The solution of the differential equation

$\left(1 + {\mathrm{y}}^2\right) + \left(\mathrm{x} - {\mathrm{e}}^{{\tan }^{-1}\left(\mathrm{y}\right)}\right)\frac{d\mathrm{y}}{d\mathrm{x}} = 0$, is

• $\left(\mathrm{x} - 2\right) = {\mathrm{Ke}}^{{\tan }^{-1}\left(\mathrm{y}\right)}$

• $2{\mathrm{xe}}^{{\tan }^{-1}\left(\mathrm{y}\right)} = {\mathrm{e}}^{2{\tan }^{-1}\left(\mathrm{y}\right)} + \mathrm{K}$

• ${\mathrm{xe}}^{{\tan }^{-1}\left(\mathrm{y}\right)} = {\tan }^{-1}\left(\mathrm{y}\right) +\hspace{0.17em}\mathrm{K}$

• ${\mathrm{xe}}^{2{\tan }^{-1}\left(\mathrm{y}\right)} = {\mathrm{e}}^{{\tan }^{-1}\left(\mathrm{y}\right)} + \mathrm{K}$

The solution of the differential equation $\frac{d\mathrm{y}}{d\mathrm{x}} = \frac{\mathrm{yf}\text{'}\left(\mathrm{x}\right) - {\mathrm{y}}^2}{\mathrm{f}\left(\mathrm{x}\right)}$ is

• f(x) = y + C

• f(x) = y(x + C)

• f(x) = x + C

• None of the above

The general solution of the differential equation (1 + y²) dx + (1 + x)dy = 0 is

• x - y = C(1 - xy)

• x - y = C(1 + xy)

• x + y = C(1 - xy)

• x + y = C(1 + xy)

84.

The order and degree of the differential equation $\mathrm{\rho } = \frac{{\left[1 + {\left({\displaystyle \frac{\mathrm{dy}}{\mathrm{dx}}}\right)}^2\right]}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}{\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}}$ are, respectively

• 2, 2

• 2, 3

• 2, 1

• None of these

The solution of the differential equation

$\frac{\mathrm{dy}}{\mathrm{dx}} + \frac{2\mathrm{yx}}{1 +\hspace{0.17em}{\mathrm{x}}^2} = \frac{1}{{\displaystyle {\left(1 +\hspace{0.17em}{\mathrm{x}}^2\right)}^2}}$ is

• $\mathrm{y}\left(1 + {\mathrm{x}}^2\right) = \mathrm{C} + {\tan }^{-1}\left(\mathrm{x}\right)$

• $\frac{\mathrm{y}}{1 +\hspace{0.17em}{\mathrm{x}}^2} = \mathrm{C} + {\tan }^{-1}\left(\mathrm{x}\right)$

• $\mathrm{y} \log \left(1 + {\mathrm{x}}^2\right) = \mathrm{C} + {\tan }^{-1}\left(\mathrm{x}\right)$

• y(1 + x²) = C + ${\sin }^{-1}\left(\mathrm{x}\right)$

The order and degree of the differential equation ${\left(1 + 4\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.} = 4\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}$ are respectively

• $1, \frac{2}{3}$

• 3, 2

• 2, 3

• $2, \frac{2}{3}$

The solution of the differential equation $\frac{\mathrm{dy}}{\mathrm{dx}} = {\left(4\mathrm{x} + \mathrm{y} + 1\right)}^2$, is

• (4x + y + 1) = tan (2x + C)

• (4x + y + 1)² = 2 tan (2x + C)

• (4x + y + 1)³ = 3 tan (2x + C)

• (4x + y + 1) = 2 tan (2x + C)

If f(x), g(x) and h(x) are three polynomials of degree 2 and $∆\left(\mathrm{x}\right) = \left|\begin{array}{ccc}\mathrm{f}\left(\mathrm{x}\right)& \mathrm{g}\left(\mathrm{x}\right)& \mathrm{h}\left(\mathrm{x}\right)\\ \mathrm{f}\text{'}\left(\mathrm{x}\right)& \mathrm{g}\text{'}\left(\mathrm{x}\right)& \mathrm{h}\text{'}\left(\mathrm{x}\right)\\ \mathrm{f}\text{'}\text{'}\left(\mathrm{x}\right)& \mathrm{g}\text{'}\text{'}\left(\mathrm{x}\right)& \mathrm{h}\text{'}\text{'}\left(\mathrm{x}\right)\end{array}\right|$, then $∆\left(\mathrm{x}\right)$ is a polynomials of degree

• 2

• 3

• 0

• atmost 3

The degree of the differential equation of all curves having normal of constant length c is

• 1

• 3

• 4

• None of these

$\frac{\mathrm{d}}{\mathrm{dx}}\left[{\mathrm{atan}}^{-1}\left(\mathrm{x}\right) + \mathrm{blog}\left(\frac{\mathrm{x} - 1}{\mathrm{x} + 1}\right)\right] = \frac{1}{{\mathrm{x}}^4 - 1}$ $\Rightarrow \mathrm{a} - 2\mathrm{b}$ is equal to

• 1

• - 1

• 0

• 2