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Answer the following.
If y = (x + √(a² + x²))^m, prove that (a² + x²)(d²y)/(dx²) + x(dy)/(dx) - m²y = 0
If $y = (x + \sqrt{a^2 + x^2})^m$, prove that $(a^2 + x^2)\frac{d^2y}{dx^2} + x\frac{dy}{dx} - m^2y = 0$
Given: $y = (x + \sqrt{a^2 + x^2})^m$
To show: $(a^2 + x^2)\frac{d^2y}{dx^2} + x\frac{dy}{dx} - m^2y = 0$
Answer
Answer
Official answer keyMethod I
$y = [x + \sqrt{a^2 + x^2}]^m$
$\frac{dy}{dx} = my\frac{1}{\sqrt{a^2+x^2}}$
$\sqrt{a^2+x^2}\frac{dy}{dx} = my$
$\sqrt{a^2+x^2}\frac{d^2y}{dx^2} + \frac{dy}{dx} \cdot \frac{1}{2} \cdot \frac{2x}{\sqrt{a^2+x^2}} = m\frac{dy}{dx}$
$(a^2+x^2)\frac{d^2y}{dx^2} + x\frac{dy}{dx} - m\sqrt{a^2+x^2}\frac{dy}{dx} = 0$
$(a^2+x^2)\frac{d^2y}{dx^2} + x\frac{dy}{dx} - m^2y = 0$
- $y = [x + \sqrt{a^2 + x^2}]^m$
- $\frac{dy}{dx} = \frac{m[x + \sqrt{a^2+x^2}]^m}{x + \sqrt{a^2+x^2}} \cdot \left[1 + \frac{1 \cdot 2x}{2\sqrt{a^2+x^2}}\right] = my\frac{1}{\sqrt{a^2+x^2}}$
- $\sqrt{a^2+x^2}\frac{dy}{dx} = my$
- $\sqrt{a^2+x^2}\frac{d^2y}{dx^2} + \frac{dy}{dx} \cdot \frac{1}{2} \cdot \frac{2x}{\sqrt{a^2+x^2}} = m\frac{dy}{dx}$
- $(a^2+x^2)\frac{d^2y}{dx^2} + x\frac{dy}{dx} - m\sqrt{a^2+x^2}\frac{dy}{dx} = 0$
- $(a^2+x^2)\frac{d^2y}{dx^2} + x\frac{dy}{dx} - m^2y = 0$