Basic Laplace Transform Test Questions

This test corresponds to a set of 10 questions with multiple choices about the Laplace transform and its inverse.

Question 1:

The Laplace Transform of f(t)=\sin(2t) is

1) F(s)=\dfrac{1}{s^2+2}

2) F(s)=\dfrac{2}{s^2+2}

3) F(s)=\dfrac{1}{s^2+4}

4) F(s)=\dfrac{2}{s^2+4}

5) F(s)=\dfrac{2}{s^2+1}

Question 2:

If f(t)=t^3 e^{4t}, then \mathscr{L}\{f(t)\} is

1) F(s)=\dfrac{3}{(s-4)^{3}}

2) F(s)=\dfrac{6}{(s-4)^{3}}

3) F(s)=\dfrac{6}{(s-4)^{4}}

4) F(s)=\dfrac{6}{(s-3)^{3}}

5) F(s)=\dfrac{6}{(s-3)^{4}}

Question 3:

The inverse Laplace transform of F(s)=\dfrac{3}{(s-5)^2+9} is given by the function f(t) such that:

1) f(t)=e^{5t}\sin(3t)

2) f(t)=e^{5t}\sin(9t)

3) f(t)=e^{5t}\cos(3t)

4) f(t)=e^{5t}\cos(9t)

5) f(t)=e^{3t}\sin(5t)

Question 4:

If F(s)=\dfrac{1}{s^{3}} then \mathscr{L}^{-1}\{F(s)\} is

1) f(t)=t^{2}

2) f(t)=t^{3}

3) f(t)=\dfrac{t^{2}}{2}

4) f(t)=\dfrac{t^{3}}{2}

5) f(t)=\dfrac{t^{2}}{6}

Question 5:

The inverse Laplace of F(s)=\frac{5}{s^2-6s+13} is

1) e^{3t}\sin(2t)

2) e^{3t}\cos(2t)

3) \dfrac{5}{2}e^{3t}\sin(2t)

4) \dfrac{5}{2}e^{3t}\cos(2t)

5) \dfrac{5}{3}e^{3t}\sin(2t)

Question 6:

Let F(s)=\dfrac{8s}{(s^2+16)^2}

1) t\sin(4t)

2) \sin(4t)

3) t\cos(4t)

4) \cos(4t)

5) t\sin(4t)\cos(4t)

Question 7:

The solution of the IVP: \left\{\begin{aligned} y''+y=e^t \\ y(0)=y'(0)=0 \end{aligned}\right., is

1) y(t)=\dfrac{1}{2}(\cos(t)-\sin(t)+e^{t})

2) y(t)=\dfrac{1}{2}(-\cos(t)+\sin(t)+e^{t})

3) y(t)=\dfrac{1}{2}(\cos(t)+\sin(t)+e^{t})

4) y(t)=\dfrac{1}{2}(-\cos(t)-\sin(t)-e^{t})

5) y(t)=\dfrac{1}{2}(-\cos(t)-\sin(t)+e^{t})

Question 8:

If y(t) is the solution of the IVP: \left\{\begin{aligned} y'+2y=f(t) \\ y(0)=0 \end{aligned}\right., where f(t) is the function which graph is given below:

then y(t) is equal to

1) \dfrac{1}{2}[e^{-2t}+1-(-e^{-2(t-1)}+1)u(t-1)]

2) \dfrac{1}{2}[-e^{-2t}+1-(e^{-2(t-1)}+1)u(t-1)]

3) \dfrac{1}{2}[-e^{-2t}+1-(-e^{-2(t-1)}+1)u(t-1)]

4) \dfrac{1}{2}[e^{-2t}+1-(-e^{-2(t-1)}-1)u(t-1)]

5) \dfrac{1}{2}[-e^{-2t}+1-(e^{-2(t-1)}+1)u(t-1)]

Question 9:

The solution of the integro-differential equation \displaystyle y(t)=t-e^{t}\int_{0}^{t}y(u)e^{-u}du is:

1) y(t)=t+\dfrac{t^2}{2}

2) y(t)=t-\dfrac{t^2}{2}

3) y(t)=t^2-\dfrac{t}{2}

4) y(t)=t^2+\dfrac{t}{2}

5) y(t)=t-t^2

Question 10:

If y(t) is the solution of the differential equation y''+4y'+13y=\delta(t-\pi)+\delta(t-3\pi) with y(0)=1,\: y'(0)=0 then

1) y(t)=-\dfrac{1}{3}e^{-2t}\sin(3t)\left[e^{2\pi}u(t-\pi)+e^{6\pi}u(t-3\pi)-3\cot(3t)-2\right]

2) y(t)=\dfrac{1}{3}e^{-2t}\cos(3t)\left[e^{2\pi}u(t-\pi)+e^{6\pi}u(t-3\pi)-3\cot(3t)-2\right]

3) y(t)=-\dfrac{1}{3}e^{-2t}\cos(3t)\left[e^{2\pi}u(t-\pi)+e^{6\pi}u(t-3\pi)-3\cot(3t)-2\right]

4) y(t)=-\dfrac{1}{3}e^{-2t}\sin(3t)\left[e^{2\pi}u(t-\pi)+e^{6\pi}u(t-3\pi)-3\cot(3t)-2\right]

5) y(t)=\dfrac{1}{3}e^{-2t}\sin(3t)\left[e^{2\pi}u(t-\pi)+e^{6\pi}u(t-3\pi)-3\cot(3t)-2\right]

Solutions

Q1: choice 4

Q2: choice 3

Q3: choice 1

Q4: choice 3

Q5: choice 3

Q6: choice 1

Q7: choice 5

Q8: choice 3

Q9: choice 2

Q10: choice 1

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