ODE: Repeated Roots and Reduction of Order

When solving the differential equations that model vibrating systems, electrical circuits, or control theory, you often encounter a snag: the characteristic equation yields the same root twice. This "repeated root" case isn't just a minor exception; it's a fundamental scenario where the standard method of finding two independent solutions fails. Mastering how to handle it—through a modified general solution and the powerful reduction of order technique—is crucial for analyzing systems with critical damping or other degenerate behaviors.

The General Solution for Repeated Roots

Consider a homogeneous, second-order linear ODE with constant coefficients: $$a{y}^{\prime \prime }+b{y}^{\prime }+cy=0$$ The standard approach is to assume a solution of the form $y=e^{rt}$, leading to the characteristic equation $a{r}^2+br+c=0$. When the discriminant $b^2-4ac$ is zero, this equation has a single repeated root $r=r_1$.

If you only used $y_1=e^{rt}$, you would have just one fundamental solution, which is insufficient to construct the general solution. To find a second, linearly independent solution, we employ a clever guess, informed by the method of reduction of order (which we will derive next). The result is that the second solution takes the form $y_2=t{e}^{rt}$. Therefore, the general solution for the repeated root case is: $$y(t)=(c_1+c_{2}t)e^{rt}$$ where $c_1$ and $c_2$ are arbitrary constants. The term $t{e}^{rt}$ is linearly independent from $e^{rt}$ because no constant multiple of $e^{rt}$ can produce the extra factor of $t$. This solution is non-oscillatory and represents a critically damped response in physical systems, where the system returns to equilibrium as quickly as possible without overshooting.

Derivation of the Reduction of Order Method

Reduction of order is a powerful, general method for finding a second solution to a homogeneous linear ODE when one solution $y_1(t)$ is already known. Its derivation is elegant and provides the theoretical backbone for the $t{e}^{rt}$ solution.

Given a second-order linear ODE in standard form: $$y^{\prime \prime }+p(t)y^{\prime }+q(t)y=0$$ and one known solution $y_1(t)$, we guess that a second solution has the form: $$y_2(t)=v(t)y_1(t)$$ where $v(t)$ is an unknown function to be determined. We substitute $y_2$ into the ODE. First, compute the derivatives: $$y_2^{\prime }=v^{\prime }{y}_1+v{y}_1^{\prime }$$ $$y_2^{\prime \prime }=v^{\prime \prime }{y}_1+2v^{\prime }{y}_1^{\prime }+v{y}_1^{\prime \prime }$$ Substituting these into $y_2^{\prime \prime }+p(t)y_2^{\prime }+q(t)y_2=0$ and grouping terms yields: $$v^{\prime \prime }{y}_1+v^{\prime }(2y_1^{\prime }+p{y}_1)+v(y_1^{\prime \prime }+p{y}_1^{\prime }+q{y}_1)=0$$ The critical observation is that the coefficient of $v$ is precisely the original ODE applied to $y_1$, which equals zero because $y_1$ is a solution. This simplifies the equation dramatically: $$y_1{v}^{\prime \prime }+(2y_1^{\prime }+p{y}_1)v^{\prime }=0$$ Now, let $w=v^{\prime }$. This transforms the equation into a first-order linear ODE for $w$: $$y_1{w}^{\prime }+(2y_1^{\prime }+p{y}_1)w=0$$ or $$w^{\prime }+\left(\frac{2y_1^{\prime }}{y_1}+p\right)w=0$$ This is a separable equation. Solving it gives us $w(t)$, and integrating $w$ gives us $v(t)$. The beauty is that the formula for $v(t)$ will always produce a second, linearly independent solution. For the constant-coefficient, repeated-root case where $y_1=e^{rt}$ and $p$ is constant, this process directly yields $v(t)=t$, confirming our earlier result $y_2=t{e}^{rt}$.

Applying Reduction of Order When One Solution is Known

The derivation provides the theory, but applying reduction of order is a straightforward, four-step process. Let's walk through a concrete example where the ODE does not have constant coefficients.

Example: Given that $y_1=t$ is a solution to $t^2{y}^{\prime \prime }+t{y}^{\prime }-y=0$ for $t>0$, find the general solution.

1. Write the ODE in standard form: Divide by $t^2$.

$$y^{\prime \prime }+\frac{1}{t}{y}^{\prime }-\frac{1}{t^2}y=0$$ Here, $p(t)=1/t$ and $q(t)=-1/t^2$.

1. Apply the formula for $v(t)$: From our derivation, we have $w^{\prime }+(2y_1^{\prime }/y_1+p)w=0$.

Since $y_1=t$, then $y_1^{\prime }=1$. Substitute $y_1$, $y_1^{\prime }$, and $p(t)$: $$w^{\prime }+\left(\frac{2(1)}{t}+\frac{1}{t}\right)w=w^{\prime }+\frac{3}{t}w=0$$

1. Solve the first-order ODE for $w$: This is separable: $dw/w=-(3/t)dt$.

Integrating: $\ln |w|=-3\ln |t|+C$. We only need one non-trivial $v$, so we choose the simplest particular solution by setting the constant $C=0$: $$w=t^{-3}$$ Since $w=v^{\prime }$, we have $v^{\prime }=t^{-3}$.

1. Integrate to find $v$ and form $y_2$:

$$v=\int t^{-3}dt=-\frac{1}{2}{t}^{-2}$$ Therefore, the second solution is $y_2=v{y}_1=(-\frac{1}{2}{t}^{-2})\cdot t=-\frac{1}{2}{t}^{-1}$. The factor of $-\frac{1}{2}$ can be absorbed into the arbitrary constant, so we simply take $y_2=t^{-1}$.

The general solution is a linear combination of the two independent solutions: $y(t)=c_{1}t+c_2{t}^{-1}$.

Abel's Formula for the Wronskian

A deep connection between the solutions of a linear ODE and the coefficients of the equation is provided by Abel's formula (also known as Abel's identity). It gives you a direct way to compute the Wronskian $W(t)$ of two solutions without even knowing the solutions explicitly, only that they exist.

For the second-order linear ODE in standard form, $y^{\prime \prime }+p(t)y^{\prime }+q(t)y=0$, if $y_1$ and $y_2$ are solutions, then their Wronskian $W(t)=y_1{y}_2^{\prime }-y_2{y}_1^{\prime }$ is given by: $$W(t)=C{e}^{-\int p(t)dt}$$ where $C$ is a constant that depends on the choice of fundamental set of solutions.

Why it matters: The Wronskian tests for linear independence ($W\ne 0$). Abel's formula proves that for these ODEs, the Wronskian is either always zero (if $C=0$, meaning the solutions are dependent) or never zero (if $C\ne 0$). It also provides a quick check on your work. In our repeated root constant-coefficient example where $p(t)$ is a constant $b/a$, Abel's formula gives $W(t)=C{e}^{-(b/a)t}$, which is indeed non-zero for $C\ne 0$, confirming the independence of $e^{rt}$ and $t{e}^{rt}$.

Common Pitfalls

1. Forgetting the factor of 't' in the general solution. When you see a repeated root, the most common error is to write $y=c_1{e}^{rt}+c_2{e}^{rt}$. This is incorrect because these two functions are linearly dependent. You must remember the modified form: $y=(c_1+c_{2}t)e^{rt}$.

1. Misapplying reduction of order by forgetting standard form. The formula $w^{\prime }+(2y_1^{\prime }/y_1+p)w=0$ requires the ODE to be in the form $y^{\prime \prime }+p(t)y^{\prime }+q(t)y=0$. If you apply it directly to $a(t)y^{\prime \prime }+b(t)y^{\prime }+c(t)y=0$, your calculation for $p(t)$ will be wrong. Always divide through by the leading coefficient first.

1. Incorrectly integrating or simplifying during reduction of order. The step from $w^{\prime }+f(t)w=0$ to $w=\exp (-\int f(t)dt)$ is straightforward but prone to sign errors and mistakes in the integration of $f(t)$. Work slowly and check that your $y_2$ indeed satisfies the original ODE.

1. Ignoring the domain when applying Abel's formula. The formula $W(t)=C{e}^{-\int p(t)dt}$ is valid on any interval where $p(t)$ is continuous. If your problem involves a $t$ value where $p(t)$ is discontinuous (like $t=0$ in an ODE with $p(t)=1/t$), you must consider solutions on separate intervals where the theory holds.

Summary

• For a constant-coefficient ODE with a repeated root $r$ of the characteristic equation, the general solution is $y(t)=(c_1+c_{2}t)e^{rt}$. The term $t{e}^{rt}$ provides the necessary second, linearly independent solution.
• The reduction of order method is a universal technique derived by substituting $y_2=v(t)y_1(t)$ into the ODE. It systematically reduces the problem to solving a first-order linear equation for $v^{\prime }$.
• When applying reduction of order, the critical workflow is: (1) Put the ODE in standard form, (2) Substitute into the derived formula for $v^{\prime }$, (3) Solve the resulting first-order equation, and (4) Integrate to find $v$ and construct $y_2$.
• Abel's formula, $W(t)=C{e}^{-\int p(t)dt}$, provides a direct link between the coefficient $p(t)$ and the Wronskian of two solutions, serving as a powerful theoretical tool and a useful check for consistency.