Differential Equations Question 244
Question: If $ y={{(x+\sqrt{1+x^{2}})}^{n}}, $ then $ (1+x^{2})\frac{d^{2}y}{dx^{2}}+x\frac{dy}{dx} $ is
Options:
A) $ n^{2}y $
B) $ -n^{2}y $
C) $ -y $
D) $ 2x^{2}y $
Show Answer
Answer:
Correct Answer: A
Solution:
[a] $ y={{(x+\sqrt{1+x^{2}})}^{n}} $
$ \frac{dy}{dx}=n{{(x+\sqrt{1+x^{2}})}^{n-1}}( 1+\frac{1}{2}{{(1+x^{2})}^{-1/2}}.2x ); $
$ \frac{dy}{dx}=n{{(x+\sqrt{1+x^{2}})}^{n-1}}\frac{(\sqrt{1+x^{2}}+x)}{\sqrt{1+x^{2}}}=\frac{n{{(\sqrt{1+x^{2}}+x)}^{n}}}{\sqrt{1+x^{2}}} $ or $ \sqrt{1+x^{2}}\frac{dy}{dx}=ny $ or $ \sqrt{1+x^{2}}y_1=ny(y_1=\frac{dy}{dx}) $ Squaring, $ (1+x^{2})y_1^{2}.n^{2}y^{2} $ Differentiating, $ (1+x^{2})2y_1y_2+y_1^{2}.2x=n^{2}.2yy_1 $
$ (Here,y_2=\frac{d^{2}y}{dx^{2}}) $ or $ (1+x^{2})y_2+xy_1=n^{2}y $
BETA
