Question

Solve the initial value problem below using the method of Laplace transforms. y"-8y'+41y=29e^{2t} ,y(0)=1 , y'(0)=7

Laplace transform

Solve the initial value problem below using the method of Laplace transforms.
$$\displaystyle{y}\text{}{8}{y}'+{41}{y}={29}{e}^{{{2}{t}}}$$
$$y(0)=1 , y'(0)=7$$

2021-09-17

The given IVP is
$$\displaystyle{y}\text{}{8}{y}+{41}{y}={29}{e}^{{{2}{t}}}$$
$$y(0)=1 , y'(0)=7$$
Taking Laplace in both side we get
$$\displaystyle{L}{\left[{y}\text{}\right]}-{8}{L}{\left[{y}'\right]}+{41}{L}{\left[{y}\right]}={29}{L}{\left[{e}^{{{2}{t}}}\right]}$$
$$\displaystyle\Rightarrow{s}^{{2}}{y}{\left({s}\right)}-{s}{y}{\left({0}\right)}-{y}'{\left({0}\right)}-{8}{\left({s}{y}{\left({s}\right)}-{y}{\left({0}\right)}\right)}+{41}{y}{\left({s}\right)}={\frac{{{29}}}{{{s}-{2}}}}$$
$$\displaystyle\Rightarrow{\left({s}^{{2}}-{8}{s}+{41}\right)}{y}{\left({s}\right)}-{s}{\left({1}\right)}-{7}-{8}{\left(-{1}\right)}={\frac{{{29}}}{{{\left({s}-{2}\right)}}}}$$
$$\displaystyle\Rightarrow{\left({s}^{{2}}-{8}{s}+{41}\right)}{y}{\left({s}\right)}-{s}-{7}+{8}={\frac{{{29}}}{{{\left({s}-{2}\right)}}}}$$
$$\displaystyle\Rightarrow{\left({s}^{{2}}-{8}{s}+{41}\right)}{y}{\left({s}\right)}-{s}+{1}={\frac{{{29}}}{{{\left({s}-{2}\right)}}}}$$
$$\displaystyle\Rightarrow{\left({s}^{{2}}-{8}{s}+{41}\right)}{y}{\left({s}\right)}={\frac{{{29}}}{{{\left({s}-{2}\right)}}}}+{s}-{1}$$
$$\displaystyle\Rightarrow{\left({s}^{{2}}-{8}{s}+{41}\right)}{y}{\left({s}\right)}={\frac{{{29}+{s}^{{2}}-{3}{s}+{2}}}{{{\left({s}-{2}\right)}}}}$$
$$\displaystyle\Rightarrow{y}{\left({s}\right)}={\frac{{{s}^{{2}}-{3}{s}+{31}}}{{{\left({s}-{2}\right)}{\left({s}^{{2}}-{8}{s}+{41}\right)}}}}$$
$$\displaystyle\Rightarrow{y}{\left({s}\right)}={\frac{{{s}^{{2}}-{3}{s}-{5}{s}+{5}{s}+{31}+{10}-{10}}}{{{\left({s}-{2}\right)}{\left({s}^{{2}}-{8}{s}+{41}\right)}}}}$$
$$\displaystyle\Rightarrow{y}{\left({s}\right)}={\frac{{{s}^{{2}}-{8}{s}+{5}{s}+{41}-{10}}}{{{\left({s}-{2}\right)}{\left({s}^{{2}}-{8}{s}+{41}\right)}}}}$$
$$\displaystyle\Rightarrow{y}{\left({s}\right)}={\frac{{{\left({s}^{{2}}-{8}{s}+{41}\right)}+{5}{\left({s}-{2}\right)}}}{{{\left({s}-{2}\right)}{\left({s}^{{2}}-{8}{s}+{41}\right)}}}}$$
or $$\displaystyle{y}{\left({s}\right)}={\frac{{{\left({s}^{{2}}-{8}{s}+{41}\right)}}}{{{\left({s}-{2}\right)}{\left({s}^{{2}}-{8}{s}+{41}\right)}}}}+{\frac{{{5}{\left({s}-{2}\right)}}}{{{\left({s}-{2}\right)}{\left({s}^{{2}}-{8}{s}+{41}\right)}}}}$$
$$\displaystyle{y}{\left({s}\right)}={\frac{{{1}}}{{{\left({s}-{2}\right)}}}}+{\frac{{{5}}}{{{\left({s}^{{2}}-{8}{s}+{41}\right)}}}}$$
or $$\displaystyle{y}{\left({s}\right)}={\frac{{{1}}}{{{\left({s}-{2}\right)}}}}+{\frac{{{5}}}{{{\left({s}^{{2}}-{8}{s}+{16}+{25}\right)}}}}$$
$$\displaystyle\Rightarrow{y}{\left({s}\right)}={\frac{{{1}}}{{{\left({s}-{2}\right)}}}}+{\frac{{{5}}}{{{\left({s}-{4}\right)}^{{2}}+{5}^{{2}}}}}{\left({1}\right)}$$
Since we know $$\displaystyle{L}{\left[{e}^{{{a}{t}}}{f{{\left({t}\right)}}}\right]}={F}{\left({s}-{a}\right)}$$ where $$\displaystyle{F}{\left({s}\right)}={L}{\left[{f{{\left({t}\right)}}}\right]}$$
i.e. we just have to put $$(s-a)$$ at place of s in F(s) (Laplace of f(t)) to find Laplace of $$\displaystyle{e}^{{{a}{t}}}\cdot{f{{\left({t}\right)}}}$$
$$\displaystyle\because\ \text{ for }\ {f{{\left({t}\right)}}}={\sin{{\left({5}{t}\right)}}}\Rightarrow{F}{\left({s}\right)}={\frac{{{5}}}{{{s}^{{2}}+{5}^{{2}}}}}$$
So for finding Laplace of $$\displaystyle{e}^{{{4}{t}}}{\sin{{\left({5}{t}\right)}}}$$ we just have to put s-4 at the place of s in F(s)
Then
$$\displaystyle{L}{\left[{e}^{{{4}{t}}}{\sin{{\left({5}{t}\right)}}}\right]}={\frac{{{5}}}{{{\left({s}-{4}\right)}^{{2}}+{5}^{{2}}}}}$$
$$\displaystyle\Rightarrow{L}^{{-{1}}}{\left[{\frac{{{5}}}{{{\left({s}-{4}\right)}^{{2}}+{5}^{{2}}}}}\right]}={e}^{{{4}{t}}}{\sin{{\left({5}{t}\right)}}}$$
Now Taking inverse Laplace of (1) we get
$$\displaystyle{y}{\left({t}\right)}={L}^{{-{1}}}{\left[{\frac{{{1}}}{{{s}-{2}}}}\right]}+{L}^{{-{1}}}{\left[{\frac{{{5}}}{{{\left({s}-{4}\right)}^{{2}}+{5}^{{2}}}}}\right]}$$
$$\displaystyle\Rightarrow{y}{\left({t}\right)}={e}^{{{2}{t}}}+{e}^{{{4}{t}}}{\sin{{\left({5}{t}\right)}}}$$
This is required solution of given IVP