While solving the ordinary differential equation using unilateral laplace transform, we consider the initial conditions of the system.

While solving an Ordinary Differential Equation using the unilateral Laplace Transform, it is possible to solve if there is no function in the right hand side of the equation in standard form and if the initial conditions are zero.

A

True

B

False

Laplace transform of the function f(t) is given by $${\text{F}}\left( {\text{s}} \right) = {\text{L}}\left\{ {{\text{f}}\left( {\text{t}} \right)} \right\} = \int_0^\infty {{\text{f}}\left( {\text{t}} \right){{\text{e}}^{ - {\text{st}}}}{\text{dt}}{\text{.}}} $$ Laplace transform of the function shown below is given by
Transform Theory mcq question image

A

$$\frac{{1 - {{\text{e}}^{ - 2{\text{s}}}}}}{{\text{s}}}$$

B

$$\frac{{1 - {{\text{e}}^{ - {\text{s}}}}}}{{2{\text{s}}}}$$

C

$$\frac{{2 - 2{{\text{e}}^{ - {\text{s}}}}}}{{\text{s}}}$$

D

$$\frac{{1 - 2{{\text{e}}^{ - {\text{s}}}}}}{{\text{s}}}$$

The unilateral Laplace transform of f(t) is $${1 \over {{s^2} + s + 1}}.$$ Which one of the following is the unilateral Laplace transform of g(t) = t.f(t)?

A

$${{ - s} \over {{{\left( {{s^2} + s + 1} \right)}^2}}}$$

B

$${{ - \left( {2s + 1} \right)} \over {{{\left( {{s^2} + s + 1} \right)}^2}}}$$

C

$${s \over {{{\left( {{s^2} + s + 1} \right)}^2}}}$$

D

$${{2s + 1} \over {{{\left( {{s^2} + s + 1} \right)}^2}}}$$

In which of the following useful signals, is the bilateral Laplace Transform different from the unilateral Laplace Transform?

A

d(t)

B

s(t)

C

u(t)

D

all of the mentioned

Laplace transform function f (t) is F(S), then how will you represent Laplace transform for differential of f (t)?

A

S⁄F(S)

B

F(S)

C

S.F(S)

D

F’(S)

The figure shows the plot of y as a function of x
Differential Equations mcq question image

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

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

Function f (t) has Laplace transform F(S). How will you represent Laplace transform of integral of f(t)?

A

S.F(S)

B

F(S)

C

F(S2)

D

F(S)⁄S

If the Laplace transform of $${{\text{e}}^{\omega {\text{t}}}}$$ is $$\frac{1}{{{\text{s}} - \omega }},$$ the Laplace transform of tcosh t is

A

$$\frac{{1 + {{\text{s}}^2}}}{{{{\left( {{{\text{s}}^2} - 1} \right)}^2}}}$$

B

$$\frac{{{\text{st}}}}{{\left( {{{\text{s}}^2} - 1} \right)}}$$

C

$$\frac{{1 - {{\text{s}}^2}}}{{{{\left( {{{\text{s}}^2} - 1} \right)}^2}}}$$

D

$$\frac{{1 + {{\text{s}}^2}}}{{1 - {{\text{s}}^2}}}$$

Laplace transform of cos (ωt) is $$\frac{{\text{s}}}{{{{\text{s}}^2} + {\omega ^2}}}.$$ The Laplace transform of e^-2t cos(4t) is

A

$$\frac{{{\text{s}} - 2}}{{{{\left( {{\text{s}} - 2} \right)}^2} + 16}}$$

B

$$\frac{{{\text{s}} + 2}}{{{{\left( {{\text{s}} - 2} \right)}^2} + 16}}$$

C

$$\frac{{{\text{s}} - 2}}{{{{\left( {{\text{s}} + 2} \right)}^2} + 16}}$$

D

$$\frac{{{\text{s}} + 2}}{{{{\left( {{\text{s}} + 2} \right)}^2} + 16}}$$

The Laplace transform of a function f(t) u(t), where f(t) is periodic with period T, is A(s) times the Laplace transform of its first period. Then

A

A(s) = s

B

$$A\left( s \right) = {1 \over {\left( {1 - \exp \left( { - {T_s}} \right)} \right)}}$$

C

$$A\left( s \right) = {1 \over {\left( {1 + \exp \left( { - {T_s}} \right)} \right)}}$$

D

A(s) = exp(T_s)

While solving the ordinary differential equation using unilateral laplace transform, we consider the initial conditions of the system.

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