If the solution curve of ((^-1y)-x),dy=(1+y^2),dx passes through (1,0), the abscissa of th…

Question

If the solution curve of $((	an^{-1}y)-x)\,dy=(1+y^2)\,dx$ passes through $(1,0)$, the abscissa of the point on the curve with ordinate $	an(1)$ is:

Accepted Answer

$$\frac{dx}{dy}+\frac{x}{1+y^2} = \frac{\tan^{-1}y}{1+y^2}.$$

I.F. $=e^{\tan^{-1}y}$. With $t=\tan^{-1}y$: $xe^t = t e^t - e^t + c$.

Using $(1,0)$: $c=2$. At $y=\tan1$ (i.e. $t=1$): $xe=e-e+2=2 \Rightarrow x=2/e$.

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