Mathematics : Differential Equations

An integrating factor of the differential equation, (1 + y + x²y)dx + (x + x³)dy = 0 is :

• $\log \left(\mathrm{x}\right)$

• x

• e^x

• $\frac{1}{\mathrm{x}}$

Let y = y(x) be the solution of the differential equation, $\left({\mathrm{x}}^2 + 1\right)\frac{\mathrm{dy}}{\mathrm{dx}} + 2\mathrm{x}\left({\mathrm{x}}^2 + 1\right)\mathrm{y} = 1$ such that y(0) = 0. If $\sqrt{\mathrm{a}}\mathrm{y}\left(1\right) = \frac{\mathrm{\pi }}{32}$, then the value of 'a' is :

• $\frac{1}{4}$

• 1

• $\frac{1}{16}$

• $\frac{1}{2}$

The solution of the differential equation $\mathrm{x}\frac{\mathrm{dy}}{\mathrm{dx}} + 2\mathrm{y} = {\mathrm{x}}^2, (\mathrm{x} \ne 0)$ with y(1)  = 1, is :

• $\mathrm{y} = \frac{4}{5}{\mathrm{x}}^3 + \frac{1}{5{\mathrm{x}}^2}$

• $\mathrm{y} = \frac{3}{4}{\mathrm{x}}^3 + \frac{1}{4{\mathrm{x}}^2}$

• $\mathrm{y} = \frac{{\mathrm{x}}^2}{4} + \frac{3}{4{\mathrm{x}}^2}$

• $\mathrm{y} = \frac{{\mathrm{x}}^3}{5} + \frac{1}{5{\mathrm{x}}^2}$

If y = y(x) is the solution of the differential equation $\frac{\mathrm{dy}}{\mathrm{dx}} = \left(\tan \left(\mathrm{x}\right) - \mathrm{y}\right){\sec }^2\left(\mathrm{x}\right), \mathrm{x} \in \left(- \frac{\mathrm{\pi }}{2}, \frac{{\displaystyle \mathrm{\pi }}}{{\displaystyle 2}}\right)$, such that y(0) = 0, then $\mathrm{y}\left(- \frac{\mathrm{\pi }}{4}\right)$ is equal to

• $2 + \frac{1}{\mathrm{e}}$

• e - 2

• $\frac{1}{2} - \mathrm{e}$

• $\frac{1}{\mathrm{e}} - 2$

Let y = y(x) be the solution of the differential equation $\frac{\mathrm{dy}}{\mathrm{dx}} + \mathrm{ytan}\left(\mathrm{x}\right) = 2\mathrm{x} +\hspace{0.17em}{\mathrm{x}}^2\tan \left(\mathrm{x}\right), \mathrm{x} \in \left(- \frac{\mathrm{\pi }}{2}, \frac{{\displaystyle \mathrm{\pi }}}{{\displaystyle 2}}\right)$ such that if y(0) = 1, then

• $\mathrm{y}\text{'}\left(\frac{\mathrm{\pi }}{4}\right) - \mathrm{y}\text{'}\left(- \frac{\mathrm{\pi }}{4}\right) = \mathrm{\pi } - \sqrt{2}$

• $\mathrm{y}\left(\frac{\mathrm{\pi }}{4}\right) - \mathrm{y}\left(- \frac{\mathrm{\pi }}{4}\right) = \sqrt{2}$

• $\mathrm{y}\left(\frac{\mathrm{\pi }}{4}\right) + \mathrm{y}\left(- \frac{\mathrm{\pi }}{4}\right) = \frac{{\mathrm{\pi }}^2}{2} + 2$

• $\mathrm{y}\text{'}\left(\frac{\mathrm{\pi }}{4}\right) + \mathrm{y}\text{'}\left(- \frac{\mathrm{\pi }}{4}\right) = - \sqrt{2}$

The solution of the differential equation $\frac{\mathrm{dx}}{\mathrm{x}} + \frac{\mathrm{dy}}{\mathrm{y}} = 0$ is

• xy = c

• x + y = c

• log(x)log(y) = c

• x² + y² = c

The differential equation obtained by eliminating arbitrary constants from y = a . e^bx, is

• $\mathrm{y}\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2} + \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

• $\mathrm{y}\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2} - \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

• $\mathrm{y}\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2} - {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^2 = 0$

• $\mathrm{y}\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2} + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^2 = 0$

f(x) is a polynomial of degree 2, f(0) = 4, f'(0) = 3 and f''(0) = 4, then f(- 1) is equal to

• 3

• - 2

• 2

• - 3

Solution of differential equation sec(x)dy - cosec(y)dx = 0 is

• cos(x) + sin(y) = c

• sin(x) + cos(y) = 0

• sin(y) - cos(x) = c

• cos(y) - sin(x) = c

The point P(9/2, 6) lies on the parabola y² = 4ax, then parameter of the point P is

• $\frac{3\mathrm{a}}{2}$

• $\frac{2}{3\mathrm{a}}$

• $\frac{2}{3}$

• $\frac{3}{2}$