Question

If $2x = {y^{\dfrac{1}{5}}} + {y^{ - \dfrac{1}{5}}}$ and $\left( {{x^2} - 1} \right)\dfrac{{{d^2}y}}{{d{x^2}}} + \lambda x\dfrac{{dy}}{{dx}} + ky = 0$, then $\lambda + k$ is equal to
(A). -23
(B). -24
(C). 26
(D). -26

Answer 1

To solve the question, at first we have to consider ${y^{\dfrac{1}{5}}}$ to be ‘p’. Then solving the quadratic equation in terms of ‘p’ we can express y in terms of x. then differentiating y with respect to x we can obtain the expression for $\dfrac{{dy}}{{dx}}$ and again differentiating with respect to x we will get second order differential equation. Finally comparing the obtained differential equation with the differential equation $\left( {{x^2} - 1} \right)\dfrac{{{d^2}y}}{{d{x^2}}} + \lambda x\dfrac{{dy}}{{dx}} + ky = 0$ we can get the values of $\lambda$ and k. Thus the value of $\lambda + k$ can be determined.

Complete step-by-step answer :
Given that
$2x = {y^{\dfrac{1}{5}}} + {y^{ - \dfrac{1}{5}}}$ ... (1)
To solve the above equation consider $p = {y^{\dfrac{1}{5}}}$, then $\dfrac{1}{p} = {y^{ - \dfrac{1}{5}}}$. So substituting these values the eq. (1) reduces to

$$\Rightarrow 2x = p + \dfrac{1}{p} \\\ \Rightarrow {p^2} - 2xp + 1 = 0 \\\$$

…………………………………… (2)
We know that the roots of a quadratic equation $a{x^2} + bx + c = 0$ is given by
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ …………………………………. (3)
Now applying this formula to eq. (2) we will get

$$p = \dfrac{{ - \left( { - 2x} \right) \pm \sqrt {{{\left( { - 2x} \right)}^2} - 4 \times 1 \times 1} }}{{2 \times 1}} \\\ = \dfrac{{2x \pm \sqrt {4{x^2} - 4 \times 1 \times 1} }}{2} \\\ = x \pm \sqrt {{x^2} - 1} \\\$$

……………………..………………. (4)
Substituting the value of $p = {y^{\dfrac{1}{5}}}$in eq. (4) we will get,

$$\Rightarrow {y^{\dfrac{1}{5}}} = x \pm \sqrt {{x^2} - 1} \\\ \Rightarrow y = {\left( {x \pm \sqrt {{x^2} - 1} } \right)^5} \\\$$

……………………………………… (5)
Now differentiating eq. (5) with respect to x on both the sides, we will get

$$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}{\left( {x \pm \sqrt {{x^2} - 1} } \right)^5} \\\ \Rightarrow \dfrac{{dy}}{{dx}} = 5{\left( {x \pm \sqrt {{x^2} - 1} } \right)^4} \cdot \dfrac{d}{{dx}}\left( {x \pm \sqrt {{x^2} - 1} } \right) \\\ \Rightarrow \dfrac{{dy}}{{dx}} = 5{\left( {x \pm \sqrt {{x^2} - 1} } \right)^4} \cdot \left( {1 \pm \dfrac{{2x}}{{2\sqrt {{x^2} - 1} }}} \right) \\\ \Rightarrow \dfrac{{dy}}{{dx}} = 5{\left( {x \pm \sqrt {{x^2} - 1} } \right)^4} \cdot \left( {\dfrac{{\sqrt {{x^2} - 1} \pm x}}{{\sqrt {{x^2} - 1} }}} \right) \\\ \Rightarrow \dfrac{{dy}}{{dx}} = 5{\left( {x \pm \sqrt {{x^2} - 1} } \right)^4} \cdot \left( {\dfrac{{\sqrt {{x^2} - 1} \pm x}}{{\sqrt {{x^2} - 1} }}} \right) \\\ \Rightarrow \dfrac{{dy}}{{dx}} = - 5{\left( {x \pm \sqrt {{x^2} - 1} } \right)^5} \cdot \left( {\dfrac{1}{{\sqrt {{x^2} - 1} }}} \right) \\\$$

…………………………………….. (6)
Substituting the value of eq. (5) in eq. (6) we will get,
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{ - 5y}}{{\sqrt {{x^2} - 1} }}$ ………..…………………………. (7)
$\Rightarrow \sqrt {{x^2} - 1} \dfrac{{dy}}{{dx}} = - 5y$ …………………………………… (8)
Differentiating eq. (7) with respect to x on both the sides, we will get

$$\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{d}{{dx}}\left( {\dfrac{{ - 5y}}{{\sqrt {{x^2} - 1} }}} \right) \\\ \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = \left( {\dfrac{{\sqrt {{x^2} - 1} \left( { - 5\dfrac{{dy}}{{dx}}} \right) - \left( { - 5y} \right)\dfrac{{2x}}{{2\sqrt {{x^2} - 1} }}}}{{{x^2} - 1}}} \right) \\\ \Rightarrow \left( {{x^2} - 1} \right)\dfrac{{{d^2}y}}{{d{x^2}}} = \left( { - 5} \right)\left( {\sqrt {{x^2} - 1} \dfrac{{dy}}{{dx}}} \right) - x \times \dfrac{{\left( { - 5y} \right)}}{{\sqrt {{x^2} - 1} }} \\\$$

…………………………………….. (9)
Substituting the value of eq. (7) and (8) in eq. (9) we will get,

$$\Rightarrow \left( {{x^2} - 1} \right)\dfrac{{{d^2}y}}{{d{x^2}}} = \left( { - 5} \right)\left( { - 5y} \right) - x\dfrac{{dy}}{{dx}} \\\ \Rightarrow \left( {{x^2} - 1} \right)\dfrac{{{d^2}y}}{{d{x^2}}} = 25y - x\dfrac{{dy}}{{dx}} \\\ \Rightarrow \left( {{x^2} - 1} \right)\dfrac{{{d^2}y}}{{d{x^2}}} + 1.x\dfrac{{dy}}{{dx}} + ( - 25)y = 0 \\\ \\\$$

…………………………………….. (10)
But given equation is
$\left( {{x^2} - 1} \right)\dfrac{{{d^2}y}}{{d{x^2}}} + \lambda x\dfrac{{dy}}{{dx}} + ky = 0$ ……………………………….. (11)
Now comparing eq. (10) and (11) we will get, the values of $\lambda$ and k which is given by
$\lambda = 1$ And $k = - 25$.
Therefore $\lambda + k = 1 - 25 = - 24$
The option (B) is correct.

Note : The formula for quotient rule of derivative is given by $\dfrac{d}{{dx}}\left[ {\dfrac{{f(x)}}{{g(x)}}} \right] = \dfrac{{g(x)f'(x) - f(x)g'(x)}}{{{{\left[ {g(x)} \right]}^2}}}$.