Question

If the solution curve of the differential equation \(((tan−1y)−x)dy=(1+y^2)dx\) passes through the point \((1, 0)\), then the abscissa of the point on the curve whose ordinate is \(tan(1)\), is

• A. \(2e\)
• B. \(\frac{2}{e}\)
• C. \(e\)
• D. \(\frac{1}{e}\)

Answer 1

The Correct Option is B

\(((tan−1y)−x)dy=(1+y^2)dx\)

\(\frac{dx}{dy}+\frac{x}{1+y^2}=\frac{tan^{−1}y}{1+y^2}\)

\(I.F.=e^{∫\frac{1}{1+y^2}dy}=e^{tan^{−1}y}\)

∴ Solution

\(x.e^{tan^{-1}y}∫\frac{e^{tan^{-1}}y^{tan^{-1}}y}{1+y^2}dy\)

Let

\(e^{tan^{−1}}y=t\)

\(\frac{e^{tan^{−1}y}}{1+y^2}=dt\)

=\(xe^{tan^{−1}y}∫ ln\; tdt =t \;ln t–t+c\)

=\(xe^{tan^{−1}y}=e^{tan^{−1}y}tan^{−1}y−e^{tan^{-1}y}+c…(i)\)

∵ It passes through \((1, 0) ⇒ c = 2\)

Now put \(y = tan1\), then

\(ex = e – e + 2\)

\(⇒x=\frac{2}{e}\)