Question

Use Laplace transform to find the solution of the IVP2y'+y=0 , y(0)=-3a) f{{left({t}right)}}

Laplace transform
ANSWERED
asked 2021-01-30

Use Laplace transform to find the solution of the IVP
\(2y'+y=0 , y(0)=-3\)
a) \(f{{\left({t}\right)}}={3}{e}^{{-{2}{t}}}\)
b)\(f{{\left({t}\right)}}={3}{e}^{{\frac{t}{{2}}}}\)
c)\(f{{\left({t}\right)}}={6}{e}^{{{2}{t}}}\)
d) \(f{{\left({t}\right)}}={3}{e}^{{-\frac{t}{{2}}}}\)

Answers (1)

2021-01-31

Step 1
Given
Use Laplace transform to find the solution of the IVP,
\(2y'+y=0, y(0)=-3\)
step 2
Solution
\(2y'+y=0\)
Taking Laplace transform on both sides
\(L(2y')+L(y)=0\)
\(2L(y')+L(y)=0\)
\({2}{\left[\Delta{L}{\left({y}\right)}-{y}{\left({0}\right)}\right]}+{L}{\left({y}\right)}={0}\)
\({2}\Delta{L}{\left({y}\right)}-{2}{y}{\left({0}\right)}+{L}{\left({y}\right)}={0}\)
\({2}\Delta{L}{\left({y}\right)}-{2}{\left(-{3}\right)}+{L}{\left({y}\right)}={0}\)
\({L}{\left({y}\right)}{\left[{2}\Delta+{1}\right]}=-{6}\)
\({L}{\left({y}\right)}=\frac{{-{6}}}{{{2}\Delta+{1}}}\)
\(=\frac{{-{6}}}{{{2}{\left(\Delta+\frac{1}{{2}}\right)}}}\)
\({Y}={L}^{ -{{1}}}{\left(\frac{{-{3}}}{{\Delta+\frac{1}{{2}}}}\right)}=-{3}{L}^{ -{{1}}}{\left(\frac{{{1}}}{{\Delta+\frac{1}{{2}}}}\right)}\)
\({Y}{\left({t}\right)}=-{3}{e}^{{-\frac{t}{{2}}}}\)
Therefore the required answer is \(f{{\left({t}\right)}}=-{3}{e}^{{-\frac{t}{{2}}}}\)

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