x²y = -x²cos x + 2x sin x + 2 cos x + \(\frac { { x }^{ 3 } }{ 3 } logx-\frac { { x }^{ 3 } }{ 9 } +c\) is the required solution.

Given, differential equation is \(\left(1+y^2\right)+\left(x-e^{\tan ^{-1} y}\right) \frac{d y}{d x}=0\)
It can be rewritten as
\(\left(1+y^2\right) \frac{d x}{d y}+x-e^{\tan ^{-1} y}=0\)
or  \(\frac{d x}{d y}+\frac{1}{\left(1+y^2\right)} x=\frac{e^{\tan ^{-1} y}}{1+y^2}\)
                [dividing both sides by (1 + y²)]
It is a linear differential equation of the form \(\frac{d x}{d y}+P x=Q\)
Here, \(P=\frac{1}{1+y^2} \text { and } Q=\frac{e^{\tan ^{-1} y}}{1+y^2}\)
Now, integrating factor, IF = \(e^{\int P d y}\)
\(=e^{\int \frac{1}{1+y^2} d y}=e^{\tan ^{-1} y}\)
\(\therefore\) The general solution of linear differential equation is given by
\(\begin{aligned} x \times \mathrm{IF} & =\int(Q \times \mathrm{IF}) d y+C \end{aligned}\)
\(\begin{aligned} \Rightarrow x \times e^{\tan ^{-1} y} & =\int \frac{e^{\tan ^{-1} y}}{1+y^2} \times e^{\tan ^{-1} y} d y+C \end{aligned}\)
\(\Rightarrow \quad x e^{\tan ^{-1} y}=\int \frac{e^{2 \tan ^{-1} y}}{1+y^2} d y+C\)    ....(i)
On putting tan^-1 y = t \(\Rightarrow \frac{1}{1+y^2} d y=d t \text { in }\)
Eq. (i), we get
\(\begin{aligned} x e^{\tan ^{-1} y} & =\int e^{2 t} d t+C \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad x e^{\tan ^{-1} y} & =\frac{e^{2 t}}{2}+C \end{aligned}\)
\(\Rightarrow \quad x e^{\tan ^{-1} y}=\frac{e^{2 \tan ^{-1} y}}{2}+C \quad\left[\because t=\tan ^{-1} y\right]\)

Equation of system of concentric circles with centre
\((1,2) \text { is }(x-1)^{2}+(y-2)^{2}=a^{2}\) 
where a is radius of circle
\(=(x-1)+(y-2) \frac{d y}{d x}=0\)

\(\frac{d y}{d x}=y+\frac{y}{x}\) 
\(y=K x e^{x}\)

The figure obtained by the given information is given below
Let the coordinate of the point P be (x, y). It is given that P is mid-point of AB. So, the coordinates of points A and are (2x, 0) and (0, 2y), respectively.
Now, slope of \(A B=\frac{0-2 y}{2 x-0}=-\frac{y}{x}\) 
Since, the segment AB is a tangent to the curve at P.
\(\therefore \frac{d y}{d x}=-\frac{y}{x} \Rightarrow \frac{d y}{y}=-\frac{d x}{x}\) 
On integrating both sides, we get
\( \log |y|=-\log |x|+\log |C| \)
\(\Rightarrow \log |y x|=\log |C| \)
\(\Rightarrow y x=C\) 
Since, the given curve passes through the point (1, 1).
\(\therefore \ 1 \cdot 1=C \Rightarrow C=1\) 
On putting C = 1in Eq. (i), we get
xy = 1