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I want to solve this equation:


I did this:

$y' = z$

$y'' = z\dfrac{dz}{dy}$








now, I'm pretty sure I did something wrong. could you please correct.

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How did you write $y'' = z dz/dy$ ? –  svenkatr Dec 8 '10 at 0:00
$y''=\dfrac{dy'}{dx}=\dfrac{dz}{dx}=\dfrac{dz}{dy}.\dfrac{dy}{dx}=z\dfrac{dz}{dy‌​}$ –  f.d Dec 8 '10 at 0:03
What method(s) have you been taught for how to do this? For example, do you know what characteristic polynomials are? –  Qiaochu Yuan Dec 8 '10 at 0:06
in this equation, since we don't have x we would consider $y'=z$ and solve the equation like a first order differential equation. –  f.d Dec 8 '10 at 0:10
Your error is from the 5th to the 6th line after "I did this", where you cancel $dz$ off the left and $dy$ off the right. –  Ross Millikan Mar 21 '11 at 0:28

8 Answers 8

up vote 10 down vote accepted

HINT $\rm\ \ \ 0\ =\ y'' + 2\ y' - 3\ y\ =\ (D^2 + 2\ D -3)\ y\ =\ (D+3)\ (D-1)\ y, \quad D = d/dx $

i.e. factor the differential operator as you would any polynomial.

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@Bill: I am sure if the OP understood what this meant he or she would have no problem with this question. –  Qiaochu Yuan Dec 8 '10 at 0:07
@Qiaochu: My experience with similar questions on sci.math and elsewhere over the decades shows that hints like that often suffice. Of course you are welcome to give your own answer if you like. –  Bill Dubuque Dec 8 '10 at 0:59
@Bill: +1 I think it is a little humorous that an answer which shouldn't make any sense was chose by the person who asked it as the best answer. =] –  Tyler Clark Dec 8 '10 at 4:43
@night - That notation is not at all uncommon. I've encountered it in 3 texts for undergrad study, and it's one of the most useful techniques discussed in the differential equations "Problem Solvers" book -- with plenty of examples given, albeit more awkwardly typed. –  Brian Vandenberg Apr 19 '11 at 15:59
@night Operator-theoretic methods in the theory of differential and difference equations are not only very natural but also very powerful and beautiful. I cannot imagine why anyone would find them "weird" or "awkward". Moreover, in my experience, they are much more easily comprehended by students than are alternative approaches. –  Bill Dubuque Apr 19 '11 at 16:11

So we see that we have a Linear Second Order Differential Equation. There are several variety of

methods to solve these type of equations which some are listed, but I think this method is the most

straight-forward one and does not require to much trickery or cleverness, its a pretty

step-by-step process.

HINT:$~~$ We know for these type of equations this must hold true:

$$A(x)\cdot y''+B(x)\cdot y'+(x)\cdot y = D(x)$$

: Dividing by $A(x)$ gives:

$$y''+p(x)\cdot y'+q(x)\cdot y = f(x)$$

$\underline {Our~~Case:}$ $~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$ Homogeneous $2^{nd}$ order differential

$$y''+p(x)\cdot y'+q(x)\cdot y = 0$$

So for our case here we have the following: $$y''+2y'-3y = 0$$

I claim that this is not too bad of a general solution to obtain:

I am assuming you learned something about the characteristic equation thus far.

So for starts, try using $\underline{y = e^{rx}}$ for our problem and try to work it out. Gonna also assume you

know what the general solution form would like that:

$$y(x) = c_{1}e^{r_{1}x}+c_{2}e^{r_{2}x}+c_{3}e^{r_{3}x}+...+c_ne^{r_{n}x}$$

Let me know how that works out.

P.S. NOTE.... $p(x)$ and $q(x)$ here is any function of the variable $x$ times the derivative of the function $y$ $(y')$ and the function $y$ itself. This means that $p(x)$ and $q(x)$ can also just be constants, which is what we have in our case, in case the notation is a bit confusing.

So now solving the equation for understanding:

Solution: Let $y=e^{rx}$, then plug this into the differential equation. This would lead us to taking derivatives of $'y'$, but for now we will just plug it in to see what it will look like. Doing so leads us to the following:



So now we see that we must take derivatives of each of the exponential terms that has primes connected to them. So doing this leads us to:

$$y(x)=e^{rx}$$ $$y'(x)=re^{rx}$$ $$y''(x)=r^{2}e^{rx}$$

So now substituting the derivatives of $y$ into the original ODE we can see that we get the following:


Now let's ask ourselves, what do we see in common in this new equation with the derivatives of our initial guess $y=e^{rx}$ to the homogeneous ODE? Well if you see it or don't see it let me point something out. All of the terms has our original guess $e^{rx}$ in them, so what we can do with that term is what we can do with any algebraic expression and that is factor it out of the equation as a common term as follows:


Now what all of this work has led us up to is something called the Characteristic Equation or better known as Characteristic Polynomial, also can be referred to as the Auxiliary Equation which is what you see inside of the parenthesis as:


To further simplify the equation we can divide out entirely by $e^{rx}$ which would just leave us with our Characteristic Polynomial being:


NOTE: All of the work we did to arrive at our Characteristic Polynomial can be saved in the future whenever we have constant coefficient ODE's, because the initial guess will always be divided out at the end. So here is the trick.

1.) Observe the order of the ODE (namely, highest derivative in the equation).

2.) Start off with the highest derivative and work your way down to the lowest term, (namely, being constant terms).

3.) Every term with a prime with have a $r^{m}$ corresponding to it, with (m = the amount of primes on that specific term) in the ODE, namely: $$y'=r^{1},~y''=r^{2},~y'''=r^{3},~ y^{(4)}=r^{4},~y^{(5)}=r^{5},~\ldots,~y^{(n)}=r^{n}.$$

4.) Any constants just multiply by the $'r'$ or sometimes used $'\lambda'$ terms.

5.) For terms not involving any primes, such as $'y'$, they will become just one. Because if we think back to the beginning, we used as our initial guess $y=e^{rx}$, so once we factored out $e^{rx}$ we will just have a one in the place where $e^{rx}$ once used to sit inside of the equation.

So for example, lets take a look at this ODE: $y^{(4)}+2y''+y=0$

The Characteristic Polynomial we seek will just be: $$r^{4}+2r^{2}+1=0$$

I know seeing is believing!, so I have some few examples for you to try out of yourself (not to fully solve) but to just see if you can carry out getting the Characteristic Polynomial by using the short method that I listed above:


$i$ ) $$s''-4s'+4s=0$$ $ii$ ) $$i''+60i'+500i=0$$ $iii$ ) $$y'''-3y''-y'+3y=0$$ $iv$ ) $$x^{(8)}-5x^{(4)}+4x'=0$$ $v$ ) $$v''-6v'+13v=0$$

So not to lose focus of what our goal here was, and that was to solve $y''+2y'-3y = 0$. So using all of our knowledge gained thus far, we can apply it now to our problem. So by doing so we can conclude that our Characteristic Polynomial will be as stated previously: $$r^{2}+2r-3=0$$

From here we want to take the following measures in this order:

1) Factor is possible.

2) If factoring is not possible, use quadractic formula. If you forgot let me remind you: $$r = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}.$$

3) If both of above if not possible (usually for cubic or higher powers) then using long division/synthetic division and usually helpful by graphing the equation first and finding at least one zero and use that factor and take the original polynomial and try long division/synthetic division. So for example if you found $2$ as a zero of the polynomial, then you would use $(r-2)$ to try and reduce the polynomial down to a quadratic if possible or to something that can be factored.

But for our problem luckily, it is factor-able as so:


With real-distinct roots being: $$\text{ Characteristic Roots are: }~ r=1,~r=-3$$

So with this known, we could put together our general solution to the ODE, which was stated above, ( namely being,$~~~~~~~~ y(x) = c_{1}e^{r_{1}x}+c_{2}e^{r_{2}x}+c_{3}e^{r_{3}x}+...+c_ne^{r_{n}x}~$).

So our general solution will be a linear combination of our homogeneous solutions of our characteristic roots $r_{1} \text{ and }r_{2}$.

$~~~~~~~~~~~~~~~~~~~~~~~~~\therefore~~~~~~~~$ $y(x)=C_{1}e^{x}+C_{2}e^{-3x}$.

No we have found our solution to the ODE, but this only a general solution because we still have undetermined coefficients in the solution (namely, $C_{1} \text{ and } C_{2}$). In order to find these, we will need to be provided with initial conditions which will then make the problem become an IVP (Initial Value Problem). So wrapping this all together, this is what we would do to solve any such $2^{nd}$ Order Linear Constant-Coefficients Homogeneous Ordinary Differential Equations, and the same applies for Higher Order.

There are more cases beyond this one of real distinct-roots. We will go over now what those other cases are.

The other type of situation that you can run into is when you have real non-distinct roots (i.e., having multiplicity of two or greater). Now for the word multiplicity, all that means is simply the following: If you get some numbers that appear more than once (meaning that they are exactly the same).

$\underline{Example~1:}$ If we had this characteristic polynomial: $~r^{2}+6r+9=0$, our roots to this polynomial will be the following: $r=-3,~r=-3$. This is a prime example of what multiplicity means. Our roots would be $r=-3$ (with multiplicity 2) because the same exact root appeared twice, this will hender having to multiply the linear combination of solution by an appropriate amount of powers of $'x'$ or $'t'$ whichever is your independent variable in the ODE. Let's take a look at a few more examples.

$\underline{Example~2:}$ If we had this characteristic polynomial: $~5r^{4}+3r^{3}=0$, our roots to this particular polynomial will be the following: $r=0,~r=0,~r=0,~r=-\dfrac{3}{5}$ This is yet another great example of when we would make the conclusion that we have yet again repeated roots being $r=0$ (now with multiplicity 3). Once again, you would have to multiply the independent homogeneous solutions by an appropriate power of your independent variable, which in our case we are using $'x'$. Let's take a look at one more example for exercise to get ourselves more comfortable with the idea. Also, by all means, please do take the time if you can and verify these characteristic roots to the polynomials to see if you get the same results and take it one step further and see if you can combine the homogeneous solutions into a linear combination of solutions to form the general solution correctly. So now, moving to our next example.

$\underline{Example~3:}$ If we had this characteristic polynomial: $~9r^{3}+12r^{2}+4r=0$, our roots to this particular polynomial will be the following: $r=0,~r=-\dfrac{2}{3},~r=-\dfrac{2}{3}$. This is once again yet another great example of when we would make the conclusion that we have found repeated roots in our polynomial being $r=-\dfrac{2}{3}$ (with multiplicity 2). Once again, you would have to multiply the independent homogeneous solutions by an appropriate power of your independent variable in the differential equation. So this pretty much covers the case of what will take place if we were to encounter such a polynomial that had non-distinct roots.

I hope this helps out. Let me know if there is something you need to be clarified a bit further. Thanks.


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Perhaps you could add the case when the (two) characteristics roots are equal. –  Américo Tavares Apr 19 '11 at 17:26
@Americo: Sure thing I will. Thanks for that reminder. I guess I could go over all the cases: Such as complex roots and repeated roots. What do you suggest? –  night owl Apr 20 '11 at 3:04
perhaps just the case of repeated real roots. –  Américo Tavares Apr 20 '11 at 10:49
@Americo: Yea you are right, I was thinking that would be too much for one to swallow in one thread. :) –  night owl Apr 20 '11 at 11:09

For this kind of differential equations Laplace transform is handy:

$$ {\cal L}[y''] + 2 {\cal L}[y'] -3{\cal L}[y] = 0 \ \Longleftrightarrow \ (s^2{\cal L}[y] -sy(0) - y'(0)) +2(s{\cal L}[y] -y(0)) -3{\cal L}[y] = 0 \ . $$

Now put $F(s) = {\cal L}[y]$ and $a= y(0), b= y'(0)$.

$$ (s^2F(s) -as - b) +2(sF(s) -a) -3F(s) = 0 \ \Longleftrightarrow \ (s^2 +2s -3)F(s) -as -2a -b = 0 \ . $$

We solve this equation for $F(s)$:

$$ F(s) = \dfrac{as + 2a +b}{s^2 +2s -3} = \dfrac{as + 2a +b}{(s-1)(s+3)} = \dfrac{A}{s-1} + \dfrac{B}{s+3} \ , $$

where $A$ and $B$ are constants depending on $a$ and $b$, which I let you the pleasure to compute :-) and we get

$$ y(t) = A{\cal L}^{-1}\left[ \dfrac{1}{s-1}\right] + B{\cal L}^{-1}\left[ \dfrac{1}{s+3} \right] = A e^t + B e^{-3t} \ . $$

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Start by assuming a solution of the form


Hence $y'(t)=ce^{ct}$ and $y''(t)=c^2e^{ct}$

Substitute into the original equation. You will find a quadratic for $c$. Solve it, and note that both solutions work, and as the ODE is linear and homogeneous, the sum of solutions also works.

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You can also write it in matrix form: $u=(y',y)$, $u'=\big(\matrix{-2 & 3 \\ \hphantom- 1 & 0}\big) u$. Find the eigenvalues and eigenvectors, turn it into a diagonal system whose solution is simple. Go back to the original coordinates.

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Start by finding the characteristic polynomial. In this case it is $x^2+2x-3$. Find its zeros; they are $x=-3, 1$. The general solution to your differential equation is $ce^{x_it}$, where $x_i$ is a root to your characteristic polynomial. So we have two linearly independent solutions; $y=e^{-3t}, y=e^{1t}$. Because this is a linear differential equation, we can take any linear combo of those two solutions, and that will also be a solution. So the general solution for $y$ is $$y(t)=c_1e^{-3t}+c_2e^{1t}$$, where $c_1$ and $c_2$ are determined by your initial conditions ($y(0)$ and $y'(0)$).

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Heh. Figured I'd try something different for this. Try this substitution:










$e^{3x}y=be^{4x}+c$, $b=\frac a4$


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