# Why is the general solution to linear homogeneous differential equation with constant coefficients different if roots are distinct or repeating?

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 $$(c_1 + c_2)e^{r_1t}$$.

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

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.