This set of 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
View Answer
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
View Answer
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)
View Answer
Explanation: Let \(f(t) = y(t) \)
\(L[f’(t)] = \int_0^∞ e^{-st} f'(t)dt \)
\( = e^{-st} f(t)(from \, 0 \, to \, \infty) – \int_0^∞ (-s) e^{-st} f(t)dt \)
\( = -f(0) + s\int_0^{\infty} e^{-st} f(t)dt \)
\( = -f(0) + sF(s) \)
\( = sY(0) -y(0) \).
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) \(2e^{-t} cos(3t)-2 \frac{te^{-t}}{3} sint(3t) \)
View Answer
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) \(e^t [Acost+A1sint+Bcos(2t)+(B1)sin(2t)] \)
View Answer
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) \(2e^{-t} \frac{t^5}{720}+e^{-t}+2e^{-t} \frac{t}{6}+4e^{-t} \frac{t^2}{24} \)
View Answer
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) \((s^3-3s^2+3s-1)Y(s)+(-as^2+(3a-b)s+(-3a-c))=\frac{2}{(s-1)^3} \)
View Answer
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) \(\frac{-1}{2} e^{t}+\frac{9}{2} e^{2t}-3e^{3t} \)
View Answer
Explanation: Taking inverse Laplace Transformation for
\(Y(s) = \frac{(s^2-s+3)}{(s+1)(s+2)(s+3)} \)
Solving the partial fractions we get,
\(Y(s) = \frac{-1}{2} \frac{1}{(s+1)}+\frac{9}{2} \frac{1}{(s+2)}-3 \frac{1}{(s+3)} \)
Therefore, \(y(t) = \frac{-1}{2} e^{-t}+\frac{9}{2} e^{-2t}-3e^{-3t}. \)
9. For the Transient analysis of a circuit with capacitors, inductors, resistors, we use bilateral Laplace Transformation to solve the equation obtained from the Kirchoff’s current/voltage law.
a) True
b) False
View Answer
Explanation: For the transient analysis of the circuit with capacitors, inductors, resistors, we have to know the initial condition of the components used. So, the unilateral Laplace Transform is used to solve the equations obtained from the Kirchoff’s current/voltage law.
10. 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
View Answer
Explanation: It is not possible to solve an equation if the input and the initial conditions are zero becase Y(s) becomes zero where Y(s) is the Laplace Transform of y(t) function.
Sanfoundry Global Education & Learning Series – Ordinary Differential Equations.
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