Question 355 of 490HardMCQ
If 2x = y\[{^{{1 \over 5}}}\] + y\[{^{ - {1 \over 5}}}\] and
(x2 - 1) \[{{{d^2}y} \over {d{x^2}}}\] + \[\lambda \]x \[{{dy} \over {dx}}\] + ky = 0,
then \[\lambda \] + k is equal to :
Solutions · 1
Official solution
It is given that
2x=y1/5+y−1/5
\[ \Rightarrow 2x = {y^{1/5}} + 1/{y^{1/5}}\]
Therefore, \[2x = a + {1 \over a} \Rightarrow {a^2} - 2ax + 1 = 0\]
\[a = {{2x \pm \sqrt {4{x^2} - 4} } \over 2}\]
\[ \Rightarrow a = {{2x \pm 2\sqrt {{x^2} - 1} } \over 2}\]
\[ \Rightarrow a = x \pm \sqrt {{x^2} - 1} \]
\[ \Rightarrow {y^{1/5}} = x \pm \sqrt {{x^2} - 1} \]
\[ \Rightarrow y = {(x \pm \sqrt {{x^2} - 1} )^5}\]
Therefore,
\[{{dy} \over {dx}} = 5{(x \pm \sqrt {{x^2} - 1} )^4}\left( {1 \pm {{2x} \over {2\sqrt {{x^2} - 1} }}} \right)\]
\[ = 5{(x + \sqrt {{x^2} - 1} )^4}\left( {{{\sqrt {{x^2} - 1} \pm x} \over {\sqrt {{x^2} - 1} }}} \right)\]
\[ \Rightarrow {{dy} \over {dx}} = {{ - 5y} \over {\sqrt {{x^2} - 1} }}\] ...... (1)
\[ \Rightarrow {{{d^2}y} \over {d{x^2}}} = {{\left[ {\sqrt {{x^2} - 1} \left( { - 5{{dy} \over {dx}}} \right) - 5( - 5y){1 \over 2}{{2x} \over {\sqrt {{x^2} - 1} }}} \right]} \over {({x^2} - 1)}}\]
Therefore, \[({x^2} - 1){{{d^2}y} \over {d{x^2}}} = - 5\sqrt {{x^2} - 1} {{dy} \over {dx}} + 5y{x \over {\sqrt {{x^2} - 1} }}\]
\[ \Rightarrow ({x^2} - 1){{{d^2}y} \over {d{x^2}}} = 25y - x{{dy} \over {dx}}\]
\[ \Rightarrow ({x^2} - 1){{{d^2}y} \over {d{x^2}}} + 1x{{dy} \over {dx}} - 25y = 0\]
Therefore, \[\lambda\] = 1, k = -25; hence,
\[\lambda + k = - 24\] Answer the question to unlock