Ordinary Differential Equations Multiple Choice Questions & Answers focuses on “Solution of DE With Constant Coefficients using the Laplace Transform”.
1. While solving the ordinary differential equation using unilateral laplace transform, we consider the initial conditions of the system.
a) True
b) False
Answer: a
Explanation: When bilateral laplace transformation is used in solving differential equations, we don’t consider the initial conditions as the transformation is from -∞ to +∞. But when we consider unilateral laplace transformation, the integral is from 0 to ∞. So, the initial conditions are considered.
2. With the help of _____________________ Mr.Melin gave inverse laplace transformation formula.
a) Theory of calculus
b) Theory of probability
c) Theory of statistics
d) Theory of residues
Answer: d
Explanation: Let f(t) be the function in time. The laplace transformation of the function is L[f(t)] = F(s). So, the inverse laplace transform of F(s) comes out to be the function f(t) in time. The formula for laplace transform is derived using the theory of residues by Mr.Melin.
3. What is the laplce tranform of the first derivative of a function y(t) with respect to t : y’(t)?
a) sy(0) – Y(s)
b) sY(s) – y(0)
c) s2 Y(s)-sy(0)-y'(0)
d) s2 Y(s)-sy'(0)-y(0)
Answer: b
Explanation: Let (f(t) = y(t) )
(L[f’(t)] = int_0^∞ e^{-st} f'(t)dt )
( = e^{-st} f(t)(from , 0 , to , infty) – int_0^∞ (-s) e^{-st} f(t)dt )
( = -f(0) + sint_0^{infty} e^{-st} f(t)dt )
( = -f(0) + sF(s) )
( = sY(0) -y(0) ).
4. Solve the Ordinary Differential Equation by Laplace Transformation y’’ – 2y’ – 8y = 0 if y(0) = 3 and y’(0) = 6.
a) (3e^t cos(3t)+tsint(3t) )
b) (3e^t cos(3t)+te^{-t} sint(3t) )
c) (2e^{-t} cos(3t)-2 frac{t}{3} sint(3t) )
d) (2e^{-t} cos(3t)-2 frac{te^{-t}}{3} sint(3t) )
Answer: a
Explanation: L[y’’ – 2y’ – 8y ] = 0
s2 Y(s) – sy(0) – y'(0) – 2sY(s) + 2y(0) – 8Y(s) = 0
(s2 – 2s – 8)Y(s) = 2s
(L[y(t)] = 2 frac{s}{(s^2-2s-8)} )
Therefore, y(t) = 3et cos(3t) + tsint(3t).
5. Solve the Ordinary Differential Equation y’’ + 2y’ + 5y = e-t sin(t) when y(0) = 0 and y’(0) = 1.(Without solving for the constants we get in the partial fractions).
a) (e^t [Acost+A1sint+Bcos(2t)+frac{(B1)}{2} sin(2t)] )
b) (e^{-t} [Acost+A1sint+Bcos(2t)+B1sin(2t)] )
c) (e^{-t} [Acost+A1sint+Bcos(2t)+frac{(B1)}{2} sin(2t)] )
d) (e^t [Acost+A1sint+Bcos(2t)+(B1)sin(2t)] )
Answer: c
Explanation: (L[y’’+2y’ +5y = e^{-t} sin(t)] )
(s^2 Y(s)-sy(0)-y'(0)+ 2sY(s) -2y(0) + 5Y(s) = frac{1}{(s+1)^2+1} )
((s^2+2s+5)Y(s)= frac{1}{(s+1)^2+1}+1 )
((s^2+2s+5)Y(s)= frac{(s^2+2s+3)}{(s^2+2s+2)} )
( Y(s) = frac{(s^2+2s+3)}{(s^2+2s+2)(s^2+2s+5)} )
( = frac{(s+1)^2+2}{((s+1)^2+1)((s+1)^2+4)} )
( y(t) = e^{-t} L^{-1} [frac{(As+A1)}{(s^2+1)}+frac{(Bs+B1)}{(s^2+4)}] )
( = e^{-t} [Acost+A1sint+Bcos(2t)+frac{(B1)}{2} sin(2t)]).
6. Solve the Ordinary Diferential Equation using Laplace Transformation y’’’ – 3y’’ + 3y’ – y = t2 et when y(0) = 1, y’(0) = 0 and y’’(0) = 2.
a) (2e^t frac{t^5}{720}+e^t+2e^t frac{t}{6}+4e^t frac{t^2}{24} )
b) (e^t frac{t^5}{720}+2e^{-t}+2e^t frac{t}{6}+4e^t frac{t^2}{24} )
c) (e^{-t} frac{t^5}{720}+e^{-t}+2e^{-t} frac{t}{6}+4e^{-t} frac{t^2}{24} )
d) (2e^{-t} frac{t^5}{720}+e^{-t}+2e^{-t} frac{t}{6}+4e^{-t} frac{t^2}{24} )
Answer: a
Explanation: L[y’’’ – 3y’’ + 3y’ – y = t2 et]
s3 Y(s) – s2 y(0) – sy'(0) – y”(0) – 3s2 Y(s) + 3sy(0) + 3y'(0) + 3sY(s) – 3y(0) – Y(s) = (frac{2}{(s-1)^3} )
(Y(s) = frac{2}{(s-1)^6} +frac{(s^2+3s+5)}{(s-1)^3} )
(y(t) = 2e^t frac{t^5}{720}+e^t+2e^t frac{t}{6}+4e^t frac{t^2}{24}).
7. Take Laplace Transformation on the Ordinary Differential Equation if y’’’ – 3y’’ + 3y’ – y = t2 et if y(0) = 1, y’(0) = b and y’’(0) = c.
a) ((s^3-3s^2+3s-1)Y(s)+(-as^2+(3a-b)s+(-3a-c))=frac{2}{(s-1)^3} )
b) ((s^3-3s^2+3s-1)Y(s)+(-as^2+(3a-b)+(-3a-c)s)=frac{2}{(s-1)^3} )
c) ((s^3-3s^2+3s)Y(s)+(-as+(3a-b)s+(-3a-c))=frac{2}{(s-1)^3} )
d) ((s^3-3s^2+3s-1)Y(s)+(-as^2+(3a-b)s+(-3a-c))=frac{2}{(s-1)^3} )
Answer: a
Explanation: L[y’’’ – 3y’’ + 3y’ – y = t2 et]
s3 Y(s) – s2 y(0) – sy'(0) – y”(0) – 3s2 Y(s) + 3sy(0) + 3y'(0) + 3sY(s) – 3y(0) – Y(s) = (frac{2}{(s-1)^3} )
((s^3-3s^2+3s-1)Y(s)+(-as^2+(3a-b)s+(-3a-c))=frac{2}{(s-1)^3}.)
8. What is the inverse Laplace Transform of a function y(t) if after solving the Ordinary Differential Equation Y(s) comes out to be (Y(s) = frac{s^2-s+3}{(s+1)(s+2)(s+3)} ) ?
a) (frac{1}{2} e^{-t}+frac{9}{2} e^{-3t}-3e^{-2t} )
b) (frac{-1}{2} e^{-t}+frac{9}{2} e^{-2t}-3e^{-3t} )
c) (frac{1}{2} e^{-t}-frac{3}{2} e^{-2t}-3e^{-3t} )
d) (frac{-1}{2} e^{t}+frac{9}{2} e^{2t}-3e^{3t} )