(2)

Let (k, k) be the centre of circle, so equation of circle with radius r is given by

${(x - k)}^2 + {(y - k)}^2 = r^2$

Now, circle is passing through origin

$\Rightarrow r^2 = 2k^2$

So, equation of circle becomes

$x^2 + y^2 -2kx - 2ky = 0$         ... (i)

On differentiating (i) w.r.t. x, we get

2x + 2yy' – 2k – 2ky' = 0

$\Rightarrow k = \frac{x + yy\text{'}}{(1 + y\text{'})}$

Substituting the value of k in (i) we get

$x^2 + y^2 - 2x\left(\frac{x + yy\text{'}}{1 + y\text{'}}\right) - 2y\left(\frac{x + yy\text{'}}{1 + y\text{'}}\right) = 0$

$\Rightarrow (x^2 + y^2)(1 + y\text{'}) - 2x^2 - 2xyy\text{'} - 2xy - 2y^{2}y\text{'} = 0$

$\Rightarrow -x^2 + y^2 + x^{2}y\text{'} - y^{2}y\text{'} - 2xyy\text{'} - 2xy = 0$

$\Rightarrow (x^2 - y^2 - 2xy)y\text{'} =x^2 - y^2 + 2xy$

$\Rightarrow (x^2 - y^2 - 2xy)dy = (x^2 - y^2 + 2xy)dx$

is the required differential equation.

(3)

Given ${\int }_0^{a}f\left(x\right)dx = e^{-a}+4a^2+a-1$

On differentiating both sides, we get $f\left(a\right) = -e^{-a}+ 8a + 1$

i.e., $f\left(x\right) = -e^{-x}+ 8x + 1$

Now, $y = c_{1}f\left(x\right) + c_2$

$\Rightarrow y\text{'} = c_{1}f\text{'}\left(x\right) = c_1(8 + e^{-x})$       ... (i)

Again, on differentiating, we get $c_1 = - e^{x}y\text{'}\text{'}$    ... (ii)

From (i) and (ii), we get

$y\text{'} = y\text{'}\text{'}{e}^x(-8 - e^{-x}) \Rightarrow (8e^x + 1)y\text{'}\text{'} +y\text{'} =0$