Linear Differential Equations

If this equation has two roots \(r_1\) and \(r_2\), then we can form an infinite number of solutions with linear combinations of \(e^{rt}\):

\begin{align} y = c_1e^{r_1t} + c_2e^{r_2t} \end{align}

Unique solutions must be linearly independent of each other. This means that for a set of funcions \(f_1, f_2, \dots, f_n\), if \(c_1f_1 + c_2f_2 + \cdots + c_nf_n = 0\), then \(c_1 = c_2 = \cdots = c_n = 0\). To check if a set of \(n\) functions \(f_1, f_n, \dots, f_n\) is linearly independent, we can check if the Wronskian is equivalent to zero:

\begin{align} W = \det\begin{bmatrix} f_1 & f_2 & \cdots & f_n \\ f_1' & f_2' & \cdots & f_n' \\ \vdots & \vdots & & \vdots \\ f_1^{(n-1)} & f_n^{(n-1)} & \cdots & f_n^{(n-1)} \end{bmatrix} \notag \end{align}

For just two functions, however, it is usually simpler to just check if they are scalar multiples of each other.

If the characteristic equation has repeated real root \(r\), then one solution is \(e^{rt}\), and the second turns out to be \(te^{rt}\). Then, the general solution is:

\begin{align} y = c_1e^{rt} + c_2te^{rt} \end{align}

More generally, if a differential equation has a repeated root \(r\) that is repeated \(n\) times, then we have solutions from \(e^{rt}\) up to \(t^{n}e^{rt}\). To see why this is, let \(D\) represent the derivative operator (i.e. \(Dy = y'\)). Realize that \(D\) fulfills the requirements to be a linear transformation, as it behaves well under addition and scalar multiplication.

We want to find the solutions to \(y^{(n)} + a_{n-1}y^{(n-1)} + \cdots + a_1y' + a_0y = 0\). If the roots of the characteristic equation are \(r_1, r_2, \dots, r_n\) (they can be repeated), then we can rewrite the differential equation as:

\begin{align} (D-r_1)(D-r_2) \cdot \cdots \cdot (D-r_n)y=0 \notag \end{align}

We will proceed by induction. First, we would like to solve the base case, when \(r\) is a repeated real root of the characteristic equation that is repeated once, we have the following term:

\begin{align} (D-r)\left[(D-r)y\right] = 0 \notag \end{align}

To find both solutions, we need \((D-r)y\) to be in the nullspace of \(D-r\). Finding the nullspace:

\begin{align} (D-r)y &= 0 \notag \\ y' - ry &= 0 \notag \\ y &= Ce^{rt} \notag \end{align}

This is our first solution, and the nullspace of \(D-r\). Then, since we want \((D-r)y\) to be in the nullspace, we can try:

\begin{align} (D-r)y &= e^{rt} \notag \\ y'-ry &= e^{rt} \notag \end{align}

Proceeding by integrating factor technique:

\begin{align} e^{-rt}(y'-ry) &= 1 \notag \\ [ye^{-rt}]' &= 1 \notag \\ ye^{-rt} &= t + C \notag \\ y &= te^{rt} + Ce^{rt} \notag \end{align}

We see that for the case of one repeated real root, the second solution is \(te^{rt}\). Now, for the inductive step: we assume that for \(n\) repeated real roots \(r\), solutions take the form \((c_nt^n + c_{n-1}t^{n-1} + \cdots + c_0)e^{rt}\). Then, for \(n+1\) repeated real roots, we can write the equation as \((D-r)^{n+2}y = 0\); we just need \((D-r)y\) to be in the nullspace of \((D-r)^{n+1}\). But that is just the solution set for the case of \(n\) repeated real roots, so we can write:

\begin{align} (D-r)y = (c_nt^n + c_{n-1}t^{n-1} + \cdots + c_0)e^{rt} \notag \end{align}

Solving this using integrating factor for \(y\), we see that we end up with a new term of \(t^{n+1}\), finishing the inductive step. Therefore, for each additional repeated root for a general differential equation, we simply multiply by \(t\).

If a characteristic equation has two nonreal roots \(\alpha \pm \beta i\) for some \(\beta \neq 0\), then we can write a general solution as:

\begin{align} y = Ce^{(\alpha + \beta i)t} + De^{(\alpha - \beta i)t} \notag \end{align}

However, it usually doesn't make sense for a solution to a differential equation to be complex. Thus, we would like to take only the real part of this solution, and then make it so that the imaginary part is set to zero. To do this, we can apply Euler's formula, \(e^{it} = \cos t + i\sin t\), which allows us to write the general formula instead as:

\begin{align} y = c_1e^{\alpha t}\cos(\beta t) + c_2 e^{\alpha t} \sin(\beta t) \end{align}

where \(c_1\) and \(c_2\) are real.

To do this, we can employ some educated guesses as to what the solution might be:

1. \(f(t)\) is polynomial, try \(y_p\) polynomial
2. \(f(t) = e^{kt}\), try \(y_p = Ae^{kt}\)
3. \(f(t) = \cos(kt)\) or \(\sin(kt)\), try \(y_p = A\cos(kt) + B\sin(kt)\)

Then, we can plug them into our differential equation and solve for the unknown constants. This is known as the method of undetermined coefficients.

Sometimes, however, the default "try" of the method of undetermined coefficients fails to work. This happens when the homogenous solution overlaps with our guess, thus making our guess always go to zero.

To fix this, we multiply the particular solution by \(t\), or if the overlap has a double root in the characteristic equation, then we multiply it by \(t^2\). Two terms overlap when their ratio is a polynomial or rational function, which results in annihilation under differentiation.

Sometimes, \(f(t)\) is in a form where we cannot use the method of undetermined coefficients (e.g. \(f(t) = \tan t\)). Then, we have to use a different method, variation of parameters.

Say \(y_1\) and \(y_2\) are the two linearly independent solutions of the complementary equation. Now, for the particular solution we try:

\begin{align} y_p = u_1(t)y_1(t) + u_2(t)y_2(t) \notag \end{align}

where \(u_1\) and \(u_2\) are unknown functions. Taking the first derivative, we get:

\begin{align} y'_p = u'_1y_1+u_1y'_1+u'_2y_2+u_2y'_2 \notag \end{align}

The key idea here is that we assume \(u'_1y_1 + u'_2y_2=0\). This gives us an extra condition to fulfill when we solve for \(u_1\) and \(u_2\) in the end, but it also allows us to simplify the derivatives greatly. Then:

\begin{align} y'_p &= u_1y'_1 + u_2y'_2 \notag \\ y''_p &= u'_1y'_1+u_1y''_1+u'_2y'_2+u_2y''_2 \notag \end{align}

Plugging this into the original equation, we get:

\begin{align} f &= a(u'_1y'_1+u_1y''_1+u'_2y'_2+u_2y''_2) + b(u_1y'_1+u_2y'_2) + c(u_1y_1+u_2y_2) \notag \\ &= a(u'_1y'_1+u'_2y'_2) + u_1(ay''_1+by'_1+cy_1) + u_2(ay''_2+by'_2+cy_2) \notag \\ &= a(u'_1y'_1+u'_2y'_2) \notag \end{align}

Thus, we end up with the following linear system:

\begin{align} \begin{bmatrix} y_1 & y_2 \\ y'_1 & y'_2 \end{bmatrix} \begin{bmatrix}u'_1 \\ u'_2\end{bmatrix} = \begin{bmatrix} 0 \\ \frac{f}{a} \end{bmatrix} \notag \end{align}

We can then solve the system with Cramer's Rule to get:

\begin{align} \boxed{u'_1 = \frac{\begin{vmatrix}0 & y_2 \\ \frac{f}{a} & y'_2\end{vmatrix}}{\begin{vmatrix}y_1 & y_2 \\ y'_1 & y'_2\end{vmatrix}} = \frac{-fy_2}{a(y_1y'_2-y'_1y_2)}} \\ \boxed{u'_2 = \frac{\begin{vmatrix}y_1 & 0 \\y'_1 & \frac{f}{a}\end{vmatrix}}{\begin{vmatrix}y_1 & y_2 \\ y'_1 & y'_2\end{vmatrix}} = \frac{fy_1}{a(y_1y'_2-y'_1y_2)}} \end{align}

Notice that the denominator is just the Wronskian. \(u_1\) and \(u_2\) can now be determined by taking integrals.