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If \( y = e^{\large a \: cos^{-1}x}, -1 \leq x \leq 1\), show that \((1- x^2)\large\frac{d^2y}{dx^2}-\normalsize x\large\frac{dy}{dx} -\normalsize a^2y \; = \; 0.\)

1 Answer

Toolbox:
  • $\large\frac{d}{dx}$$(\cos^{-1}x)=\large\frac{-1}{\sqrt{1-x^2}}$
Step 1:
We have $y=e^{\large a\cos^{-1}x}$
$y_1=e^{\large a\cos^{-1}x}.\large\frac{-a}{\sqrt{1-x^2}}$
$\sqrt{1-x^2}y_1=-ae^{\large a\cos^{-1}x}$
Step 2:
$\Rightarrow \sqrt{1-x^2}y_2+y_1.\large\frac{1}{2}.\frac{1}{\sqrt{1-x^2}}$$(-2x)=\large\frac{a^2e^{\Large a\cos^{-1}x}}{\sqrt{1-x^2}}$
$(1-x^2)y_2-xy_1=a^2y$
$(1-x^2)y_2-xy_1-a^2y=0$
Hence proved.
answered May 15, 2013 by sreemathi.v
 

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