Question

Solve the following initial value problem by using Laplace transform
\(y''+4y=4t+12 , y(0)=4 , y'(0)=0\)

Answer 1

\(y''+4y=4t+12 , y(0)=4 , y'(0)=0\)
Taking the Laplace transform on both side
\(L\left\{y"\right\}+4L\left\{y\right\}=4L\left\{t\right\}+12L\left\{1\right\}\)
\(s^2L\left\{y\right\}-sy(0)-y'(0)+4L\left\{y\right\}=4\frac{1}{s^2}+12\frac{1}{s}\)
\((s^2+4)L\left\{y\right\}-4s-0=4\left(\frac{3s+1}{s}\right)\)
\((s^2+4)L\left\{y\right\}=4\left(\frac{3s+1}{s}+s\right) \)
\((s^2+4)L\left\{y\right\}=4\left(\frac{3s+1+s^2}{s}\right)\)
\(L\left\{y\right\}=4\left(\frac{s^2+3s+1}{s(s^2+4)}\right)\)
\(\displaystyle={4}{\left({\frac{{{s}^{{2}}+{4}+{3}{s}-{3}}}{{{s}{\left({s}^{{2}}+{4}\right)}}}}\right)}\)
\(\displaystyle={4}{\left[{\frac{{{1}}}{{{s}}}}+{\frac{{{3}{s}-{3}}}{{{s}{\left({s}^{{2}}+{4}\right)}}}}\right]}\)
\(\displaystyle={4}{\left[{\frac{{{1}}}{{{s}}}}+{\frac{{{1}}}{{{4}}}}{\left({\frac{{{12}{s}-{12}}}{{{s}{\left({s}^{{2}}+{4}\right)}}}}\right)}\right]}\)
\(\displaystyle={4}{\left[{\frac{{{1}}}{{{s}}}}+{\frac{{{1}}}{{{4}}}}{\left({\frac{{{3}{s}^{{2}}-{3}{s}^{{2}}+{12}{s}-{12}}}{{{s}{\left({s}^{{2}}+{4}\right)}}}}\right)}\right]}\)
\(\displaystyle={4}{\frac{{{1}}}{{{s}}}}+{\frac{{{s}{\left({3}{s}+{12}\right)}-{3}{\left({s}^{{2}}+{4}\right)}}}{{{s}{\left({s}^{{2}}+{4}\right)}}}}\)
\(L\left\{y\right\}=\frac{4}{s}+\frac{3s+12}{s^2+4}-\frac{3}{s}\)
\(L\left\{y\right\}=\frac{1}{s}+3\frac{s}{s^2+4}+\frac{12}{s^2+4}\)
\(y(t)=L^{-1}\left\{\frac{1}{s}\right\}+3L^{-1}\left\{\frac{s}{s^2+4}\right\}+6L^{-1}\left\{\frac{2}{s^2+4}\right\}\)
\(\displaystyle{y}{\left({t}\right)}={1}+{3}{\cos{{\left({2}{t}\right)}}}+{6}{\sin{{\left({2}{t}\right)}}}\)