Differential Equations - Packet 07

Linear equations in higher order

The simplest second-order equations can be solved by directed integration performed twice, as we saw in Packet 02. To solve the IVP:

$$y^{\prime \prime }(t)=f(t),y(0)=a_0,y^{\prime }(0)=b_0,$$

we can integrate twice:

$$y(t)=\mathrm{\Phi }(t)+c_{1}t+c_2,$$

where ${\mathrm{\Phi }}^{\prime }(t)=F(t)$ and $F^{\prime }(t)=f(t)$. At each stage in the integration, a new constant is produced, and the new constant covers all possibilities. (All possibilities because: any two antiderivatives differ by a constant. This fact comes from the Mean Value Theorem.)

From this we learn to expect that any second-order differential equation will determine a family of solution curves involving two parameters. Later we will consider a good argument for this expectation in the case of linear equations. For right now, an intuitive argument in the general case is that a second-order equation involves 3 functions: $y,y^{\prime },y^{\prime \prime }$, and the ODE determines any one of them in terms of the other two. Initial conditions are therefore fully specified by giving starting values of two of the functions, e.g. $y(0)$ and $y^{\prime }(0)$, since the equation can be used to solve for the third. These two components of the initial condition correspond to the two free parameters in the complete family of solutions.

Second order equations are called linear when the items $y,y^{\prime },y^{\prime \prime }$ appear only in first-degree powers. A more precise and complete definition would require the definition of a second-order linear differential operator $L$, and we avoid the details for that right now. The ODE is determined from such an operator by setting $L[y]=0$ (homogeneous case) or $L[y]=f(t)$ (inhomogeneous case) for some forcing function $f(t)$.

Second-order, constant coefficients

Second-order linear equation with constant coefficients:

• Homogeneous: $a{y}^{\prime \prime }+b{y}^{\prime }+cy=0$
• Inhomogeneous: $a{y}^{\prime \prime }+b{y}^{\prime }+cy=d$

In this usage, the word ‘homogeneous’ means “no constant term”. This is a different meaning from ‘homogeneous’ as we have seen previously, although the two meanings are connected.

Homogeneous case It is easy to guess a system of solutions for the homogeneous case, inspired by the use of $e^{rt}$ for first-order linear equations. Suppose $y=e^{rt}$ is a solution. Then:

$$\begin{array}{cc}a{(e^{rt})}^{\prime \prime }+b{(e^{rt})}^{\prime }+c(e^{rt})& =0\\ a{r}^2{e}^{rt}+br{e}^{rt}+c{e}^{rt}& =0\\ (a{r}^2+br+c)e^{rt}& =0\\ a{r}^2+br+c& =0.\end{array}$$

• We conclude: any solution of the form $y=e^{rt}$ must have $r$ a solution to $a{r}^2+br+c=0$.

• In reverse: if $r$ satisfies $a{r}^2+br+c=0$, then $y=e^{rt}$ is a solution to the differential equation.

• The polynomial equation $a{r}^2+br+c=0$ is sometimes called the characteristic equation of the linear ODE.

• Notice also: if $y_1(t)$ and $y_2(t)$ are solutions, then $\alpha y_1(t)+\beta y_2(t)$ are also solutions, for any constants $\alpha ,\beta$. This is the practical meaning of linearity. It is sometimes also called the superposition principle.

• More abstractly: the homogeneous linear ODE has a vector space of solutions. Concretely: solutions can be added and scaled to form new solutions.

Normally, the quadratic equation $a{r}^2+br+c=0$ has two distinct solutions, namely $r_1=\frac{-b+\sqrt{b^2-4ac}}{2a}$ and $r_2=\frac{-b-\sqrt{b^2-4ac}}{2a}$. These two numbers provide a $2$D vector space of solutions spanned by $y_1(t)=e^{r_{1}t}$ and $y_2(t)=e^{r_{2}t}$. Namely:

$$y(t)=C_1{e}^{r_{1}t}+C_2{e}^{r_{2}t}$$

is a solution for any choice of $C_1,C_2$ combo. This gives a two-parameter family of solutions.

There are two potential complications for this method:

• What happens if $r_1=r_2$?
• What happens if $r_1$ and $r_2$ are complex numbers?

Repeated roots What happens if $r_1=r_2$?

This means $b^2-4ac=0$ and $r_1=r_2=-\frac{b}{2a}$. Now watch this: plug in $t{e}^{rt}$ to the equation:

$$\begin{array}{cc}a(t{e}^{rt}{)}^{\prime \prime }+b(t{e}^{rt}{)}^{\prime }+c(t{e}^{rt})& =0\\ a(2r+r^{2}t)e^{rt}+b(1+rt)e^{rt}+ct{e}^{rt}& =0\\ ((a{r}^2+br+c)t+(2ar+b))e^{rt}& =0.\end{array}$$

When $r=r_1=r_2$, certainly we have $a{r}^2+br+c=0$. But moreover, we also have $2ar+b=0$.

The extra terms coming out of the product rule applied to $t{e}^{rt}$ exactly cancel from the condition $r=-\frac{b}{2a}$ of repeated roots. So $t{e}^{rt}$ is a new solution if-and-only-if $r$ is a repeated root.

Example

Repeated roots

Problem: Solve the ODE: $y^{\prime \prime }-6y^{\prime }+9y=0$.

Solution: We first solve $r^2-6r+9={(r-3)}^2=0$. So we have $r_1=r_2=3$. Then $y_1(t)=e^{3t}$ is one solution, and $y_2(t)=t{e}^{3t}$ is another solution. The complete family of solutions:

$$y(t)=C_1{e}^{3t}+C_{2}t{e}^{3t}.$$

Check this complete family by plugging it into the ODE!

Exercise 07B-01 [During class.]

Second-order linear: distinct roots, repeated roots

Find the 2-parameter family of solution curves for the ODEs:

• (a) $2y^{\prime \prime }+6y^{\prime }+2y=0$
• (b) $4y^{\prime \prime }-12y^{\prime }+9y=0$

(It is a fact, a magical fact from this point of view, that in higher order, for an ODE like $a_n{y}^{(n)}+a_{n-1}{y}^{(n-1)}+\cdots +a_{0}y=0$, the same phenomenon generalizes: we will have $e^{rt},t{e}^{rt},t^2{e}^{rt},\dots ,t^{k-1}{e}^{rt}$ spanning the solutions precisely when $r$ is a $k$-times repeated root of $a_n{r}^n+a_{n-1}{r}^{n-1}+\cdots +a_0=0$.)

Imaginary roots What happens if $r_1$ and $r_2$ have imaginary components?

In this case, we are naturally led to consider complex-valued solutions. It turns out, however, that by doing so, we will discover a family of pure real solutions. The pure real family is produced by applying a complex change of basis to the complex solution space.

Let us write $r_1=h+ki$ and $r_2=h-ki$. (Since $r_1,r_2$ solve a quadratic, we know they must come in a conjugate pair like this with $\pm ki$.) We can write two complex solutions:

$$z_1(t)=e^{(h+ki)t},z_2(t)=e^{(h-ki)t}.$$

In light of the Euler formula $e^{x+yi}=e^x(\cos y+i\sin y)$, we make a judicious choice of linear combination of these solutions. It is designed to cancel away the imaginary parts:

$$\begin{array}{cc}{y}_1(t)& =\frac{1}{2}{z}_1(t)+\frac{1}{2}{z}_2(t)=e^{ht}\cos (kt)\\ y_2(t)& =\frac{1}{2i}{z}_1(t)-\frac{1}{2i}{z}_2(t)=e^{ht}\sin (kt).\end{array}$$

(These solution are given by applying the matrix $\left(\begin{array}{cc}1/2& -i/2\\ 1/2& i/2\end{array}\right)$ to the original set. Notice that the columns are perpendicular, so the matrix is a ‘complex rotation’ plus overall scaling.)

Exercise 07B-02 [During class.]

Cancelling imaginary parts

By substituting the formulas for $z_1(t)$ and $z_2(t)$, verify that $y_1(t)=e^{ht}\cos (kt)$ and $y_2(t)=e^{ht}\sin (kt)$.

Exercise 07B-03 [During class.]

Verifying trigonometric solutions

Verify that $y_1(t)=e^{ht}\cos (kt)$ and $y_2(t)=e^{ht}\sin (kt)$ solve the original ODE.

To do this, first find the correct ODE: expand $a{r}^2+br+c=(r-(h+ki))(r-(h-ki))$ in order to find the appropriate $a,b,c$ for the general case of complex roots.

Then check that $e^{ht}\cos (kt)$ and $e^{ht}\sin (kt)$ are solutions to the ODE $a{y}^{\prime \prime }+b{y}^{\prime }+cy=0$.

Summary: because the repeated roots solutions have the form $r=h\pm ki$, we can always use the above method, namely judicious linear combinations, to reveal a pair of solutions $y_1(t)$ and $y_2(t)$ that do not involve imaginary numbers.

Real vs. complex solutions

Even first-order linear ODEs have complex solutions. For example, $y^{\prime }=ky$ has real solution $e^{kt}$ but it also has the imaginary solution $i{e}^{kt}$, and every combination of them. In the second-order case, we could have simply guessed the real solutions $e^{ht}\cos (kt)$ and $e^{ht}\sin (kt)$. But the natural procedure using $e^{rt}$ leads us first to the complex solutions, and from these we can extract purely real solutions using a complex combination.

Example

Second-order linear, complex roots, pure real solutions

Problem: Solve the ODE: $y^{\prime \prime }-2y^{\prime }+5y=0$.

Solution: We first solve $r^2-2r+5=0$ and find $r_1=1+2i$ and $r_2=1-2i$. We therefore have complex solutions

$$z_1(t)=e^t(\cos (2t)+i\sin (2t)),z_2(t)=e^t(\cos (2t)-i\sin (2t))$$

and real solutions

$$y_1(t)=e^t\cos (2t),y_2(t)=e^t\sin (2t).$$

The complete family of solutions is therefore:

$$y(t)=C_1{e}^t\cos (2t)+C_2{e}^t\sin (2t).$$

Problems due Tuesday 12 Mar 2024 by 11:59pm

Problem 07-01

Solving second-order linear ODEs

Find the complete family of solutions to the following ODEs:

• (a) $y^{\prime \prime }+y^{\prime }-6y=0$
• (b) $y^{\prime \prime }+4y^{\prime }+5y=0$
• (c) $y^{\prime \prime }+y^{\prime }=0$
• (d) $16y^{\prime \prime }-8y^{\prime }+y=0$

Problem 07-02

Solving second-order linear IVPs

Solve the IVP:

• $y^{\prime \prime }-5y^{\prime }+6y=0$, $y(1)=e^2$, $y^{\prime }(1)=3e^2$
• $y^{\prime \prime }+4y^{\prime }+5y=0$, $y(0)=1$, $y^{\prime }(0)=0$

Problem 07-03

Asymptotic behavior

Consider the ODE: $y^{\prime \prime }+b{y}^{\prime }+cy=0$. Let $y(t)$ be a solution.

Show that the following statements are equivalent:

• $b>0$ and $c>0$
• $y(t)\to 0$ as $t\to +\infty$

Problem 07-04

Extending methods to a third-order ODE

Find a complete family of solutions to the third-order ODE $y^{\prime \prime \prime }-3y^{\prime \prime }+4y^{\prime }-2y=0$ using the methods discussed in this Packet.

Problem 07-05

Backwards reasoning: find the ODE

Find the ODE that has the following family of solutions: $C_1{e}^{-4t}\cos t+C_2{e}^{-4t}\sin t$.