Consider the second order linear homogeneous differential equation:

$$ax'' + bx' + c = 0$$

If we start from the assumption that the solution has the form $$ x(t) = e^{rt}$$ Then, if we solve the characteristic polynomial, then we will have 2 values for $r$, so the following solutions are valid: $$x(t) = e^{r_1t}$$ and $$x(t) = e^{r_2t}$$ So, I would expect the general solution to look like this: $$x(t) = c_1e^{r_1t} + c_2e^{r_2t}$$ But, if the roots are repeating (e.g. $r_1=r_2$), we write the solution as: $$x(t) = c_1e^{r_1t} + c_2te^{r_1t}$$ But, why? I know if we have repeating roots, then, we can factor out the constant, and we end up with a single constant multiplied by an exponential e.g., $(c_1 + c_2)e^{r_1t}$, but isn't that form still valid? I mean, can we just use that form? Why use $x(t) = c_1e^{r_1t} + c_2te^{r_1t}$ instead of $(c_1 + c_2)e^{r_1t}$.

  • $\begingroup$ The solution for repeated roots should be $x(t) = c_1 e^{r_1 t} + c_2 \mathbf{t} e^{r_1 t}$ tho. $\endgroup$
    – Yuki.F
    Commented Jul 26, 2020 at 3:14
  • $\begingroup$ @Yuki.F Right. I edited my question. Still I have the same question. $\endgroup$
    – Ralff
    Commented Jul 26, 2020 at 3:57

2 Answers 2


You are looking for the general solution. That is why only using solutions $ce^{rt}$, when $r$ is a double root, is insufficient. (Writing $(c_1 + c_2)e^{rt}$ conveys no knowledge beyond solutions of the form $ce^{rt}$.) If “solve” only meant find some solution then just give the solution $0$. Do you agree that is useless?

Remember that you want the general solution in order to find the solution fitting some initial conditions. A solution where $y(0) = 0$ and $y’(0) = 1$ can’t be $ce^{rt}$. But $te^{rt}$ is a solution fitting those initial conditions. If you refuse to consider solutions like $te^{rt}$ when $r$ is a double root then you'll never be able to solve that ODE when $y(0) = 0$ and $y'(0) = 1$.

There are many situations in math where multiple roots behave differently than distinct roots. An example in basic calculus is partial fraction decompositions. If $a \not= b$ then $$ \frac{1}{(x-a)(x-b)} = \frac{c}{x-a} - \frac{c}{x-b} $$ where $c = 1/(a-b)$, but this is not valid when $a = b$. The partial fraction decomposition for $1/(x-a)^2$ is, well, itself. There is nothing to do in that case.

Every linear second-order ODE with constant coefficients has a $2$-dimensional solution space. That property is true whether or not the roots of the quadratic polynomial are equal or distinct. But concrete formulas for a basis of the solution space are different in the cases of distinct roots and repeated roots.

Perhaps you don't understand how someone could discover the extra solution $te^{rt}$ when $r$ is a double root. Here is some motivation. In the case of distinct roots $r_1$ and $r_2$, you have solutions $c_1e^{r_1t} + c_2e^{r_2t}$. In particular, $(e^{r_1t} - e^{r_2t})/(r_1-r_2)$ is a solution. Now let $r_2 \to r_1$. By L'Hopital's rule, $$ \lim_{r_2 \to r_1} \frac{e^{r_1t} - e^{r_2t}}{r_1-r_2} = te^{r_1t}. $$ That suggests that when $r_2 = r_1$ we should check if $te^{r_1t}$ fits the ODE, and you can check it really does. Another way to think about this is that when $r_1 \not= r_2$, the function $y(t) = (e^{r_1t} - e^{r_2t})/(r_1-r_2)$ satisfies $y(0) = 0$ and $y'(0) = 1$. For a single $r_1$, $y(0) = te^{r_1t}$ also satisfies $y(0) = 0$ and $y'(0) = 1$.


$$ax'' + bx' + c = 0$$ If the differential equation has double roots then since you already have a solution you can apply the method of reduction of order and solve the differential equation. If $y_1=e^{r_1t}$ is a solution of the DE then to for the second solution you try $y=v(t)e^{r_1t}$ then you find the second solution. And you find that $y_2=te^{r_1t}$ is another solution.


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