Question

Use the Laplace transform to solve the given initial-value problem. displaystyle{y}{''}+{8}{y}'+{41}{y}=delta{left({t}-piright)}+delta{left({t}-{3}piright)}, {y}{left({0}right)}={1}, {y}'{left({0}right)}={0} {y}{left({t}right)}=?

Laplace transform
ANSWERED
asked 2020-12-06
Use the Laplace transform to solve the given initial-value problem.
\(\displaystyle{y}{''}+{8}{y}'+{41}{y}=\delta{\left({t}-\pi\right)}+\delta{\left({t}-{3}\pi\right)},\ \ \ \ {y}{\left({0}\right)}={1},\ \ \ \ {y}'{\left({0}\right)}={0}\ \ {y}{\left({t}\right)}=?\)

Answers (1)

2020-12-07
Step 1
Given initial value problem,
\(\displaystyle{y}{''}+{8}{y}'+{41}{y}=\delta{\left({t}-\pi\right)}+\delta{\left({t}-{3}\pi\right)}\),
\(y(0) = 1\),
\(y'(0) = 0\)
Solve this problem using Laplace transform method.
Step 2
Now
\(\displaystyle{y}{''}+{8}{y}'+{41}{y}=\delta{\left({t}-\pi\right)}+\delta{\left({t}-{3}\pi\right)}\)
\(\displaystyle{L}{\left[{y}{''}+{8}{y}'+{41}{y}\right]}={L}{\left[\delta{\left({t}-\pi\right)}+\delta{\left({t}-{3}\pi\right)}\right]}\)
\(\displaystyle{L}{\left[{y}{''}\right]}+{8}{L}{\left[{y}'\right]}+{41}{L}{\left[{y}\right]}={L}{\left[\delta{\left({t}-\pi\right)}\right]}+{L}{\left[\delta{\left({t}-{3}\pi\right)}\right]}\)
Use the formula such that
\(\displaystyle{L}{\left[{y}{''}\right]}={s}^{2}{L}{\left[{y}\right]}-{s}{y}{\left({0}\right)}-{y}'{\left({0}\right)}\)
\(\displaystyle{L}{\left[{y}'\right]}={s}{L}{\left[{y}\right]}-{y}{\left({0}\right)}\)
\(\displaystyle{L}{\left[\delta{\left({t}-{a}\right)}\right]}={e}^{{-{a}{s}}}\)
Step 3
Then,
\(\displaystyle{s}^{2}{L}{\left[{y}\right]}-{s}{y}{\left({0}\right)}-{y}'{\left({0}\right)}+{8}{\left[{s}{L}{\left[{y}\right]}-{y}{\left({0}\right)}\right]}+{41}{L}{\left[{y}\right]}={e}^{{-\pi{s}}}+{e}^{{-{3}\pi{s}}}\)
\(\displaystyle{s}^{2}{L}{\left[{y}\right]}-{s}\cdot{0}-{0}+{8}{\left[{s}{L}{\left[{y}\right]}-{0}\right]}+{41}{L}{\left[{y}\right]}={e}^{{-\pi{s}}}+{e}^{{-{3}\pi{s}}}\)
\(\displaystyle{s}^{2}{L}{\left[{y}\right]}+{8}{s}{L}{\left[{y}\right]}+{41}{L}{\left[{y}\right]}={e}^{{-\pi{s}}}+{e}^{{-{3}\pi{s}}}\)
\(\displaystyle{\left[{s}^{2}+{8}{s}+{41}\right]}{L}{\left[{y}\right]}={e}^{{-\pi{s}}}+{e}^{{-{3}\pi{s}}}\)
\(\displaystyle{L}{\left[{y}\right]}=\frac{{{e}^{{-\pi{s}}}+{e}^{{-{3}\pi{s}}}}}{{{s}^{2}+{8}{s}+{41}}}\)
Taking inverse Laplace transform of both sides,
\(\displaystyle{y}={L}^{ -{{1}}}{\left[\frac{{{e}^{{-\pi{s}}}+{e}^{{-{3}\pi{s}}}}}{{{s}^{2}+{8}{s}+{41}}}\right]}\)
\(\displaystyle={L}^{ -{{1}}}{\left[\frac{{{e}^{{-\pi{s}}}}}{{{s}^{2}+{8}{s}+{41}}}\right]}+{L}^{ -{{1}}}{\left[\frac{{{e}^{{-{3}\pi{s}}}}}{{{s}^{2}+{8}{s}+{41}}}\right]}\ldots{\left({1}\right)}\)
Step 4
Use the formula such that
If \(\displaystyle{L}^{ -{{1}}}{\left[{F}{\left({s}\right)}\right]}= f{{\left({t}\right)}}\), then
\(\displaystyle{L}^{ -{{1}}}{\left[{e}^{{-{a}{s}}}{F}{\left({s}\right)}\right]}={H}{\left({t}-{a}\right)} f{{\left({t}-{a}\right)}}\)
Here H(t) is a Heaviside step function.
Now
\(\displaystyle{L}^{ -{{1}}}{\left[\frac{1}{{{s}^{2}+{8}{s}+{41}}}\right]}={L}^{ -{{1}}}{\left[\frac{1}{{{\left({s}+{4}\right)}^{2}+{25}}}\right]}\)
\(\displaystyle={e}^{{-{4}{t}}}{L}^{ -{{1}}}{\left[\frac{1}{{{s}^{2}+{25}}}\right]}\)
\(\displaystyle=\frac{{{e}^{{-{4}{t}}}}}{{5}}{L}^{ -{{1}}}{\left[\frac{5}{{{s}^{2}+{5}^{2}}}\right]}\)
\(\displaystyle=\frac{{{e}^{{-{4}{t}}}}}{{5}} \sin{{\left({5}{t}\right)}}\)
So,\(\displaystyle{L}^{ -{{1}}}{\left[\frac{{{e}^{{-\pi{s}}}}}{{{s}^{2}+{8}{s}+{41}}}\right]}={H}{\left({t}-\pi\right)}\frac{{e}^{{-{4}{\left({t}-\pi\right)}}}}{{5}} \sin{{\left({5}{\left({t}-\pi\right)}\right)}}\)
Step 5
and
\(\displaystyle{L}^{ -{{1}}}{\left[\frac{{{e}^{{-{3}\pi{s}}}}}{{{s}^{2}+{8}{s}+{41}}}\right]}={H}{\left({t}-{3}\pi\right)}\frac{{e}^{{-{4}{\left({t}-{3}\pi\right)}}}}{{5}} \sin{{\left({5}{\left({t}-{3}\pi\right)}\right)}}\)
From (1),
\(\displaystyle{y}={L}^{ -{{1}}}{\left[\frac{{{e}^{{-\pi{s}}}}}{{{s}^{2}+{8}{s}+{41}}}\right]}+{L}^{ -{{1}}}{\left[\frac{{{e}^{{-{3}\pi{s}}}}}{{{s}^{2}+{8}{s}+{41}}}\right]}\)
\(\displaystyle={H}{\left({t}-\pi\right)}\frac{{e}^{{-{4}{\left({t}-\pi\right)}}}}{{5}} \sin{{\left({5}{\left({t}-\pi\right)}\right)}}+{H}{\left({t}-{3}\pi\right)}\frac{{e}^{{-{4}{\left({t}-{3}\pi\right)}}}}{{5}} \sin{{\left({5}{\left({t}-{3}\pi\right)}\right)}}\)
\(\displaystyle=\frac{1}{{5}}{\left[{H}{\left({t}-\pi\right)}{e}^{{-{4}{\left({t}-\pi\right)}}} \sin{{\left({5}{\left({t}-\pi\right)}\right)}}+{H}{\left({t}-{3}\pi\right)}\frac{{e}^{{-{4}{\left({t}-{3}\pi\right)}}}}{ \sin{{\left({5}{\left({t}-{3}\pi\right)}\right)}}}\right]}\)
Step 6
Hence, the solution of the given initial value problem is
\(\displaystyle{y}{\left({t}\right)}=\frac{1}{{5}}{\left[{H}{\left({t}-\pi\right)}{e}^{{-{4}{\left({t}-\pi\right)}}} \sin{{\left({5}{\left({t}-\pi\right)}\right)}}+{H}{\left({t}-{3}\pi\right)}\frac{{e}^{{-{4}{\left({t}-{3}\pi\right)}}}}{ \sin{{\left({5}{\left({t}-{3}\pi\right)}\right)}}}\right]}\)
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