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Differential Equations

Multiple Choice Questions

361.

The solution of the differential equation $\frac{dy}{dx} = \sin (x + y) \tan (x + y) - 1 is$

$\csc (x + y) + \tan (x + y) = x + c$
$x + \csc (x + y) = c$
$x + \tan (x + y) = c$
$x + \sec (x + y) = c$

362.

$The differential equation of the family y = {ae}^{x} + {bxe}^{x} + {cx}^{2} e^{x} of curves, where a, b, c are arbitrary constants, is$

$y''' + 3 y'' + 3' + y = 0$
$y''' + 3 y'' - 3' - y = 0$
$y''' - 3 y'' - 3' + y = 0$
$y''' - 3 y'' + 3' - y = 0$

363.

$The solution of \tan (y) \frac{dy}{dx} = \sin (x + y) + \sin (x - y) is$

sec(y) = 2cos(x) + c
sec(y) = - 2cos(x) + c
tan(y) = - 2cos(x) + c
sec²(y) = - 2cos(x) + c

364.

A family of curves has the differential equation $xy \frac{dy}{dx} = 2 y^{2} - x^{2}$ . Then, the family of curves is

$y^{2} = {cx}^{2} + x^{3}$
$y^{2} = {cx}^{4} + x^{3}$
$y^{2} = x + {cx}^{4}$
$y^{2} = x^{2} + {cx}^{4}$

365.

$The solution of the differential equation \frac{dy}{dx} = \frac{y}{x} + \frac{ϕ (\frac{y}{x})}{ϕ' (\frac{y}{x})} is$

$xϕ (\frac{y}{x}) = k$
$ϕ (\frac{y}{x}) = kx$
$yϕ (\frac{y}{x}) = k$
$ϕ (\frac{y}{x}) = ky$

366.

If y = y(x) is the solution of the differential equation $(\frac{2 + \sin (x)}{y + 1}) \frac{dy}{dx} + \cos (x) = 0$ then y( $\frac{π}{2}$ )is equal to

$\frac{1}{3}$
$\frac{2}{3}$
1
$\frac{4}{3}$

367.
$If u = f (r), where r^{2} = x^{2} + y^{2}, then (\frac{\partial u}{\partial x^{2}} + \frac{\partial^{2} u}{\partial y^{2}}) = ?$

f''(r)

f''(r) +f'(r)

f''(r) + $\frac{1}{r}$ f'(r)

f''(r) + rf'(r)

f''(r) + $\frac{1}{r}$ f'(r)

$Given, u = f (r) and r^{2} = x^{2} + y^{2} ∴ u_{x} = f' (r) \frac{δr}{δx} = f' (r) \frac{x}{r} \Rightarrow u_{xx} = f'' (r) \times \frac{x^{2}}{r^{2}} + f' (r) \frac{r^{2} - x \times x}{r^{2}} = f'' (r) \times \frac{x^{2}}{r^{2}} + f' (r) \frac{r^{2} - x^{2}}{r^{3}} u_{y} = f' (r) \frac{δr}{δy} = f' (r) \frac{2 y}{2 \sqrt{x^{2} + y^{2}}} = f' (r) (\frac{y}{r}) and u_{yy} = f'' (r) \frac{y^{2}}{r^{2}} + f' (r) \frac{r^{2} - y^{2}}{r^{3}} ∴ u_{xx} + u_{yy} = f'' (r) (\frac{x^{2}}{r^{2}} + \frac{y^{2}}{r^{2}}) + f' (r) (\frac{r^{2} - x^{2}}{r^{3}} + \frac{r^{2} - y^{2}}{r^{3}}) = f'' (r) (\frac{r^{2}}{r^{2}}) + f' (r) \frac{2 r^{2} - (x^{2} + y^{2})}{r^{3}} = f'' (r) + f' (r) \frac{r^{2}}{r^{3}} = f'' (r) + \frac{f' (r)}{r}$