Question
If $$y = {\left( {x + \sqrt {1 + {x^2}} } \right)^n},$$ then $$\left( {1 + {x^2} } \right)\frac{{{d^2}y}}{{d{x^2}}} + x\frac{{dy}}{{dx}}{\text{ is :}}$$
A.
$${n^2}y$$
B.
$$ - {n^2}y$$
C.
$$ - y$$
D.
$$2{x^2}y$$
Answer :
$${n^2}y$$
Solution :
$$\eqalign{
& y = {\left( {x + \sqrt {1 + {x^2}} } \right)^n} \cr
& \frac{{dy}}{{dx}} = n{\left( {x + \sqrt {1 + {x^2}} } \right)^{n - 1}}\left( {1 + \frac{1}{2}{{\left( {1 + {x^2}} \right)}^{ - \frac{1}{2}}},2x} \right)\,; \cr
& \frac{{dy}}{{dx}} = n{\left( {x + \sqrt {1 + {x^2}} } \right)^{n - 1}}\frac{{\left( {\sqrt {1 + {x^2}} + x} \right)}}{{\sqrt {1 + {x^2}} }} = \frac{{n{{\left( {\sqrt {1 + {x^2}} + x} \right)}^n}}}{{\sqrt {1 + {x^2}} }} \cr
& {\text{or }}\sqrt {1 + {x^2}} \frac{{dy}}{{dx}} = ny \cr
& {\text{or }}\sqrt {1 + {x^2}} {y_1} = ny\,\,\,\,\,\,\,\,\left( {{y_1} = \frac{{dy}}{{dx}}} \right) \cr
& {\text{Squaring, }}\left( {1 + {x^2}} \right)y_1^2 = {n^2}{y^2} \cr
& {\text{Differentiating,}} \cr
& \left( {1 + {x^2}} \right)2{y_1}{y_2} + y_1^2.2x = {n^2}.2y{y_1} \cr
& \left( {{\text{Here }}{y_2} = \frac{{{d^2}y}}{{d{x^2}}}} \right) \cr
& {\text{or }}\left( {1 + {x^2}} \right){y_2} + x{y_1} = {n^2}y \cr} $$