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When we are solving a second order ODE with constant coefficients, and when the roots of the characteristic equation are equal, we let $y_1(t)=e^{rt}$ then let $y_2(t)=v(t)y_1(t)$, and find $v(t)$. Why can we let the solution in such way? Is there any reason behind such guessing?

Another question is similar, which is about method of variation of parameters. Why can we guess the particular solution in such a from $\psi=v_1y_1+v_2y_2$? Is there any reason behind such guessing? Also why can we impose a condition that $v_1'y_1+v_2'y_2=0$ It seems that this condition comes from nowhere. What if we let the R.H.S of $v_1'y_1+v_2'y_2=0$ to be some constant other than $0$, or even some functions $\phi(t)$? Can we still get the same result?

Thank you!

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You can at least sketch an answer to the "multiply $t,t^2,\dots,t^{m-1}$ where $m$ is the multiplicity" question by considering a family of IVPs where two roots are approaching one another. Consider the IVPs

$$y''-(1+a)y'+ay,y(0)=0,y'(0)=1$$

where $a$ is approaching $1$. Let us solve this whenever $a$ is not $1$. The roots of the characteristic polynomial are $1$ and $a$, so the general solution is $c_1 e^{t} + c_2 e^{at}$. To solve the IVP we have to solve the system of equations

$$\begin{bmatrix} 1 & 1 \\ 1 & a \end{bmatrix} \begin{bmatrix} c_1 \\ c_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \end{bmatrix}.$$

The solution to this system is $\begin{bmatrix} \frac{-1}{a-1} \\ \frac{1}{a-1} \end{bmatrix}$. Thus the solution to this IVP is $\frac{e^{at} - e^{ t}}{a-1}$. Now compute $\lim_{a \to 1}$ of that. You'll find the familiar $te^t$. Regular perturbation theory then tells us that $te^t$ solves the IVP $y''-2y'+y=0,y(0)=0,y'(0)=1$, as you can of course check directly.

Of course you would not want to do this in the general situation. But this hints at trying it. Once you see that, to see why it actually works in the general situation, you can note that when you apply a linear differential operator with constant coefficients to a polynomial of degree $d$ times an exponential, you get a polynomial times that same exponential. The degree of the resulting polynomial just turns out to be $d-m$. So if you go up to polynomials of degree $d+m$ you can make the resulting polynomial be an arbitrary polynomial of degree $d$.

Another perspective is offered by linear algebra. Here the factors of $t$ arise when you take the matrix exponential of a Jordan block. For example, one can explicitly calculate $e^{At}$ where $A=\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$ using the series expansion. You get $\begin{bmatrix} e^t & t e^t \\ 0 & e^t \end{bmatrix}$. Now the matrix associated to $y''-2y'+y=0$ is $\begin{bmatrix} 0 & 1 \\ -1 & 2 \end{bmatrix}$ which has the Jordan form above. You can do the same for higher order Jordan blocks.

I'm not so sure how to give a sketch of why variation of parameters should work, though. One thing that might help a little bit is to realize that the only ad hoc part of variation of parameters is the idea of multiplying the homogeneous solutions by variable coefficients. Everything else follows by just following through the algebra/calculus. Typically in textbooks this "everything else" is taught as part of the method itself because this algebra is actually a little messy. That's great for ease of just making the method work, not so great for understanding where it came from.

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There actually is a very convincing reason for the Variation of parameters guess. Let's have a look at a the first order homogenous problem: $$ y ' + p y = 0 $$ You definitely learned the separation of variables method to solve an ODE like this (if $p $ is nice enough), and you found $$ y(t) = A \exp \left (- \int p dt \right)$$ Then consider the first order non-homogeneous problem: $$y' + py = g$$ using an integrating factor $$ \mu(t) = \exp \left ( \int p dt \right) $$ you are able to write $$ y(t) = \frac{1}{\mu(t)} \int _{t_0}^t \mu(s) g(s) ds= \left [\int _{t_0}^t \mu(s) g(s) ds\right ] * \exp \left (- \int p dt \right) $$ If you compare the two, we see the only difference is $$ A = \left [\int _{t_0}^t \mu(s) g(s) ds\right ]$$ Eventually you'll deal with first order systems, and it's more or less the exact same computation as above. Let's check. Assume we have the homogeneous linear system (with $M$ nice enough $n \times n$ matrix of functions only depending on $t$) (you may not have reached this yet, let me know if you'd like clarification ) $$ \dot x = M x $$ You'll find $n$ solutions, and be able to call them $$ x(t) = c_1 x_1+ \ldots +c_n x_n =X(t) c$$ where $c = (c_1,\ldots,c_n)$ and $X(t) = ( x_1, \ldots,x_n)$. Now assume you'd like to solve $$\dot x = Mx + g$$ Just as in the first order case, we now vary the parameter $$ x(t) = X(t) c(t) $$ Thus $$ \dot x = \dot X c + X \dot c = MX c + X\dot c = MXc + g \implies X \dot c = g \implies c(t) = \int X^{-1} g dt $$ basically your textbook is showing you weird results to get to the punchline without having some of the more natural tools. In fact, notice that $\det X(t) = $ Wronskain, that one always seemed weird to introduce without using systems. Also, once you have the wronskian, notice if you have $n-1$ of the solutions it gives you an ODE to solve for the last one. Which is related to the above method. For example, for $y'' + py' + qy =0$ and if $y_1$ is known, then by Abel's fact $W' +pW =0$ so $$ W = y_1 y_2 ' - y_2 y_1' = A \exp ( - \int p dt ) \implies \left (\frac{ y_2}{y_1} \right) '=\frac{y_2'}{y_1} - \frac{ y_2 y_1 '}{y_1^2} = \frac{A}{y_1^2}\exp ( - \int p dt ) $$ $$\implies y_2(t) = y_1 \int \frac{ \exp ( - \int p dt ) }{y_1^2} dt$$

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  • $\begingroup$ I get lost in the content of homogeneous linear system since I haven't learnt this part yet, would you try to elaborate more? I have learnt how to use Abel's theorem to find another solution $y_2$ when another solution $y_1$ is given, but it seems like the method of reduction of order and Abel's theorem are two completely different things. And yeah, the introduction of Wronskain is quite weird since it comes from nowhere again. Is this just something mathematicians find accidentally, or is there actually a smooth logic behind to introduce this determinant? $\endgroup$ – mathshungry Jun 6 '16 at 17:02
  • $\begingroup$ The Wronskian comes from the question: Are the solutions to my system linearly independent? You can group the $n$ solutions (the reason you can find $n$ solutions is due to linear algebra) as $X(t)$ and people usually call it the fundamental matrix to the system. Thus if $\det X \neq 0$, the solutions are linearly independent. $\endgroup$ – Jeb Jun 7 '16 at 4:58

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