Practicing Success
If $y=\left(x+\sqrt{1+x^2}\right)^m$ then $\left(1+x^2\right) y_2+x y_1-m^2 y=$ |
0 1 -1 2 |
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$y_1=m\left[x+\sqrt{1+x^2}\right]^{m-1} . (\left.1+\frac{1 . 2 x}{2 \sqrt{1+x^2}}\right)$ $=\frac{m y}{\sqrt{1+x^2}}$ $\Rightarrow y_1{ }^2\left(1+x^2\right)=m^2 y^2$ Differentiating w.r.t. x, $2 y_1 y_2\left(1+x^2\right)+y_1^2(2 x)=2 m^2 y y_1$ Canceling 2y1, $\left(1+x^2\right) y_2+x y_1=m^2 y$ Hence (1) is correct answer. |