\left( {{x^2} - 1} \right)\dfrac{{{d^2}y}}{{d{x^2}}} + 4x\dfrac{{dy}}{{dx}} - y = 0$ which is the required equation.

Note: In this question, you may get confused at many steps as there are numerous formulae used such as differentiation of $\sqrt x$ and other rules of differentiation (product rule and quotient rule). Be careful while substituting the values of $\dfrac{{dy}}{{dx}}$ in the obtained equation of $\dfrac{{{d^2}y}}{{d{x^2}}}$ and further simplification can be tricky as well as we have manipulated various terms for obtaining our desired result.