Question
If $\text{y}=\text{cosec}^{-1}\text{x},\text{x}>1$ prove that $\text{x}(\text{x}^2-1)\frac{\text{d}^2\text{y}}{\text{dx}^2}+(2\text{x}^2-1)\frac{\text{dy}}{\text{dx}}=0.$
✓
Answer
Here,
$\text{y}=\text{cosec}^{-1}\text{x}$
Differentiating w.r.t.x, we get
$\frac{\text{dy}}{\text{dx}}=\frac{-1}{\text{x}\sqrt{{}\text{x}^2-1}}$
Differentiating w.r.t.x, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\sqrt{\text{x}^2-1}+\frac{\text{x}^2}{\sqrt{\text{x}^2-1}}}{\text{x}^2(\text{x}^2-1)}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{x}^2-1+\text{x}^2}{\text{x}^2(\text{x}^2-1)\sqrt{\text{x}^2-1}}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{2\text{x}^2-1}{\text{x}^2(\text{x}^2-1)\sqrt{\text{x}^2-1}}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{2\text{x}}{(\text{x}^2-1)\sqrt{\text{x}^2-1}}-\frac{1}{\text{x}^2(\text{x}^2-1)\sqrt{\text{x}^2-1}}$
$\Rightarrow(\text{x}^2-1)\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{2\text{x}}{\sqrt{\text{x}^2-1}}-\frac{1}{\sqrt{\text{x}^2-1}}$
$\Rightarrow(\text{x}^2-1)\frac{\text{d}^2\text{y}}{\text{dx}^2}=-2\text{x}\frac{\text{dy}}{\text{dx}}+\frac{1}{\text{x}}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow(\text{x}^2-1)\frac{\text{d}^2\text{y}}{\text{dx}^2}=-2\text{x}\frac{\text{dy}}{\text{dx}}+\frac{1}{\text{x}}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow(\text{x}^2-1)\frac{\text{d}^2\text{y}}{\text{dx}^2}=-(2\text{x}^2-1)\frac{\text{dy}}{\text{dx}}$
$\Rightarrow(\text{x}^2-1)\frac{\text{d}^2\text{y}}{\text{dx}^2}=+(2\text{x}^2-1)\frac{\text{dy}}{\text{dx}}=0$
Hence proved
$\text{y}=\text{cosec}^{-1}\text{x}$
Differentiating w.r.t.x, we get
$\frac{\text{dy}}{\text{dx}}=\frac{-1}{\text{x}\sqrt{{}\text{x}^2-1}}$
Differentiating w.r.t.x, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\sqrt{\text{x}^2-1}+\frac{\text{x}^2}{\sqrt{\text{x}^2-1}}}{\text{x}^2(\text{x}^2-1)}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{x}^2-1+\text{x}^2}{\text{x}^2(\text{x}^2-1)\sqrt{\text{x}^2-1}}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{2\text{x}^2-1}{\text{x}^2(\text{x}^2-1)\sqrt{\text{x}^2-1}}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{2\text{x}}{(\text{x}^2-1)\sqrt{\text{x}^2-1}}-\frac{1}{\text{x}^2(\text{x}^2-1)\sqrt{\text{x}^2-1}}$
$\Rightarrow(\text{x}^2-1)\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{2\text{x}}{\sqrt{\text{x}^2-1}}-\frac{1}{\sqrt{\text{x}^2-1}}$
$\Rightarrow(\text{x}^2-1)\frac{\text{d}^2\text{y}}{\text{dx}^2}=-2\text{x}\frac{\text{dy}}{\text{dx}}+\frac{1}{\text{x}}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow(\text{x}^2-1)\frac{\text{d}^2\text{y}}{\text{dx}^2}=-2\text{x}\frac{\text{dy}}{\text{dx}}+\frac{1}{\text{x}}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow(\text{x}^2-1)\frac{\text{d}^2\text{y}}{\text{dx}^2}=-(2\text{x}^2-1)\frac{\text{dy}}{\text{dx}}$
$\Rightarrow(\text{x}^2-1)\frac{\text{d}^2\text{y}}{\text{dx}^2}=+(2\text{x}^2-1)\frac{\text{dy}}{\text{dx}}=0$
Hence proved
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