I have this second-order ode equation: y''-4y'+13y=0Z

maduregimc 2021-12-16 Answered
I have this second-order ode equation:
\(\displaystyle{y}{''}-{4}{y}'+{13}{y}={0}\)
I've identified it as x missing case as \(\displaystyle{y}{''}={f{{\left({y}',{y}\right)}}}={4}{y}'-{13}{y}\), so I'm substituting witth:
\(\displaystyle{y}'={P},\ {y}{''}={P}{\frac{{{\left.{d}{y}\right.}^{{2}}}}{{{d}^{{2}}{x}}}}={f{{\left({P},{y}\right)}}}={4}{P}-{13}{y}\)
At this point I have \(\displaystyle{P}{\frac{{{d}{p}}}{{{\left.{d}{y}\right.}}}}={4}{P}-{13}{y}\), which seems a non-linear first-order ODE. This is currently beyond the scope of my course, so I'm unsure if I should continue and search online for solving techniques, or did I already do something wrong?

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movingsupplyw1
Answered 2021-12-17 Author has 5612 answers
We have auxilliary equation:
\(\displaystyle{m}^{{2}}-{4}{m}+{13}={0}\to{m}={2}\pm{3}{i}\)
Thus the general solution is:
\(\displaystyle{y}={e}^{{{2}{x}}}{\left({A}{\cos{{\left({3}{x}\right)}}}+{B}{\sin{{\left({3}{x}\right)}}}\right)}\)
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movingsupplyw1
Answered 2021-12-18 Author has 5612 answers
What do you mean as a x missing case ? This is a second order linear differential equation, not a first order one.
Assume a solution is proportional to \(\displaystyle{e}^{{\lambda{x}}}\) for some constant \(\displaystyle\lambda\). Substitute \(\displaystyle{y}{\left({x}\right)}={e}^{{\lambda{x}}}\) into the ODE :
\(\displaystyle\lambda^{{2}}{e}^{{\lambda{x}}}-{4}\lambda{e}^{{\lambda{x}}}+{13}{e}^{{\lambda{x}}}={0}\Rightarrow{e}^{{\lambda{x}}}{\left(\lambda^{{2}}-{4}\lambda+{13}\right)}\)
\(\displaystyle={0}\Rightarrow\lambda={2}\pm{3}{i}\)
The roots \(\displaystyle\lambda={2}\pm{3}{i}\) give \(\displaystyle{y}_{{1}}{\left({x}\right)}={c}_{{1}}{e}^{{{\left({2}+{3}{i}\right)}{x}}}\) and \(\displaystyle{y}_{{2}}{\left({x}\right)}={e}^{{{\left({2}-{3}{i}\right)}{x}}}\). The general solution is the sum of the above equations :
\(\displaystyle{y}{\left({x}\right)}={y}_{{1}}{\left({x}\right)}+{y}_{{2}}{\left({x}\right)}\Rightarrow{y}_{{g{{\left({x}\right)}}}}={c}_{{1}}{e}^{{{\left({2}+{3}{i}\right)}{x}}}+{c}_{{2}}{e}^{{{\left({2}-{3}{i}\right)}{x}}}\)
Now, by applying Euler's Identity
\(\displaystyle{e}^{{{a}+{i}{b}}}={e}^{{a}}{\cos{{b}}}+{i}{e}^{{a}}{\sin{{b}}}\) and regrouping terms, you can yield a final form for the solution :
\(\displaystyle{y}{\left({x}\right)}={c}_{{1}}{e}^{{{2}{x}}}{\cos{{\left({3}{x}\right)}}}+{c}_{{2}}{e}^{{{2}{x}}}{\sin{{\left({3}{x}\right)}}}\)
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RizerMix
Answered 2021-12-29 Author has 9349 answers

In order to make your problem solvable in the manner in which you seek I think a substitution of \(y=e^{2t}w\) is required. Here w is the new dependent variable. We calculate,
\(y'=e^{2t}(w'+2w)\)
\(y''=e^{2t}(w''+4w'+4w)\)
substituting into \(t''+4y'+9y=0\) yields:
\(e^{2t}(w''+9w)=0\)
hence solve the much easier problem \(w''+9w=0\) by a technique like the one you mention. Personally, I prefer the following trick \(w''=v\frac{dv}{dw}\ where\ v=\frac{dw}{dt}\)
Thus, \(w''+9w=0\) yields
\(v\frac{dv}{dw}+9w=0\)
or
\(vdv+9wdw=d(v^2/2+9w^2/2)=0\)
hence \(v^2+9w^2=9C^2\) is the solution. Solve for \(v=\pm\sqrt{9C^2-9w^2}\) But \(v=\frac{dw}{dt}\) hence we reduce to the quadrature:
\(t=\pm\int\frac{dw}{\sqrt{(C^2-9w^2}}=\frac{\pm}{3|C|}\int\frac{dw}{\sqrt{1-w^2/C^2}}\)
Let \(u=w/C\) hence \(du=dw/C\) and
\(\int\frac{du}{\sqrt{1-u^2}}=\sin^{-1}(u)+c_1\). Consequently,
\(t=\frac{\pm C}{3|C|}(\sin^{-1}(u)+c_1)\)
Or,
\(\sin^{-1}(w/C)=\pm3t-c_1\)
Yielding,
\(w=C\sin(\pm3t-c_1)=A\sin(3t+\phi)\)
Since \(y=e^{2t}w\) we conclude,
\(y=Ae^{2t}\sin(3t+\phi)=c_2e^{2t}\sin(3t)+c_3e^{2t}\cos(3t)\)

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