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If $y=e^{\tan ^{-1} x}$ then
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Verified Answer
The correct answer is:
$\left(1+x^2\right) y_2+(2 x-1) y_1=0$
$\frac{d y}{d x}=e^{\tan ^{-1} x} \times \frac{1}{1+x^2}$
$\Rightarrow\left(1+x^2\right) y^{\prime}=y$
or $2 x y^{\prime}+\left(1+x^2\right) y^{\prime \prime}=y^{\prime}$
or $\left(1+x^2\right) y^{\prime \prime}+y^{\prime}(2 x-1)=0$
$\Rightarrow\left(1+x^2\right) y^{\prime}=y$
or $2 x y^{\prime}+\left(1+x^2\right) y^{\prime \prime}=y^{\prime}$
or $\left(1+x^2\right) y^{\prime \prime}+y^{\prime}(2 x-1)=0$
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