This set of Ordinary Differential Equations Multiple Choice Questions & Answers (MCQs) focuses on “Convolution”.
1. Find the \(L^{-1} (\frac{1}{s(s^2+4)})\).
a) \(\frac{1-sin(t)}{4}\)
b) \(\frac{1-cos(t)}{4}\)
c) \(\frac{1-sin(2t)}{4}\)
d) \(\frac{1-cos(2t)}{4}\)
View Answer
Explanation: In the given question
Let p1(s) = \(\frac{1}{s^2+4}\) and p2(s) = \(\frac{1}{s}\) \(f_1 (t) = L^{-1} (\frac{1}{s^2+4}) = \frac{sin(2t)}{2}\)
\(f_2 (t) = L^{-1} (\frac{1}{s}) = 1\)
By Convolution Theorem,
\(L^{-1} (p_1(s)×p_2(s)) = \int_{0}^{t}f_1(u) f_2(t-u) dt\)
\(L^{-1} (\frac{1}{s(s^2+4)}) = \int_{0}^{t}\frac{1}{2} sin(2u) du\)
=\(\frac{1-cos(2t)}{4}\)
Thus, the answer is \(\frac{1-cos(2t)}{4}\).
2. Find the \(L^{-1} (\frac{1}{s(s+4)^\frac{1}{2}})\), give the answer in terms of error function.
a) \(\frac{1}{2} erf(2t)\)
b) \(\frac{1}{2} erf(\sqrt{t})\)
c) \(\frac{1}{2} erf(2\sqrt{t})\)
d) \(\frac{1}{2} erf(4\sqrt{t})\)
View Answer
Explanation: In the given question,
Let \(p_1(s) = \frac{1}{(s+4)^{\frac{1}{2}}}\) and \(p_2(s) = \frac{1}{s}\)
\(f_1 (t) = L^{-1} (\frac{1}{(s+4)^{\frac{1}{2}}}) = \frac{e^{-4t}×\sqrt{t}}{\sqrt\pi}\)
\(f_2 (t) = L^{-1} \frac{1}{s} = 1\)
By Convolution Theorem,
\(L^{-1} (p_1(s)×p_2(s)) = \int_{0}^{t}f_1(u) f_2(t-u) dt\)
\(L^{-1} (\frac{1}{s(s+4)^{\frac{1}{2}}}) = \int_{0}^{t}\frac{\sqrt{u}}{\sqrt\pi} e^{-4u} \,du\)
\(=\frac{1}{\sqrt\pi} \int_{0}^{t}\sqrt{u} e^{-4u} du\)
We know that error function is given by –
\(erf(x)=\frac{2}{\sqrt\pi} \int_{0}^{x}e^{-z^2} dz\)
Applying, 4u=z2 And setting the limits of u in terms of z, we get
\(L^{-1} (\frac{1}{s(s+4)^{\frac{1}{2}}}) = \frac{1}{2} erf(2\sqrt{t})\)
Thus, the answer is \(\frac{1}{2} erf(2\sqrt{t})\).
3. Find the \(L^{-1} \frac{s}{(s^2+4)^2}\).
a) \(\frac{1}{4} tcos(2t)\)
b) \(\frac{1}{4} tsin(t)\)
c) \(\frac{1}{4} tsin(2t)\)
d) \(\frac{1}{2} tsin(2t)\)
View Answer
Explanation: In the given question,
Let \(p_1(s) = \frac{1}{s^2+4}\) and \(p_2(s) = \frac{1}{s}\)
\(f_1 (t) = L^{-1} (\frac{1}{s^2+4}) = \frac{sin(2t)}{2}\)
\(f_2 (t) = L^{-1} (\frac{s}{s^2+4}) = cos(2t)\)
By Convolution Theorem,
\(L^{-1} (p_1(s)×p_2(s)) = \int_{0}^{t} f_1(u) f_2(t-u) dt\)
\(L^{-1} \left (\frac{s}{(s^2+4^2)^2}\right )=\int_{0}^{t} sin(2u)×\frac{1}{2}×cos(2(t-u))du\)
=\(\frac{1}{4}[tsin(2t)-\frac{cos(2t)}{4}+\frac{cos(2t)}{4}]\)
=\(\frac{1}{4} tsin(2t)\)
Thus, the correct answer is \(\frac{1}{4} tsin(2t)\)
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Sanfoundry Global Education & Learning Series – Ordinary Differential Equations.
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