Question
If $\text{y}=(\sin^{-1}\text{x})^2,$ prove that $(1-\text{x}^2)\text{y}_2-\text{xy}_1-2=0$
✓
Answer
Given,
$\text{y}=(\sin^{-1}\text{x})^2\dots\text{ eq.1}$
To prove: $(1-\text{x}^2)\text{y}_2-\text{xy}_1-2=0$
Let's find $\frac{\text{d}^2\text{y}}{\text{dx}^2}$
As, $\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{d}}{\text{dx}}\Big(\frac{\text{dy}}{\text{dx}}\Big)$
So, lets first find $\frac{\text{dy}}{\text{dx}}$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\sin^{-1}\text{x})^2$
Using chain rule we will differentiate the above expression:
Let $\text{t}=\sin^{-1}\text{x}$
$\Rightarrow\frac{\text{dt}}{\text{dx}}=\frac{1}{\sqrt{(1-\text{x}^2)}}$ $[$using formula for derivative of $\sin^{-1}\text{x}]$
And $y = t^2$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{dy}}{\text{dt}}\frac{\text{dt}}{\text{dx}}$
$\frac{\text{dy}}{\text{dx}}=2\text{t}\frac{1}{\sqrt{(1-\text{x}^2)}}=2\sin^{-1}\text{x}\frac{1}{\sqrt{(1-\text{x}^2)}}\dots\text{ eq. 2}$
Again differentiating with respect to x applying product rule:
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=2\sin^{-1}\text{x}\frac{\text{d}}{\text{dx}}\Big(\frac{1}{\sqrt{1-\text{x}^2}}\Big)+\frac{2}{\sqrt{(1-\text{x}^2)}}\frac{\text{d}}{\text{dx}}\sin^{-1}\text{x}$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\frac{2\sin^{-1}\text{x}}{2(1-\text{x}^2)\sqrt{1-\text{x}^2}}(-2\text{x})+\frac{2}{(1-\text{x}^2)}$ $\bigg[\text{using }\frac{\text{d}}{\text{dx}}(\text{x}^\text{n})=\text{nx}^{\text{n}-1}\frac{\text{d}}{\text{dx}}\sin^{-1}\text{x}=\frac{1}{\sqrt{(1-\text{x}^2)}}\bigg]$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{2\text{x}\sin^{-1}\text{x}}{(1-\text{x}^2)\sqrt{1-\text{x}^2}}+\frac{2}{(1-\text{x}^2)}$
$(1-\text{x}^2)\frac{\text{d}^2\text{y}}{\text{dx}^2}=2+\frac{2\text{x}\sin^{-1}\text{x}}{\sqrt{1-\text{x}^2}}$
Using eq. 2:
$(1-\text{x}^2)\frac{\text{d}^2\text{y}}{\text{dx}^2}=2+\frac{\text{dy}}{\text{dx}}$
$\therefore(1-\text{x}^2)\text{y}_2-\text{xy}_1-2=0\dots\text{ proved.}$
$\text{y}=(\sin^{-1}\text{x})^2\dots\text{ eq.1}$
To prove: $(1-\text{x}^2)\text{y}_2-\text{xy}_1-2=0$
Let's find $\frac{\text{d}^2\text{y}}{\text{dx}^2}$
As, $\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{d}}{\text{dx}}\Big(\frac{\text{dy}}{\text{dx}}\Big)$
So, lets first find $\frac{\text{dy}}{\text{dx}}$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\sin^{-1}\text{x})^2$
Using chain rule we will differentiate the above expression:
Let $\text{t}=\sin^{-1}\text{x}$
$\Rightarrow\frac{\text{dt}}{\text{dx}}=\frac{1}{\sqrt{(1-\text{x}^2)}}$ $[$using formula for derivative of $\sin^{-1}\text{x}]$
And $y = t^2$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{dy}}{\text{dt}}\frac{\text{dt}}{\text{dx}}$
$\frac{\text{dy}}{\text{dx}}=2\text{t}\frac{1}{\sqrt{(1-\text{x}^2)}}=2\sin^{-1}\text{x}\frac{1}{\sqrt{(1-\text{x}^2)}}\dots\text{ eq. 2}$
Again differentiating with respect to x applying product rule:
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=2\sin^{-1}\text{x}\frac{\text{d}}{\text{dx}}\Big(\frac{1}{\sqrt{1-\text{x}^2}}\Big)+\frac{2}{\sqrt{(1-\text{x}^2)}}\frac{\text{d}}{\text{dx}}\sin^{-1}\text{x}$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\frac{2\sin^{-1}\text{x}}{2(1-\text{x}^2)\sqrt{1-\text{x}^2}}(-2\text{x})+\frac{2}{(1-\text{x}^2)}$ $\bigg[\text{using }\frac{\text{d}}{\text{dx}}(\text{x}^\text{n})=\text{nx}^{\text{n}-1}\frac{\text{d}}{\text{dx}}\sin^{-1}\text{x}=\frac{1}{\sqrt{(1-\text{x}^2)}}\bigg]$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{2\text{x}\sin^{-1}\text{x}}{(1-\text{x}^2)\sqrt{1-\text{x}^2}}+\frac{2}{(1-\text{x}^2)}$
$(1-\text{x}^2)\frac{\text{d}^2\text{y}}{\text{dx}^2}=2+\frac{2\text{x}\sin^{-1}\text{x}}{\sqrt{1-\text{x}^2}}$
Using eq. 2:
$(1-\text{x}^2)\frac{\text{d}^2\text{y}}{\text{dx}^2}=2+\frac{\text{dy}}{\text{dx}}$
$\therefore(1-\text{x}^2)\text{y}_2-\text{xy}_1-2=0\dots\text{ proved.}$
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