The solution of the equation $\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2} = {\mathrm{e}}^{- 2\mathrm{x}}$ is

• $\mathrm{y} = \frac{1}{4}{\mathrm{e}}^{- 2\mathrm{x}} + \frac{\mathrm{cx}}{2} +\hspace{0.17em}\mathrm{d}$

• $\mathrm{y} = \frac{1}{4}{\mathrm{e}}^{- 2\mathrm{x}} + \mathrm{cx} +\hspace{0.17em}\mathrm{d}$

• $\mathrm{y} = \frac{1}{4}{\mathrm{e}}^{- 2\mathrm{x}} + {\mathrm{cx}}^2 +\hspace{0.17em}\mathrm{d}$

• $\mathrm{y} = \frac{1}{4}{\mathrm{e}}^{- 2\mathrm{x}} + {\mathrm{cx}}^3 +\hspace{0.17em}\mathrm{d}$

The solution of the differential equation $\frac{\mathrm{dy}}{\mathrm{dx}} = \mathrm{y} \tan \left(\mathrm{x}\right) - 2 \sin \left(\mathrm{x}\right)$, is

• $\mathrm{y} \sin \left(\mathrm{x}\right) = \mathrm{c} + \sin \left(2\mathrm{x}\right)$

• $\mathrm{y} \cos \left(\mathrm{x}\right) = \mathrm{c} + \frac{1}{2}\sin \left(2\mathrm{x}\right)$

• $\mathrm{y} \cos \left(\mathrm{x}\right) = \mathrm{c} - \sin \left(2\mathrm{x}\right)$

• $\mathrm{y} \cos \left(\mathrm{x}\right) = \mathrm{c} + \frac{1}{2}\cos \left(2\mathrm{x}\right)$

The differential equation of system of concentric circles with centre (1, 2) is :

• $\left(\mathrm{x} - 2\right) + \left(\mathrm{y} - 1\right)\frac{\mathrm{dy}}{\mathrm{dx}} = 0$

• $\left(\mathrm{x} - 1\right) + \left(\mathrm{y} - 2\right)\frac{\mathrm{dy}}{\mathrm{dx}} = 0$

• $\left(\mathrm{x} + 1\right)\frac{\mathrm{dy}}{\mathrm{dx}} + \left(\mathrm{y} - 2\right) = 0$

• $\left(\mathrm{x} + 2\right)\frac{\mathrm{dy}}{\mathrm{dx}} + \left(\mathrm{y} - 1\right) = 0$

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 :

• y(1 + x²) = c + tan^-1(x)

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

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

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

The solution of the differential equation $\mathrm{xdy} - \mathrm{ydx} = \sqrt{{\mathrm{x}}^2 + {\mathrm{y}}^2}\mathrm{dx}$ is :

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

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

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

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

The solution of the differential equation $\frac{\mathrm{dy}}{\mathrm{dx}} = {\mathrm{e}}^{\mathrm{x} - \mathrm{y}} + {\mathrm{x}}^2{\mathrm{e}}^{- \mathrm{y}}$ is :

• y = e^{x - y} + x²e^{- y} + c

• ${\mathrm{e}}^{\mathrm{y}} - {\mathrm{e}}^{\mathrm{x}} = \frac{1}{3}{\mathrm{x}}^3 + \mathrm{c}$

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

• ${\mathrm{e}}^x - {\mathrm{e}}^y = \frac{1}{3}{\mathrm{x}}^3 + \mathrm{c}$

577.

The integrating factor of the differential equation $\frac{\mathrm{dy}}{\mathrm{dx}} + \frac{1}{\mathrm{x}}\mathrm{y} = 3\mathrm{x}$ is :

• x

• in x

• 0

• $\infty$

The solution of the differential equation sec²(x)tan(y))dx + sec²(y)tan(x))dy = 0 is :

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

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

• $\frac{{\tan }^2\left(\mathrm{x}\right) }{\tan \left(\mathrm{y}\right)}= \mathrm{c}$

• None of these

The differential equation of all straight lines passing through origin is :

• $\mathrm{y} = \sqrt{\mathrm{x}}\frac{\mathrm{dy}}{\mathrm{dx}}$

• $\frac{\mathrm{dy}}{\mathrm{dx}} = \mathrm{y} + \mathrm{x}$

• $\frac{\mathrm{dy}}{\mathrm{dx}} = \mathrm{y} - \mathrm{x}$

• None of these

To reduce the differential equation $\frac{\mathrm{dy}}{\mathrm{dx}} = \mathrm{Py} = \mathrm{Q}\left(\mathrm{x}\right) . {\mathrm{y}}^{\mathrm{n}}$ to the linear form, the substitution is :

• $\mathrm{v} = \frac{1}{{\mathrm{y}}^{\mathrm{n}}}$

• $\mathrm{v} = \frac{1}{{\mathrm{y}}^{\mathrm{n} - 1}}$

• v = yⁿ

• v = y^{n - 1}