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If y = x+\(\sqrt{a^2+x^2}\), where a is a constant,prove that \(\frac{a^2}{x^2}\frac{d^2y}{dx^2}=\frac{a^2}{\sqrt{a^2+x^2}}\)(a2+x^2) d2y/dx2+xdy/dx-y=0 |
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Answer» y = x + \(\sqrt{a^2 + x^2}\) \(\frac{dy}{dx}\) = 1 + \(\frac{2x}{2\sqrt{a^2+x^2}}\) = 1 + \(\frac{x}{\sqrt{a^2+x^2}}\) \(\frac{d^2y}{dx^2}\) = \(\frac{\sqrt{a^2+x^2}\times 1-x \times 2x/(2\sqrt{a^2+x^2})}{(a^2+x^2)}\) = \(\frac{a^2+x^2-x^2}{(a^2+x^2)^{3/2}}\) = \(\frac{a^2}{(a^2+x^2)^{1/2}(a^2+x^2)}\) ⇒ (a2 + x2) \(\frac{d^2y}{dx^2}\) = \(\frac{a^2}{\sqrt{a^2+x^2}}\) Now, (a2 + x2) \(\frac{d^2y}{dx^2}\) + x\(\frac{dy}{dx}\) - y = \(\frac{a^2}{\sqrt{a^2+x^2}}\) + x + \(\frac{x^2}{\sqrt{a^2+x^2}}\) - (x + \(\sqrt{a^2+x^2)}\) = \(\frac{a^2+x^2}{\sqrt{a^2+x^2}}\) - \(\sqrt{a^2+x^2}\) = \(\sqrt{a^2+x^2}\) - \(\sqrt{a^2+x^2}\) = 0 = R.H.S Hence proved. |
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