How to solve this second order ODE question? 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?
 A: We have auxilliary equation:
$$m^2-4m+13=0\to m=2\pm3i$$
Thus the general solution is:
$y=e^{2x}(A\cos(3x)+B\sin(3x))$
A: 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 :
$$y(x) =c_1e^{2x}\cos(3x) + c_2e^{2x}\sin(3x)$$
A: 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. 
