The solution of the differential equation $$\frac{{{{\text{d}}^2}{\text{y}}}}{{{\text{d}}{{\text{x}}^2}}} = 0$$ with boundary conditions $${\text{i}}{\text{.}}\,\frac{{{\text{dy}}}}{{{\text{dx}}}} = 1{\text{ at x}} = 0;\,{\text{ii}}{\text{.}}\,\frac{{{\text{dy}}}}{{{\text{dx}}}} = 1{\text{ at x}} = 1{\text{ is}}$$

The figure shows the plot of y as a function of x

The function shown is the solution of the differential equation (assuming all initial conditions to be zero) is

A

$$\frac{{{{\text{d}}^2}{\text{y}}}}{{{\text{d}}{{\text{x}}^2}}} = 1$$

B

$$\frac{{{\text{dy}}}}{{{\text{dx}}}} = {\text{x}}$$

C

$$\frac{{{\text{dy}}}}{{{\text{dx}}}} = - {\text{x}}$$

D

$$\frac{{{\text{dy}}}}{{{\text{dx}}}} = \left| {\text{x}} \right|$$

Singular solution of a differential equation is one that cannot be obtained from the general solution gotten by the usual method of solving the differential equation.

A

True

B

False

Consider the following second-order differential equation: y" - 4y' + 3y = 2t - 3t²
The particular solution of the differential equation is

A

-2 - 2t - t2

B

-2t - t2

C

2t - t2

D

-2 - 2t - 3t2

Consider the differential equation $$\left( {{{\text{t}}^2} - 81} \right)\frac{{{\text{dy}}}}{{{\text{dt}}}} + 5{\text{ty}} = \sin \left( {\text{t}} \right)$$ with y(1) = 2π. Thereexists a unique solution for this differential equation when t belongs to the interval

A

(-2, 2)

B

(-10, 10)

C

(-10, 2)

D

(0, 10)

A differential equation is given as
$${{\text{x}}^2}\frac{{{{\text{d}}^2}{\text{y}}}}{{{\text{d}}{{\text{x}}^2}}} - 2{\text{x}}\frac{{{\text{dy}}}}{{{\text{dx}}}} + 2{\text{y}} = 4$$
The solution of differential equation in terms of arbitrary constant C₁ and C₂ is

A

$${\text{y}} = \frac{{{{\text{C}}_1}}}{{{{\text{x}}^2}}} + {{\text{C}}_2}{\text{x}} + 2$$

B

$${\text{y}} = {{\text{C}}_1}{{\text{x}}^2} + {{\text{C}}_2}{\text{x}} + 4$$

C

$${\text{y}} = {{\text{C}}_1}{{\text{x}}^2} + {{\text{C}}_2}{\text{x}} + 2$$

D

$${\text{y}} = \frac{{{{\text{C}}_1}}}{{{{\text{x}}^2}}} + {{\text{C}}_2}{\text{x}} + 2$$

Consider the differential equation $$\frac{{{\text{dy}}}}{{{\text{dx}}}} = 1 + {{\text{y}}^2}.$$
Which one of the following can be a particular solution of this differential equation?

A

y = tan(x + 3)

B

y = tan x + 3

C

x = tan(y + 3)

D

x = tan y + 3

The differential equation $$\frac{{{\text{dy}}}}{{{\text{dx}}}} + 4{\text{y}} = 5$$ is valid in the domain 0 ≤ x ≤ 1 with y(0) = 2.25. The solution of differential equation is

A

y = e-4x + 1.25

B

y = e-4x + 5

C

y = e4x + 5

D

y = e4x + 1.25

Consider the differential equation $${{\text{x}}^2}\frac{{{{\text{d}}^2}{\text{y}}}}{{{\text{d}}{{\text{x}}^2}}} + {\text{x}}\frac{{{\text{dy}}}}{{{\text{dx}}}} - {\text{y}} = 0.$$ Which of the following is a solution to this differential equation for x > 0 ?

A

ex

B

x2

C

$$\frac{1}{{\text{x}}}$$

D

$$l$$n x

A function n(x) satisfies the differential equation $$\frac{{{{\text{d}}^2}{\text{n}}\left( {\text{x}} \right)}}{{{\text{d}}{{\text{x}}^2}}} - \frac{{{\text{n}}\left( {\text{x}} \right)}}{{{{\text{L}}^2}}} = 0$$ where L is a constant. The boundary conditions are: n(0) = K and n($$\infty $$) = 0. The solution to this equation is

A

n(x) = K exp(x/L)

B

n(x)= K exp(-x/√L)

C

n(x) = K2 exp(-x/L)

D

n(x) = K exp(-x/L)

The nature of normal grain growth in the presence of a second phase deserves special consideration. The moving boundaries will be attached to the particles exert a pulling force on the boundary restricting its motion. Therefore if the boundary intersects the particle surface at 90° the particle will feel a pull of (2πrγcosΘ)*sinΘ. This will be counterbalanced by an equal and opposite force acting on the boundary. As the boundary moves over the particle surface Θ changes and the drag reaches a maximum value, this happens when Θ becomes______

A

90

B

45

C

60

D

30