Question
If $\log\text{y}=\tan^{-1}$ show that $(1+\text{x}^2)\text{y}_2+(2\text{x}-1)\text{y}_1=0$
$\log\text{y}=\tan^{-1}$
Differentiating w.r.t.x, we get
$\frac{1}{\text{y}}\times\text{y}_1=\frac{1}{1+\text{x}^2}$ $\Rightarrow(1+\text{x}^2)\text{y}_1=\text{y}$ $\Rightarrow(1+\text{x}^2)\text{y}_2+2\text{xy}_1=\text{y}_1$ $\Rightarrow(1+\text{x}^2)\text{y}_2+2\text{xy}_1-\text{y}_1=0$$\Rightarrow(1+\text{x}^2)\text{y}_2+(25\text{x}-1)\text{y}_1=0$
hence proved
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