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+1 vote
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in Differential Equations by (20 points)
edited by

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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1 Answer

+1 vote
by (34.5k points)
edited by

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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