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I have this second-order ode equation:

$y''-4y'+13y=0$

I've identified it as a x missing case as $y''=f(y',y)=4y'-13y$, so I'm substituting with: $y'=P, y''=P\frac{dy^2}{d^2x}=f(P,y)=4P-13y$.

At this point I have $P\frac{dP}{dy}=4P-13y$, 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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  • $\begingroup$ This is a second order linear ODE with constant coefficents --- see here and this search, if needed, for some internet references to what the answers here are telling you. $\endgroup$ Jun 16, 2018 at 16:40
  • $\begingroup$ What have you learned in your course? What do you know about solving second-order equations? What do you know about linear equations with constant coefficients? Without knowing your background, it's hard to give an answer that works best for you. $\endgroup$
    – Dylan
    Jun 16, 2018 at 16:47
  • $\begingroup$ I know how to solve $y''+y=0$ via integration. Your problem requires a bit more thinking... $\endgroup$ Jun 16, 2018 at 16:50

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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, $$ y' = e^{2t}(w'+2w) \qquad \& \qquad y'' = e^{2t}(w''+4w'+4w) $$ substituting into $y''+4y'+9y=0$ 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: $$ t = \pm \int \frac{dw}{\sqrt{ 9C^2-9w^2}} = \frac{\pm}{3|C|} \int \frac{dw}{\sqrt{ 1-w^2/C^2}}$$ Let $u = w/C$ hence $du = dw/C$ and $\displaystyle \int \frac{du}{\sqrt{1-u^2}} = \sin^{-1}(u)+c_1$. Consequently, $$ t = \frac{\pm C}{3|C|}\left(\sin^{-1}(u)+c_1 \right)$$ Or, $$ \sin^{-1}(w/C) = \pm 3t-c_1 $$ Yielding, $$ w = C \sin(\pm 3t-c_1) = A\sin(3t+ \phi)$$ Since $y=e^{2t}w$ we conclude, $$ y = Ae^{2t}\sin(3t+\phi) = c_2e^{2t} \sin(3t)+c_3e^{2t} \cos(3t)$$ This argument can be easily modified to solve any problem of the form $[(D-\alpha)^2+\beta^2][y]=0$. In fact, if we are willing to chase a few more signs then we also may use an argument such as this to derive solutions to $[(D-\alpha)^2-\beta^2][y]=0$ of the form $y = c_1e^{\alpha t}\cosh(\beta t)+c_2e^{\alpha t}\sinh(\beta t)$. Naturally, in-between these we face problems of the form $(D-\alpha)^2[y] = 0$ and the substitution of $y = e^{\alpha t}w$ shows $(D-\alpha)^2[y] = (D-\alpha)^2[e^{\alpha t}w] = e^{\alpha t}w''=0$ which means $w''=0$. Integrate twice, $w= c_1+c_2t = e^{-\alpha t}y$ hence $y = c_1e^{\alpha t}+c_2te^{\alpha t}$. Most introductory DEqns courses do not emphasize these arguments since there are more economical algebraic methods to simply derive the needed solutions. The other answers indicate the algebraic methods, but I think your course probably wants something more like I show here.

Also, I also got stuck on a non-linear first order problem when I directly applied my trick, so, I tend to think you are not wrong. Usually, if I ask students to solve second order ODEs by a substitution which reduces order then I make sure the problem is simple enough to see through the needed integration etc. In some sense, this is just integration so remember all these techniques boil down to educated guessing at some level. For that reason, don't dismiss the algebraic techniques. Just because we don't work something out from first or second semester calculus that doesn't make it any less valid. It just makes it less familiar.

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  • $\begingroup$ Thanks for your thorough explanation and the nice substitution. My course mentioned both the quick solution using auxiliary equation and the substitution technique. Somehow I didn't immediately realise the former is faster considering the problem is just a constant coefficients ODE. $\endgroup$
    – neilli1992
    Jun 16, 2018 at 21:46
  • $\begingroup$ glad to help, I'm curious which textbook is your course using? $\endgroup$ Jun 16, 2018 at 22:42
  • $\begingroup$ We're not using a specific textbook. I'm taking a course targeting in quantitative finance domain, most candidates are from various backgrounds (IT, bank, etc), not necessarily with a strong math background. We're using a set of slides compiled by the lecturers themselves. Speaking of this, do you have any recommendation on differentiation books, but focusing on introducing solving techniques / practical applications, rather than deriving all the underlying theories? $\endgroup$
    – neilli1992
    Jun 17, 2018 at 13:17
  • $\begingroup$ I think that describes almost all the books these days. Certainly an old edition of Zill, or Nagle Saff and Snider, or Boyce and Diprima are all worth the $10 or less dollars you can spend to get these on your shelf. My notes are at supermath.info/DEqns2017.pdf as well as a bunch of other stuff at supermath.info/DEqns.html $\endgroup$ Jun 17, 2018 at 13:51
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We have auxilliary equation: $$m^2-4m+13=0\to m=2\pm3i$$ Thus the general solution is: $y=e^{2x}(A\cos(3x)+B\sin(3x))$

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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 $e^{\lambda x}$ for some constant $\lambda$. Substitute $y(x) = e^{\lambda x}$ into the ODE :

$$\lambda^2e^{\lambda x} - 4\lambda e^{\lambda x} + 13e^{\lambda x}=0 \Rightarrow e^{\lambda x}(\lambda^2-4\lambda+13) = 0 \implies \lambda = 2\pm 3i$$

The roots $\lambda = 2 \pm 3i$ give $y_1(x) = c_1e^{(2+3i)x}$ and $y_2(x) = e^{(2-3i)x}$. The general solution is the sum of the above equations :

$$y(x) = y_1(x) + y_2(x) \Rightarrow y_g(x) = c_1e^{(2+3i)x} + c_2 e^{(2-3i)x}$$

Now, by applying Euler's Identity $e^{a+ib} = e^a\cos b + ie^a \sin b$ and regrouping terms, you can yield a final form for the solution :

$$y(x) =c_1e^{2x}\cos(3x) + c_2e^{2x}\sin(3x)$$

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    $\begingroup$ "$x$ missing" means an autonomous differential equation which has nothing to do with order $\endgroup$
    – Dylan
    Jun 16, 2018 at 16:45
  • $\begingroup$ Hi @Dylan, x missing, it's a substitution technique meant to reduce the second-order ODE to a simpler first-order ODE. I found a page on this: copingwithcalculus.com/reduction-of-order.html. My problem was similar to example 4 on that page, but that example reduced to a linear first-order equation, which is simpler to solve. Unfortunately my problem became a non-linear first-order equation, it actually gets harder. Admittedly I didn't immediately realise the original ODE is a constant coefficient one, in which case using the auxiliary equation is much simpler... $\endgroup$
    – neilli1992
    Jun 16, 2018 at 21:37

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