If y=(x+ 1+x^ 2 )^ n , then (1+x^ 2 ) d^ 2 y d x^ 2 +x (d y)/(d x) is

Correct answer: A. n^ 2 y y=(x+ 1+x^ 2 )^ n array c (d y)/(d x)= n( x+ 1+x^ 2 )^ n-1 (1+(1)/(2)(1+x^ 2 )^ -1 / 2 ·2 x) array c (d y)/(d x)= n( x+ 1+x^ 2 )^ n-1 ( 1+x^ 2 +x) 1+x^ 2 = n( 1+x^ 2 +x)^ n 1+x^ 2 array array or 1+x^ 2 (d y)/(d x)=n y or 1+x^ 2 y_ 1 =n y (y_ 1 =(d y)/(d x))Squaring, (1+x^ 2 ) y_ 1 ^ 2 =n^ 2 y^ 2 Differentiating, (1+x^ 2 ) 2 y_ 1 y_ 2 +y_ 1 ^ 2 ·2 x=n^ 2 ·2 y y_ 1 or (1+x^ 2 ) y_ 2 +x y_ 1 =n^ 2 y

BITSAT 2015

Exam: BITSAT
Year: 2015
Subject: Mathematics
Chapter: Differential Equations

BITSAT 2015 · Mathematics · Differential Equations

If y=(x+ 1+x^ 2 )^ n , then (1+x^ 2 ) d^ 2 y d x^ 2 +x (d y)/(d x)...

y=\left(x+\sqrt{1+x^{2}}\right)^{n}, then \left(1+x^{2}\right) \frac{d^{2} y}{d x^{2}}+x \frac{d y}{d x}

Options

A $n^{2} y$
B $-n^{2} y$
C $-y$
D $2 x^{2} y$

Correct answer

A. $n^{2} y$

Step-by-step solution

y=\left(x+\sqrt{1+x^{2}}\right)^{n}

\begin{array}{c}\frac{d y}{d x}= n\left( x+\sqrt{1+x^{2}}\right)^{n-1}\left(1+\frac{1}{2}\left(1+x^{2}\right)^{-1 / 2}\cdot2 x\right) \\ \qquad \begin{array}{c}\frac{d y}{d x}= 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}}} \end{array}

\end{array}

\sqrt{1+x^{2}} \frac{d y}{d x}=n y

\sqrt{1+x^{2}} y_{1}=n y \left(y_{1}=\frac{d y}{d x}\right)Squaring, \left(1+x^{2}\right) y_{1}^{2}=n^{2} y^{2}

Differentiating,

\left(1+x^{2}\right) 2 y_{1} y_{2}+y_{1}^{2} \cdot2 x=n^{2} \cdot2 y y_{1} or \left(1+x^{2}\right) y_{2}+x y_{1}=n^{2} y

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