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5.4 Differential Equations: Class Notes

Definitions

A differential equation is an equation that contains the derivatives of one or more unknown functions.

The order of a differential equation is the order of the highest derivative it contains.

A differential equation of the form $y\prime =ay$ has a general solution of the form $y=c{e}^{ax}$.

Check: if $y=c{e}^{ax}$, then $y\prime =ac{e}^{ax}=ay$.

More formally, separate the variables and integrate:

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle \frac{dy}{dx}}& {\displaystyle =ay}\hfill \\ \hfill {\displaystyle \int \frac{dy}{y}}& {\displaystyle =\int adx}\hfill \\ \hfill {\displaystyle \ln |y|}& {\displaystyle =ax+C_1}\hfill \\ \hfill {\displaystyle |y|}& {\displaystyle =e^{ax+C_1}=e^{ax}{e}^{C_1}}\hfill \\ \hfill {\displaystyle y}& {\displaystyle =c{e}^{ax}}\hfill \end{array}}$$

where $c=\pm e^{C_1}$.

A condition which specifies the value of the general solution at a point is called an initial condition, and the problem of solving a differential equation subject to an initial condition is called an initial-value problem.

If we know, for instance, that $y(0)=P_0$, then $y=P_0{e}^{ax}$.

A constant coefficient first-order homogeneous linear system has the form

$$\begin{array}{cc}\hfill {\displaystyle y_1\prime }& {\displaystyle =a_{11}{y}_1+a_{12}{y}_2+\dots +a_{1n}{y}_n}\hfill \\ \hfill {\displaystyle y_2\prime }& {\displaystyle =a_{21}{y}_1+a_{22}{y}_2+\dots +a_{2n}{y}_n}\hfill \\ \hfill {\displaystyle ⋮}& {\displaystyle ⋮}\hfill \\ \hfill {\displaystyle y_n\prime }& {\displaystyle =a_{n1}{y}_1+a_{n2}{y}_2+\dots +a_{nn}{y}_n}\hfill \end{array}$$

where $y_i=f_i(x)$ are functions to be determined, and the $a_{ij}$’s are constants.

This can be written in matrix notation as

$$\left[\begin{array}{c}{y}_1\prime \\ y_2\prime \\ ⋮\\ y_n\prime \end{array}\right]=\left[\begin{array}{cccc}{a}_{11}& a_{12}& \cdots & a_{1n}\\ a_{21}& a_{22}& \cdots & a_{2n}\\ ⋮& ⋮& & ⋮\\ a_{n1}& a_{n2}& \cdots & a_{nn}\end{array}\right]\left[\begin{array}{c}{y}_1\\ y_2\\ ⋮\\ y_n\end{array}\right]$$

or $𝐲\prime =A𝐲$.

The trivial solution to the above differential equation is $y_1=y_2=\dots =y_n=0$.

If we can find a matrix $P$ that diagonalizes $A$, then we can use the diagonal matrix in solving the system. If $P$ diagonalizes $A$, then form $\overrightarrow{y}=P\overrightarrow{u}$, where $\overrightarrow{u}$ is an unknown vector. Then

$$\overrightarrow{y}\prime =A\overrightarrow{y}\Rightarrow P\overrightarrow{u}\prime =AP\overrightarrow{u}\Rightarrow \overrightarrow{u}\prime =P^{-1}AP\overrightarrow{u}$$

This has the form $\overrightarrow{u}\prime =D\overrightarrow{u}$, where $D$ is diagonal. This has the form of a system whose solutions contain exponentials.

Recap: Imagine an unknown vector $\overrightarrow{u}$ exists. Then find $\overrightarrow{u}$ and use what we find.

Example

a. Solve the system of differential equations.

$$\begin{array}{cc}\hfill {\displaystyle y_1\prime }& {\displaystyle =2y_1+3y_2}\hfill \\ \hfill {\displaystyle y_2\prime }& {\displaystyle =2y_1+y_2}\hfill \end{array}$$

Solution

The process: $A\to {\lambda }_i$ $\to D$ $\to \overrightarrow{u}\prime =D\overrightarrow{u}$ $\to {\overrightarrow{p}}_i$ $\to P$ $\to \overrightarrow{y}=P\overrightarrow{u}$.

a. Find the eigenvalues and eigenvectors of $A$.

$$A=\left[\begin{array}{cc}2& 3\\ 2& 1\end{array}\right]\Rightarrow \left|\begin{array}{cc}\lambda -2& -3\\ -2& \lambda -1\end{array}\right|=0$$

$${\lambda }^2-3\lambda +2-6=0\Rightarrow {\lambda }^2-3\lambda -4=0\Rightarrow (\lambda -4)(\lambda +1)=0\Rightarrow \lambda =4,-1$$

The eigenvalues are distinct, so $A$ is diagonalizable, with $D=\left[\begin{array}{cc}4& 0\\ 0& -1\end{array}\right]$.

$$\lambda =4:\left[\begin{array}{cc}2& -3\\ -2& 3\end{array}\right]\to \left[\begin{array}{cc}2& -3\\ 0& 0\end{array}\right]\Rightarrow {\overrightarrow{p}}_1=\left[\begin{array}{c}3\\ 2\end{array}\right]t$$

$$\lambda =-1:\left[\begin{array}{cc}-3& -3\\ -2& -2\end{array}\right]\to \left[\begin{array}{cc}1& 1\\ 0& 0\end{array}\right]\Rightarrow {\overrightarrow{p}}_2=\left[\begin{array}{c}1\\ -1\end{array}\right]t$$

$${\displaystyle P=\left[\begin{array}{cc}3& 1\\ 2& -1\end{array}\right],P^{-1}=-\frac{1}{5}\left[\begin{array}{cc}-1& -1\\ -2& 3\end{array}\right]=\left[\begin{array}{cc}\frac{1}{5}& \frac{1}{5}\\ \frac{2}{5}& -\frac{3}{5}\end{array}\right]}$$

Then $P^{-1}AP=D$, and

$$\overrightarrow{u}\prime =D\overrightarrow{u}\Rightarrow \left[\begin{array}{c}{u}_1\prime \\ u_2\prime \end{array}\right]=\left[\begin{array}{cc}4& 0\\ 0& -1\end{array}\right]\left[\begin{array}{c}{u}_1\\ u_2\end{array}\right]=\left[\begin{array}{c}4u_1\\ -u_2\end{array}\right]$$

Each equation has the form $y\prime =ay$:

$$u_1\prime =4u_1\Rightarrow u_1=c_1{e}^{4x},u_2\prime =-u_2\Rightarrow u_2=c_2{e}^{-x}$$

$$\overrightarrow{y}=P\overrightarrow{u}=\left[\begin{array}{cc}3& 1\\ 2& -1\end{array}\right]\left[\begin{array}{c}{c}_1{e}^{4x}\\ c_2{e}^{-x}\end{array}\right]=\left[\begin{array}{c}3c_1{e}^{4x}+c_2{e}^{-x}\\ 2c_1{e}^{4x}-c_2{e}^{-x}\end{array}\right]$$

b. Find the solution that satisfies the initial conditions $y_1(0)=25$, $y_2(0)=5$.

b. When $x=0$, $y_1=25$ and $y_2=5$:

$$\begin{array}{cc}\hfill {\displaystyle 25}& {\displaystyle =3c_1+c_2}\hfill \\ \hfill {\displaystyle 5}& {\displaystyle =2c_1-c_2}\hfill \end{array}$$

Adding the equations gives $30=5c_1$, so $c_1=6$, and then $c_2=7$. Or use Cramer’s rule:

$$D=\left|\begin{array}{cc}3& 1\\ 2& -1\end{array}\right|=-5,D_{c_1}=\left|\begin{array}{cc}25& 1\\ 5& -1\end{array}\right|=-30,D_{c_2}=\left|\begin{array}{cc}3& 25\\ 2& 5\end{array}\right|=-35$$

$${\displaystyle c_1=\frac{D_{c_1}}{D}=6,c_2=\frac{D_{c_2}}{D}=7}$$

#7

Sometimes it is possible to solve a single higher-order linear differential equation with constant coefficients by expressing it as a system and applying the methods of this section. For the differential equation $y″-y\prime -6y=0$, show that the substitutions $y_1=y$ and $y_2=y\prime$ lead to the system

$$\begin{array}{cc}\hfill {\displaystyle y_1\prime }& {\displaystyle =y_2}\hfill \\ \hfill {\displaystyle y_2\prime }& {\displaystyle =6y_1+y_2}\hfill \end{array}$$

Solve this system, and use the result to solve the original differential equation.

Solution

$y_1=y\Rightarrow y_1\prime =y\prime$. Since $y_2=y\prime$, we have $y_1\prime =y_2$.

If $y_2=y\prime$, then $y_2\prime =y″$. So

$$y″-y\prime -6y=0\Rightarrow y_2\prime -y_2-6y_1=0\Rightarrow y_2\prime =6y_1+y_2$$

Thus

$$\begin{array}{cc}\hfill {\displaystyle y_1\prime }& {\displaystyle =y_2}\hfill \\ \hfill {\displaystyle y_2\prime }& {\displaystyle =6y_1+y_2}\hfill \end{array}\text{or}\overrightarrow{y}\prime =\left[\begin{array}{cc}0& 1\\ 6& 1\end{array}\right]\overrightarrow{y}$$

Now solve the system.

$$A=\left[\begin{array}{cc}0& 1\\ 6& 1\end{array}\right]\Rightarrow \left|\begin{array}{cc}\lambda & -1\\ -6& \lambda -1\end{array}\right|=0\Rightarrow {\lambda }^2-\lambda -6=0\Rightarrow (\lambda -3)(\lambda +2)=0$$

So $\lambda =-2,3$.

$$\lambda =-2:\left[\begin{array}{cc}-2& -1\\ -6& -3\end{array}\right]\Rightarrow {\overrightarrow{p}}_1=\left[\begin{array}{c}1\\ -2\end{array}\right]t,\lambda =3:\left[\begin{array}{cc}3& -1\\ -6& 2\end{array}\right]\Rightarrow {\overrightarrow{p}}_2=\left[\begin{array}{c}1\\ 3\end{array}\right]t$$

$$P=\left[\begin{array}{cc}1& 1\\ -2& 3\end{array}\right],D=\left[\begin{array}{cc}-2& 0\\ 0& 3\end{array}\right]\Rightarrow \overrightarrow{u}\prime =\left[\begin{array}{cc}-2& 0\\ 0& 3\end{array}\right]\overrightarrow{u}$$

$$u_1\prime =-2u_1\Rightarrow u_1=c_1{e}^{-2x},u_2\prime =3u_2\Rightarrow u_2=c_2{e}^{3x}$$

$$\overrightarrow{y}=P\overrightarrow{u}=\left[\begin{array}{cc}1& 1\\ -2& 3\end{array}\right]\left[\begin{array}{c}{c}_1{e}^{-2x}\\ c_2{e}^{3x}\end{array}\right]=\left[\begin{array}{c}{c}_1{e}^{-2x}+c_2{e}^{3x}\\ -2c_1{e}^{-2x}+3c_2{e}^{3x}\end{array}\right]$$

We only need $y_1=y$:

Tip: Note that $y_2=-2c_1{e}^{-2x}+3c_2{e}^{3x}$ is $y_1\prime$, as it should be.