Consider the DE
$$ay'' + by' + cy = 0,$$where $a$, $b$, $c$ are constants. Then the characteristic roots are given by
$$r_{1,2} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.$$Then
It remains to prove the case of repeated roots $r_1 = r_2 = -\dfrac{b}{2a}$.
We follow an idea due originally to Frobenius (1849–1917). Let $r$ be a variable, and let
$$L = a\frac{d^2}{dt^2} + b\frac{d}{dt} + c = aD^2 + bD + c.$$Then
\begin{align} L(e^{rt}) &= \Big(a\frac{d^2}{dt^2} + b\frac{d}{dt} + c\Big)e^{rt} \\[6pt] &= a\frac{d^2}{dt^2}e^{rt} + b\frac{d}{dt}e^{rt} + ce^{rt} = (ar^2 + br + c)e^{rt} \\[6pt] &= (ar^2 + br + c)e^{rt} - (ar_1^2 + br_1 + c)e^{rt} \\[6pt] &= a(r - r_1)^2 e^{rt} \end{align}where we have used $r_1 = -\dfrac{b}{2a}$. Clearly $L(e^{r_1 t}) = 0$ which reconfirms that $y = e^{r_1 t}$ is a solution.
Let us differentiate the $L(e^{rt})$ with respect to $r$:
$$\frac{\partial L(e^{rt})}{\partial r} = 2a(r - r_1)L(e^{rt}) + a(r - r_1)^2 tL(e^{rt}).$$Substituting $r = r_1$ yields
$$L(te^{r_1 t}) = \frac{\partial L(e^{rt})}{\partial r}\Big|_{r=r_1} = 2a(r_1 - r_1)L(e^{r_1 t}) + a(r_1 - r_1)^2 te^{r_1 t} = 0,$$thus showing that the function $y = te^{r_1 t}$ is also a solution to the DE.
It remains to show that the $e^{r_1 t}$, $te^{r_1 t}$ are linearly independent:
$$W\big(te^{-\frac{b}{2a}t},\, e^{-\frac{b}{2a}t}\big) = \begin{vmatrix} e^{-\frac{b}{2a}t} & te^{-\frac{b}{2a}t} \\[4pt] -\frac{b}{2a}e^{-\frac{b}{2a}t} & \big(1 - \frac{bt}{2a}\big)e^{-\frac{b}{2a}t} \end{vmatrix} = e^{-\frac{b}{2a}t}$$which never vanishes. So according to an earlier theorem about Wronskians, the $e^{r_1 t}$, $te^{r_1 t}$ are linearly independent and so they form a fundamental set of solutions to the original DE.
Let
$$y(t) = v(t)y_1(t) = v(t)\,e^{-\frac{b}{2a}t}.$$Suppose the $y$ so defined above is a solution to the original DE. Then
$$y'(t) = \Big(v'(t) - \frac{b}{2a}v(t)\Big)e^{-\frac{b}{2a}t}$$and similarly,
$$y''(t) = \Big(v''(t) - \frac{b}{a}v'(t) + \frac{b^2}{(2a)^2}v(t)\Big)e^{-\frac{b}{2a}t}.$$Substitute them into the original equation yields:
\begin{align} ay'' + by' + cy &= a\Big(v''(t) - \frac{b}{a}v'(t) + \frac{b^2}{(2a)^2}v(t)\Big)e^{-\frac{b}{2a}t} \\[6pt] &\quad + b\Big(v'(t) - \frac{b}{2a}v(t)\Big)e^{-\frac{b}{2a}t} + cv(t)\,e^{-\frac{b}{2a}t} \\[6pt] &= \Big[av''(t) + 0 \cdot v'(t) + \Big(\frac{4ac - b^2}{4a}\Big)v(t)\Big]e^{-\frac{b}{2a}t} \\[6pt] &= \Big[av''(t) + \Big(\frac{4ac - b^2}{4a}\Big)v(t)\Big]e^{-\frac{b}{2a}t} \end{align}But we have $b^2 - 4ac = 0$. So the above equation reduces to
$$v''(t) = 0,$$and thus
$$v(t) = At + B.$$Without loss of generality, we may choose $A = 1$ so that $v(t) = t + B$. Hence
$$y(t) = (t + B)\,e^{-\frac{b}{2a}t} = te^{-\frac{b}{2a}t} + Be^{-\frac{b}{2a}t}.$$But the expression $Be^{-\frac{b}{2a}t} = By_1(t)$ and so it can be combined with the original $y_1$. The expression $te^{-\frac{b}{2a}t}$ is really new which we denote by
The reduction of order method introduced for linear second order equations with constant coefficients also works on more general coefficients. Let $y_1(t)$ be a solution of
$$y'' + p(t)y' + q(t)y = 0.$$Let
$$y = v(t)y_1.$$Then
$$y' = vy_1' + v'y_1,$$and
$$y'' = vy_1'' + 2v'y_1' + v''y_1.$$Substitute $y'$ and $y''$ into the DE:
$$y_1 v'' + (2y_1' + p(t)y_1)v' + [y_1'' + p(t)y_1' + q(t)y_1]v = 0.$$That is,
$$y_1 v'' + (2y_1' + p(t)y_1)v' = 0.$$Let $u = v'$. Then the equation becomes
$$y_1 u' + (2y_1' + p(t)y_1)u = 0.$$We now solve this first order DE in $u$. Then solve for $v$ to get $vy_1$.
Example 1. Solve the IVP $y'' + 2y' + y = 0$, $y(0) = 1$, $y'(0) = 0$.
We have the characteristic equation
$$(r + 1)^2 = 0,$$so that we have repeated characteristic root $r = -1$. Thus the first solution is $y_1(t) = e^{-t}$, and according to the above derivation the second linearly independent solution is $y_2(t) = te^{-t}$.
Let $y = Ae^{-t} + Bte^{-t}$. Taking into consideration of the IC,
$$1 = y(0) = A,$$and
$$y'(t) = -e^{-t} + Be^{-t} - Bte^{-t},$$hence
$$0 = y'(0) = -1 + B,$$so that $B = 1$. We deduce
Example 2. Solve $y'' - y' + 0.25y = 0$, $y(0) = 2$, $y'(0) = 1/3$.
We have the characteristic equation
$$r^2 - r + \frac{1}{4} = \Big(r - \frac{1}{2}\Big)^2 = 0,$$so we have repeated root $r = \frac{1}{2}$. We have linearly independent solutions
$$e^{t/2}, \quad te^{t/2}.$$The IC implies
Solve the following DEs:
Solve the following IVPs:
Solve the initial value problem:
$$y'' + 2y' + y = 0, \quad y(0) = 1, \quad y'(0) = 0$$The characteristic equation is $r^2 + 2r + 1 = 0$. What type of roots does it have?
For a repeated root $r$, what form does the general solution take?
The repeated root is $r = -1$. What is the general solution?
From $y(0) = 1$: $c_1 e^0 + c_2(0)e^0 = 1$. What is $c_1$?
We need $y'$. Differentiating $y = c_1 e^{-t} + c_2 te^{-t}$:
From $y'(0) = 0$ with $c_1 = 1$: $-1 + c_2 = 0$. What is $c_2$?
The solution to the IVP is: