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Question:

If $y=\left(\tan ^{-1} x\right)^{2}$, show that $\left(x^{2}+1\right)^{2} y_{2}+2 x\left(x^{2}+1\right) y_{1}=2$

Solution:

The given relationship is $y=\left(\tan ^{-1} x\right)^{2}$

Then,

$y_{1}=2 \tan ^{-1} x \frac{d}{d x}\left(\tan ^{-1} x\right)$

$\Rightarrow y_{1}=2 \tan ^{-1} x \cdot \frac{1}{1+x^{2}}$

$\Rightarrow\left(1+x^{2}\right) y_{1}=2 \tan ^{-1} x$

Again differentiating with respect to $x$ on both the sides, we obtain

$\left(1+x^{2}\right) y_{2}+2 x y_{1}=2\left(\frac{1}{1+x^{2}}\right)$

$\Rightarrow\left(1+x^{2}\right)^{2} y_{2}+2 x\left(1+x^{2}\right) y_{1}=2$

Hence, proved.

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