Solve the ordinary differential equation with initial conditions using Laplace transform.y"+4y'+13y=20e^(-t) , y_0=1 , y'_0=3

Burhan Hopper 2021-09-30 Answered

Solve the ordinary differential equation with initial conditions using Laplace transform.
\(y''+4y'+13y=20e^{-t} , y_0=1 , y'_0=3\)
The solution of the above equation is given as
\(\displaystyle{y}={A}{e}^{{{B}{x}}}+{e}^{{{C}{x}}}{\sin{{3}}}{t}-{e}^{{{D}{x}}}{\cos{{3}}}{t}\)

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Expert Answer

likvau
Answered 2021-10-01 Author has 4596 answers

Step 1
The given equation is \(y''+4y'+13y=20e^{-t}\)
The initial conditions are \(\displaystyle{y}{\left({0}\right)}={1},{y}'{\left({0}\right)}={3}\)
Taking Laplace transform on both sides
\(L\left\{y''+4y'+13y\right\}=L\left\{20e^{-t}\right\}\)
Apply formula Laplace transform for the derivatives
\(L\left\{f'(t)\right\}=sL\left\{f(t)\right\}-f(0)\)
\(L\left\{f''(t)\right\}=s^2L\left\{f(t)\right\}-sf(0)-f'(0)\)
\(L\left\{e^{-t}\right\}=\frac{1}{s+1}\)
Step 2
Using the formulas
\(L\left\{y"+4y'+13y\right\}=L\left\{20e^{-t}\right\}\)
\(\left[s^2L\left\{y\right\}-sy(0)-y'(0)\right]+4\left[sL\left\{y\right\}-y(0)\right]+13L\left\{y\right\}=20L\left(e^{-t}\right)\)
\(\left[s^2L\left\{y\right\}-s(1)-3\right]+4\left[sL\left\{y\right\}-1\right]+13L\left\{y\right\}=20\frac{1}{s+1}\)
\(s^2L\left\{y\right\}-s(1)-3+4sL\left\{y\right\}-4+13L\left\{y\right\}=20\frac{1}{s+1}\)
\(s^2L\left\{y\right\}-s+4sL\left\{y\right\}+13L\left\{y\right\}-7=20\frac{1}{s+1}\)
\(s^2L\left\{y\right\}+4sL\left\{y\right\}+13L\left\{y\right\}-s-7=20\frac{1}{s+1}\)
\(L\left\{y\right\}\left\{s^2+4s+13\right\}-s-7=\frac{20}{s+1}\)
Step 3
\(L\left\{y\right\}\left\{s^2+4s+13\right\}=\frac{20}{s+1}+s+7\)
\(L\left\{y\right\}=\frac{1}{s^2+4s+13}\left(\frac{20+s^2+7s+s+7}{s+1}\right)\)
\(L\left\{y\right\}=\frac{1}{s^2+4s+13}\left(\frac{s^2+8s+27}{s+1}\right)\)
\(L\left\{y\right\}=\frac{s^2+8s+27}{(s+1)(s^2+4s+13)}\)
Take inverse Laplace
\(L^{-1}\left\{L\left\{y\right\}\right\}=L^{-1}\left\{\frac{s^2+8s+27}{(s+1)(s^2+4s+13)}\right\}\)
\(y=L^{-1}\left\{\frac{s^2+8s+27}{(s+1)(s^2+4s+13)}\right\} \dots (1)\)
Step 4
Making the partial fractions
\(\displaystyle{\frac{{{s}^{{2}}+{8}{s}+{27}}}{{{\left({s}+{1}\right)}{\left({s}^{{2}}+{4}{s}+{13}\right)}}}}={\frac{{{a}}}{{{\left({s}+{1}\right)}}}}+{\frac{{{b}{s}+{c}}}{{{\left({s}^{{2}}+{4}{s}+{13}\right)}}}}\)
Multiply both sides by \(\displaystyle{\left({s}+{1}\right)}{\left({s}^{{2}}+{4}{s}+{13}\right)}\)
\(\displaystyle{\frac{{{s}^{{2}}+{8}{s}+{27}}}{{{\left({s}+{1}\right)}{\left({s}^{{2}}+{4}{s}+{13}\right)}}}}={\frac{{{a}}}{{{\left({s}+{1}\right)}}}}+{\frac{{{b}{s}+{c}}}{{{\left({s}^{{2}}+{4}{s}+{13}\right)}}}}\)
\(\displaystyle{\frac{{{\left({s}^{{2}}+{8}{s}+{27}\right)}{\left({s}+{1}\right)}{\left({s}^{{2}}+{4}{s}+{13}\right)}}}{{{\left({s}+{1}\right)}{\left({s}^{{2}}+{4}{s}+{13}\right)}}}}={\frac{{{a}{\left({s}+{1}\right)}{\left({s}^{{2}}+{4}{s}+{13}\right)}}}{{{\left({s}+{1}\right)}}}}+{\frac{{{\left({b}{s}+{c}\right)}{\left({s}+{1}\right)}{\left({s}^{{2}}+{4}{s}+{13}\right)}}}{{{\left({s}^{{2}}+{4}{s}+{13}\right)}}}}\)
\(\displaystyle{\left({s}^{{2}}+{8}{s}+{27}\right)}={a}{\left({s}^{{2}}+{4}{s}+{13}\right)}+{\left({b}{s}+{c}\right)}{\left({s}+{1}\right)}\)
\(\displaystyle{\left({s}^{{2}}+{8}{s}+{27}\right)}={a}{\left({s}^{{2}}+{4}{s}+{13}\right)}+{\left({b}{s}^{{2}}+{c}{s}+{b}{s}+{c}\right)}\)
\(\displaystyle{\left({s}^{{2}}+{8}{s}+{27}\right)}={\left({a}+{b}\right)}{s}^{{2}}+{\left({4}{a}+{b}+{c}\right)}{s}+{13}{a}+{c}\)
Step 5
Equating both sides to get a+b=1
\(4a+b+c=8\)
\(13a+c=27\)
Solve for a, b and c to get the following values
a=2
b=-1
c=1
Therefore, \(\displaystyle{\frac{{{s}^{{2}}+{8}{s}+{27}}}{{{\left({s}+{1}\right)}{\left({s}^{{2}}+{4}{s}+{13}\right)}}}}={\frac{{{2}}}{{{\left({s}+{1}\right)}}}}+{\frac{{{\left(-{1}\right)}{s}+{1}}}{{{\left({s}^{{2}}+{4}{s}+{13}\right)}}}}\)
Equation (1) reduces to \(L^{-1}\left\{\frac{s^2+8s+27}{(s+1)(s^2+4s+13)}\right\}=L^{-1}\left\{\frac{2}{(s+1)}+\frac{(-1)s+1}{(s^2+4s+13)}\right\}\)
Step 6
\(L^{-1}\left\{\frac{2}{(s+1)}+\frac{(-1)s+1}{(s^2+4s+13)}\right\}=L^{-1}\left\{\frac{2}{(s+1)}\right\}+L^{-1}\left\{\frac{-s+1}{(s^2+4s+13)}\right\} \dots (2)\)
Again Rewrite \(\displaystyle{\frac{{-{s}+{1}}}{{{\left({s}^{{2}}+{4}{s}+{13}\right)}}}}=-{\frac{{{s}-{1}}}{{{\left({s}^{{2}}+{4}{s}+{4}+{9}\right)}}}}\)
\(\displaystyle=-{\frac{{{s}+{2}}}{{{\left({\left({s}+{2}\right)}^{{2}}+{9}\right)}}}}+{\frac{{{3}}}{{{\left({\left({s}+{2}\right)}^{{2}}+{9}\right)}}}}\)
Equation (2) reduces to
\(L^{-1}\left\{\frac{2}{(s+1)}+\frac{(-1)s+1}{(s^2+4s+13)}\right\}=L^{-1}\left\{\frac{2}{(s+1)}\right\}+L^{-1}\left\{-\frac{s+2}{(s+2)^2+9}\right\}+L^{-1}\left\{\frac{3}{(s+2)^2+9}\right\}\)
\(=L^{-1}\left\{\frac{2}{(s+1)}\right\}+L^{-1}\left\{-\frac{s+2}{(s+2)^2+9}\right\}+3L^{-1}\left\{\frac{1}{(s+2)^2+9}\right\}\)
\(\displaystyle={2}{e}^{{-{t}}}-{e}^{{-{2}{t}}}{\cos{{\left({3}{t}\right)}}}+{3}{\frac{{{1}}}{{{3}}}}{e}^{{-{2}{t}}}{\sin{{\left({3}{t}\right)}}}\)
\(\displaystyle={2}{e}^{{-{t}}}-{e}^{{-{2}{t}}}{\cos{{\left({3}{t}\right)}}}+{e}^{{-{2}{t}}}{\sin{{\left({3}{t}\right)}}}\)
Step 7
comparing with the given equation
The answer is :
A=2
B=-2
C=-2
D=-2

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