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

maduregimc 2021-12-16 Answered
I have this second-order ode equation:
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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Answered 2021-12-17 Author has 5612 answers
We have auxilliary equation:
Thus the general solution is:
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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 :
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 :
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 :
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,
substituting into \(t''+4y'+9y=0\) yields:
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
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:
Let \(u=w/C\) hence \(du=dw/C\) and
\(\int\frac{du}{\sqrt{1-u^2}}=\sin^{-1}(u)+c_1\). Consequently,
\(t=\frac{\pm C}{3|C|}(\sin^{-1}(u)+c_1)\)
Since \(y=e^{2t}w\) we conclude,


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