 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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We have auxilliary equation:
$$\displaystyle{m}^{{2}}-{4}{m}+{13}={0}\to{m}={2}\pm{3}{i}$$
Thus the general solution is:
$$\displaystyle{y}={e}^{{{2}{x}}}{\left({A}{\cos{{\left({3}{x}\right)}}}+{B}{\sin{{\left({3}{x}\right)}}}\right)}$$
Not exactly what you’re looking for? movingsupplyw1
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 :
$$\displaystyle{y}{\left({x}\right)}={c}_{{1}}{e}^{{{2}{x}}}{\cos{{\left({3}{x}\right)}}}+{c}_{{2}}{e}^{{{2}{x}}}{\sin{{\left({3}{x}\right)}}}$$ RizerMix

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,
$$y=Ae^{2t}\sin(3t+\phi)=c_2e^{2t}\sin(3t)+c_3e^{2t}\cos(3t)$$