Question

How to solve this equation \(y'''-4y''+2y'-16y=4x+1\) using Method of Undetermined Coefficient, Variation of Parameters and Laplace Transformation

Answer 1

Solve:
The equation is \(y'''-4y''+2y'-16y=4x+1\)
Variation of Parameters:
The homogeneous equation is \(y'''-4y''+2y'-16y=0\)
The homogeneous solution is \(\displaystyle{y}_{{h}}={c}_{{1}}{e}^{{{4.37796}{x}}}+{e}^{{-{0.18898}{x}}}{\left({c}_{{2}} \cos{{\left({1.90236}{x}\right)}}+{c}_{{3}} \sin{{\left({1.90236}{x}\right)}}\right)}\)
Obtain the Wro
ians:
\(\displaystyle{W}={\left|\begin{matrix}{e}^{{{a}{x}}}&{e}^{{{b}{x}}} \cos{{c}}{x}&{e}^{{{b}{x}}} \sin{{c}}{x}\\{a}{e}^{{{a}{x}}}&{b}{e}^{{{b}{x}}} \cos{{c}}{x}-{c}{e}^{{{b}{x}}} \sin{{c}}{x}&{b}{e}^{{{b}{x}}} \sin{{c}}{x}-{c}{e}^{{{b}{x}}} \cos{{c}}{x}\\{a}^{2}{e}^{{{a}{x}}}&{\left({b}^{2}-{c}^{2}\right)}{e}^{{{b}{x}}} \cos{{c}}{x}-{2}{b}{c}{e}^{{{b}{x}}} \sin{{c}}{x}&{\left({b}^{2}-{c}^{2}\right)}{e}^{{{b}{x}}} \sin{{c}}{x}-{2}{b}{c}{e}^{{{b}{x}}} \cos{{c}}{x}\end{matrix}\right|}\)
\(\displaystyle{W}_{{1}}={\left|\begin{matrix}{0}&{e}^{{{b}{x}}} \cos{{c}}{x}&{e}^{{{b}{x}}} \sin{{c}}{x}\\{0}&{b}{e}^{{{b}{x}}} \cos{{c}}{x}-{c}{e}^{{{b}{x}}} \sin{{c}}{x}&{b}{e}^{{{b}{x}}} \sin{{c}}{x}-{c}{e}^{{{b}{x}}} \cos{{c}}{x}\\{1}&{\left({b}^{2}-{c}^{2}\right)}{e}^{{{b}{x}}} \cos{{c}}{x}-{2}{b}{c}{e}^{{{b}{x}}} \sin{{c}}{x}&{\left({b}^{2}-{c}^{2}\right)}{e}^{{{b}{x}}} \sin{{c}}{x}-{2}{b}{c}{e}^{{{b}{x}}} \cos{{c}}{x}\end{matrix}\right|}\)
The values of \(\displaystyle{a}={4.37796},{b}=-{0.18898}\ \text{ and }\ {c}={1.90236}\)
Further we have
\(\displaystyle{W}_{{2}}={\left|\begin{matrix}{e}^{{{a}{x}}}&{0}&{e}^{{{b}{x}}} \sin{{c}}{x}\\{a}{e}^{{{a}{x}}}&{0}&{b}{e}^{{{b}{x}}} \sin{{c}}{x}-{c}{e}^{{{b}{x}}} \cos{{c}}{x}\\{a}^{2}{e}^{{{a}{x}}}&{1}&{\left({b}^{2}-{c}^{2}\right)}{e}^{{{b}{x}}} \sin{{c}}{x}-{2}{b}{c}{e}^{{{b}{x}}} \cos{{c}}{x}\end{matrix}\right|}\)
\(\displaystyle{W}_{{3}}={\left|\begin{matrix}{e}^{{{a}{x}}}&{e}^{{{b}{x}}} \cos{{c}}{x}&{0}\\{a}{e}^{{{a}{x}}}&{b}{e}^{{{b}{x}}} \cos{{c}}{x}-{c}{e}^{{{b}{x}}} \sin{{c}}{x}&{0}\\{a}^{2}{e}^{{{a}{x}}}&{\left({b}^{2}-{c}^{2}\right)}{e}^{{{b}{x}}} \cos{{c}}{x}-{2}{b}{c}{e}^{{{b}{x}}} \sin{{c}}{x}&{1}\end{matrix}\right|}\)
The particular solution of the equation is
\(\displaystyle{y}_{{p}}={e}^{{{4.37796}{x}}}\int\frac{{{\left({4}{x}+{1}\right)}{W}_{{1}}}}{{W}}{\left.{d}{x}\right.}+{e}^{{-{0.18898}{x}}}{\left( \cos{{\left({1.90236}{x}\right)}}\int\frac{{{\left({4}{x}+{1}\right)}{W}_{{2}}}}{{W}}{\left.{d}{x}\right.}+ \sin{{\left({1.90236}{x}\right)}}\int\frac{{{\left({4}{x}+{1}\right)}{W}_{{3}}}}{{W}}{\left.{d}{x}\right.}\right)}\)
Therefore, \(\displaystyle{y}={c}_{{1}}{e}^{{{4.37796}{x}}}+{e}^{{-{0.18898}{x}}}{\left({c}_{{2}} \cos{{\left({1.90236}{x}\right)}}+{c}_{{3}} \sin{{\left({1.90236}{x}\right)}}\right)}+{e}^{{{4.37796}{x}}}\int\frac{{{\left({4}{x}+{1}\right)}{W}_{{1}}}}{{W}}{\left.{d}{x}\right.}+{e}^{{-{0.18898}{x}}}{\left( \cos{{\left({1.90236}{x}\right)}}\int\frac{{{\left({4}{x}+{1}\right)}{W}_{{2}}}}{{W}}{\left.{d}{x}\right.}+ \sin{{\left({1.90236}{x}\right)}}\int\frac{{{\left({4}{x}+{1}\right)}{W}_{{3}}}}{{W}}{\left.{d}{x}\right.}\right)}\)
Laplace Transformation:
Apply Laplace transform:
\(\displaystyle{L}{\left\lbrace{y}{'''}-{4}{y}{''}+{2}{y}'-{16}{y}\right\rbrace}={L}{\left\lbrace{4}{x}+{1}\right\rbrace}\)
\(\displaystyle{s}{3}^{Y}{\left({s}\right)}-{s}^{2}{y}{\left({0}\right)}-{s}{y}'{\left({0}\right)}-{y}{''}{\left({0}\right)}-{4}{s}^{2}{Y}{\left({s}\right)}+{4}{s}{y}{\left({0}\right)}+{4}{s}{y}'{\left({0}\right)}+{2}{s}{Y}{\left({s}\right)}-{y}{\left({0}\right)}-{16}{Y}{\left({s}\right)}=\frac{4}{{s}^{2}}+\frac{1}{{s}}\)
In order to simplify above equation, we need the initial values y(0), y'(0), y''(0) so that we obtain a function of s on which we can apply inverse Laplace transform easily.
Thus, a solution of the equation \(\displaystyle{y}{'''}-{4}{y}{''}+{2}{y}'-{16}{y}={4}{x}+{1}\) is obtained.