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DE & LT CAT I -MCQ

Total questions: 10

Worksheet time: 30mins

Name
Class
Date
1.

What is the Laplace Transform for  f(t)=t3f\left(t\right)=t^3  ?

a)

 1s3\frac{1}{s^3}  

b)

 2s3\frac{2}{s^3}  

c)

 6s4\frac{6}{s^4}  

d)

 3s3\frac{3}{s^3}  

2.


 Laplace transform of e3t12Laplace\ transform\ of\ e^{3t}-\frac{1}{2}  

a)

 1s+312s\frac{1}{s+3}-\frac{1}{2s}  

b)

 1s32s\frac{1}{s-3}-\frac{2}{s}  

c)

 1s312s\frac{1}{s-3}-\frac{1}{2s}  

d)

 12(s3)\frac{1}{2\left(s-3\right)}  

3.

 f(t)=te2tf\left(t\right)=te^{-2t}  

a)
b)
c)
d)
4.

 L(t cos 6t)L\left(t\ \cos\ 6t\right)  Which Laplace property is use to solve this?

a)

Linearity Property

b)

Derivative of Laplace Transform

c)

First Shifting Property

d)

Second Shifting Property

5.

Inverse Laplace transform, transform

a)

f(t) to F(s)f\left(t\right)\ to\ F\left(s\right)

b)

F(s) to f(t)F\left(s\right)\ to\ f\left(t\right)

c)

f(t ) to f(t)f'\left(t\ \right)\ to\ f\left(t\right)

d)

f(t) to f(t)f\left(t\right)\ to\ f'\left(t\right)

6.

 L1(6(s)4ss2+25)L^{-1}\left(\frac{6}{\left(s\right)}-\frac{4s}{s^2+25}\right)  What is the property use to find the Inverse Laplace Transform of this equation?

a)

Second Shifting Property

b)

First Shifting Property

c)

Linearity Property

d)

Convolution theorem

7.

 L1(1(s3)2)L^{-1}\left(\frac{1}{\left(s-3\right)^2}\right)  

a)

 e2tte^{2t}t  

b)

 ett2e^tt^2  

c)

 e3tte^{3t}t  

8.

 L1(s+2s22s5)L^{-1}\left(\frac{s+2}{s^2-2s-5}\right)  

a)

 f(t)=et(cosh6 t+36sinh6 t)f(t)=e^t\left(\cosh\sqrt{6}\ t+\frac{3}{\sqrt{6}}\sinh\sqrt{6\ }t\right)  

b)

 f(t)=et(cosh6 t+36sinh6 t)f(t)=e^{-t}\left(\cosh\sqrt{6}\ t+\frac{3}{\sqrt{6}}\sinh\sqrt{6\ }t\right)  

c)

 f(t)=et(cosh2 t+32sinh2t)f(t)=e^t\left(\cosh2\ t+\frac{3}{2}\sinh2t\right)  

9.

 (D2+1)x=sin2t, then \left(D^2+1\right)x=\sin2t,\ then\   with initial conditions as zero then,

a)

 L(y(t))=2(s2+1)(s2+4)L\left(y\left(t\right)\right)=\frac{2}{\left(s^2+1\right)\left(s^2+4\right)}  

b)

 L(x(t))=2(s2+1)(s2+4)L\left(x\left(t\right)\right)=\frac{2}{\left(s^2+1\right)\left(s^2+4\right)}  

10.

GIVEN

 (D25D+6)y=0,  y(0)=1,  y(0)=1\left(D^2-5D+6\right)y=0,\ \ y\left(0\right)=1,\ \ y'\left(0\right)=-1  

a)

 L(y)=s+4s25s+6L\left(y\right)=\frac{s+4}{s^2-5s+6}  

b)

 L(y)=s6s25s+6L\left(y\right)=\frac{s-6}{s^2-5s+6}  

c)

 L(y)=s+5s25s+6L\left(y\right)=\frac{s+5}{s^2-5s+6}  

d)

 L(y)=s5s25s+6L\left(y\right)=\frac{s-5}{s^2-5s+6}