Subjective Type

If $$y=e^{m sin^{-1}x}$$, then show that $$(1-x^2)\dfrac{d^2y}{dx^2}-x\dfrac{dy}{dx}-m^2y=0$$.

Solution

This problem requires the use of chain rule in evaluating derivatives:
$$y={ e }^{ m\sin ^{ -1 }{ x } }\\ \dfrac { dy }{ dx } ={ e }^{ m\sin ^{ -1 }{ x } }.\dfrac { m }{ \sqrt { 1-{ x }^{ 2 } } } =\dfrac { my }{ \sqrt { 1-{ x }^{ 2 } } } \\ \dfrac { { d }^{ 2 }y }{ { dx }^{ 2 } } =\dfrac { d }{ dx } (\dfrac { dy }{ dx } )=\dfrac { m }{ 1-{ x }^{ 2 } } .(\sqrt { 1-{ x }^{ 2 } } .\dfrac { dy }{ dx } +\dfrac { xy }{ \sqrt { 1-{ x }^{ 2 } } } )\\=\dfrac { m }{ 1-{ x }^{ 2 } } .(\sqrt { 1-{ x }^{ 2 } } .\dfrac { my }{ \sqrt { 1-{ x }^{ 2 } } } +\dfrac { xy }{ \sqrt { 1-{ x }^{ 2 } } } )=\dfrac { m }{ 1-{ x }^{ 2 } } .(my+\dfrac { xy }{ \sqrt { 1-{ x }^{ 2 } } } )\\ \Rightarrow x\dfrac { dy }{ dx } =\dfrac { mxy }{ \sqrt { 1-{ x }^{ 2 } } } ,\quad (1-{ x }^{ 2 })\dfrac { { d }^{ 2 }y }{ { dx }^{ 2 } } =m.(my+\dfrac { xy }{ \sqrt { 1-{ x }^{ 2 } } } )={ m }^{ 2 }y+\dfrac { mxy }{ \sqrt { 1-{ x }^{ 2 } } } \\ \Rightarrow (1-{ x }^{ 2 })\dfrac { { d }^{ 2 }y }{ { dx }^{ 2 } } ={ m }^{ 2 }y+x\dfrac { dy }{ dx } \Rightarrow (1-{ x }^{ 2 })\dfrac { { d }^{ 2 }y }{ { dx }^{ 2 } } -x\dfrac { dy }{ dx } -{ m }^{ 2 }y=0$$

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