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

Hint: 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 \[\dfrac{d}{{dx}}\left[ {\dfrac{{f(x)}}{{g(x)}}} \right] = \dfrac{{g(x)f'(x) - f(x)g'(x)}}{{{{\left[ {g(x)} \right]}^2}}}\].