91.

The solution of the differential equation $\frac{d\mathrm{y}}{d\mathrm{x}} = \sin \left(\mathrm{x} + \mathrm{y}\right)\tan \left(\mathrm{x} +\hspace{0.17em}\mathrm{y}\right) - 1$ is

• $\csc \left(\mathrm{x} +\hspace{0.17em}\mathrm{y}\right) + \tan \left(\mathrm{x} + \mathrm{y}\right) = \mathrm{x} +\hspace{0.17em}\mathrm{c}$

• $\mathrm{x} + \csc \left(\mathrm{x} + \mathrm{y}\right) = \mathrm{c}$

• $\mathrm{x} + \tan \left(\mathrm{x} + \mathrm{y}\right) = \mathrm{c}$

• $\mathrm{x} + \sec \left(\mathrm{x} + \mathrm{y}\right) = \mathrm{c}$

The differential equation of all straight lines touching the circle x² + y² = a² is

• ${\left(\mathrm{y} - \frac{\mathrm{dy}}{\mathrm{dx}}\right)}^2 = {\mathrm{a}}^2\left[1 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^2\right]$

• ${\left(\mathrm{y} - \mathrm{x}\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^2 = {\mathrm{a}}^2\left[1 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^2\right]$

• $\left(\mathrm{y} - \mathrm{x}\frac{\mathrm{dy}}{\mathrm{dx}}\right) = {\mathrm{a}}^2\left[1 + \frac{\mathrm{dy}}{\mathrm{dx}}\right]$

• $\left(\mathrm{y} - \frac{\mathrm{dy}}{\mathrm{dx}}\right) = {\mathrm{a}}^2\left[1 - \frac{\mathrm{dy}}{\mathrm{dx}}\right]$

The differential equation $\left|\frac{\mathrm{dy}}{\mathrm{dx}}\right| + \left|\mathrm{y}\right| + 3 = 0$ admits

• infinite number of solutions

• no solutions

• a unique solution

• many solutions

Solution of the differential equation $\mathrm{xdy} - \mathrm{ydx} - \sqrt{{\mathrm{x}}^2 + {\mathrm{y}}^2}\mathrm{dx} = 0$

• $\mathrm{y} - \sqrt{{\mathrm{x}}^2 + {\mathrm{y}}^2} = {\mathrm{cx}}^2$

• $\mathrm{y} + \sqrt{{\mathrm{x}}^2 + {\mathrm{y}}^2} = {\mathrm{cx}}^2$

• $\mathrm{y} + \sqrt{{\mathrm{x}}^2 + {\mathrm{y}}^2} = {\mathrm{cy}}^2$

• $\mathrm{x} - \sqrt{{\mathrm{x}}^2 + {\mathrm{y}}^2} = {\mathrm{cy}}^2$

If $\mathrm{xsin}\left(\frac{\mathrm{x}}{\mathrm{y}}\right)\mathrm{dy} = \left[\mathrm{ysin}\left(\frac{\mathrm{y}}{\mathrm{x}}\right) - \mathrm{x}\right]\mathrm{dx}$ and $\mathrm{y}\left(1\right) = \frac{\mathrm{\pi }}{2}$, then the value of $\cos \left(\frac{\mathrm{y}}{\mathrm{x}}\right)$ is equal to :

• x

• $\frac{1}{\mathrm{x}}$

• $\log \left(\mathrm{x}\right)$

• e^x

The differential equation of the system of all circles of radius r in the xy plane is :

• ${\left[1 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^3\right]}^2 = {\mathrm{r}}^2{\left(\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}\right)}^2$

• ${\left[1 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^3\right]}^2 = {\mathrm{r}}^2{\left(\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}\right)}^3$

• ${\left[1 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^2\right]}^3 = {\mathrm{r}}^2{\left(\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}\right)}^2$

• ${\left[1 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^2\right]}^3 = {\mathrm{r}}^2{\left(\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}\right)}^3$

The general solution of the differential equation

$\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2} + 2\frac{d\mathrm{y}}{d\mathrm{x}} + \mathrm{y} = 2{\mathrm{e}}^{3\mathrm{x}}$ is given by

• $\mathrm{y} = \left({\mathrm{c}}_1 + {\mathrm{c}}_2\mathrm{x}\right){\mathrm{e}}^{\mathrm{x}} + \frac{{\mathrm{e}}^{3\mathrm{x}}}{8}$

• $\mathrm{y} = \left({\mathrm{c}}_1 + {\mathrm{c}}_2\mathrm{x}\right){\mathrm{e}}^{- \mathrm{x}} + \frac{{\mathrm{e}}^{- 3\mathrm{x}}}{8}$

• $\mathrm{y} = \left({\mathrm{c}}_1 + {\mathrm{c}}_2\mathrm{x}\right){\mathrm{e}}^{- \mathrm{x}} + \frac{{\mathrm{e}}^{3\mathrm{x}}}{8}$

• $\mathrm{y} = \left({\mathrm{c}}_1 + {\mathrm{c}}_2\mathrm{x}\right){\mathrm{e}}^{\mathrm{x}} + \frac{{\mathrm{e}}^{- 3\mathrm{x}}}{8}$

The solution of the differential ydx + (x - y³)dy = 0 is:

• xy = $\frac{1}{3}{\mathrm{y}}^3 + \mathrm{c}$

• xy = y⁴ + c

• y⁴ = 4xy + c

• 4y = y³ + c

If the distances covered by a particle in time t is proportional to the cube root of its velocity, then the acceleration is

• a constant

• $\propto {\mathrm{s}}^3$

• $\propto \frac{1}{{\mathrm{s}}^3}$

• $\propto {\mathrm{s}}^5$

The solution of the differential equation $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{y}}{\mathrm{x}} + \frac{\mathrm{\varphi }\left({\displaystyle \frac{\mathrm{y}}{\mathrm{x}}}\right)}{\mathrm{\varphi }\text{'}\left(\frac{\mathrm{y}}{\mathrm{x}}\right)}$ is

• $\mathrm{x\varphi }\left(\frac{\mathrm{y}}{\mathrm{x}}\right) = \mathrm{k}$

• $\mathrm{\varphi }\left(\frac{\mathrm{y}}{\mathrm{x}}\right) = \mathrm{kx}$

• $\mathrm{y\varphi }\left(\frac{\mathrm{y}}{\mathrm{x}}\right) = \mathrm{k}$

• $\mathrm{\varphi }\left(\frac{\mathrm{y}}{\mathrm{x}}\right) = \mathrm{ky}$