1.3 - Classification of Differential Equations

Categorizing differential equations by different types will inform our solution techniques. The most important features are

• The number of dependent/independent variables,
• The highest-order derivative present,
• Whether the equation is linear.

For the first feature, we have the following definitions.

Definition \(\PageIndex{1}\): Ordinary Differential Equation (ODE)

A differential equation with a single independent variable and a single dependent variable.

Example

\[ \notag
\frac{d y}{d t}=a y+b
\]

Definition \(\PageIndex{2}\): Partial Differential Equation (PDE)

A differential equation with several independent variables and a single dependent variable.

Example

\[ \notag
\frac{\partial x y}{\partial x}=\frac{\partial x y}{\partial y}
\]

Definition \(\PageIndex{3}\): System of Ordinary Differential Equations

A collection of differential equations with one independent variable and several dependent variables.

Example

\[ \notag
\begin{align*}
\frac{dx_1}{dt} &= 3x_1 - 2x_2, \\
\frac{dx_2}{dt} &= -2x_1 + x_2.
\end{align*}
\]
Note that \(x_1(t)\) and \(x_2(t)\) are functions of \(t\).

 Toward the second feature, we have the important definition below.

Definition \(\PageIndex{4}\): Order

The order of a differential equation is the order of the highest derivative present in the equation.

Examples

• \(\frac{dv}{dt} = g - \frac{\gamma}{m}v \quad \) is a \(1\)-st order (ordinary) differential equation.
• \(m\frac{d^2x}{dt^2} + kx = 0 \quad \) is a \(2\)-nd order (ordinary) differential equation.
• \(\frac{\partial u}{\partial t} = \alpha \frac{\partial^2 u}{\partial x^2} \quad \) is a \(2\)-nd order (partial) differential equation called the Heat Equation.
• \(y^{(n)}=f\left(t, y, \ldots, y^{(n-1)}\right) \quad \) is the general expression of an \(n\)-th order (ordinary) differential equation.

The last feature is the biggest factor toward our solution techniques, and borrows a concept from Linear Algebra.

Definition \(\PageIndex{5}\): Linearity

An ordinary differential equation is linear if it can be written as a linear combination involving \(y, y', \dots, y^{(n)}\). A general linear \(n\)-th order differential equation has the form

\[ \notag
a_n(t)y^{(n)} + a_{n-1}(t)y^{(n-1)} + \cdots + a_1(t)y' + a_0(t)y = g(t),
\]

for some coefficient functions \(a_i(t)\). If an equation cannot be written in this way, it is non-linear.

Examples

Below are examples of linear equations:

• \(2\frac{d^2x}{dt^2} + 2t\frac{dx}{dt} + 5 = 10\cos(t)\).
• \(t^3y' + 3t^2y = 4\).
• \(\frac{dv}{dt} = g - \frac{\gamma}{m}v\).

Below are examples of non-linear equations:

• \(\frac{dy}{dt} = 0.5\left(1-\frac{y}{3}\right)y \quad \) is non-linear because of the \(y^2\) term after distributing.
• \(\sin\left(\frac{dy}{dx}\right) = xy \quad \) is non-linear (only) because of the \(\sin\left(\frac{dy}{dx}\right)\) term.

Exercise \(\PageIndex{6}\)

Determine the linearity of the following differential equations.

1) \(ty'' - t^3y' - 3y = \sin(t)\)

2) \(2y'' - 3y' - 3y^2 = 0\)

Answer

Both equations are linear since only \(y\) and its derivatives \(y'\) and \(y''\) are present, and not functions of them.

Exercise \(\PageIndex{7}\)

Q1) For which value of \(r\) is \(y(t)=e^{rt}\) a solution of the linear \(1\)-st order differential equation \(2 y^{\prime}-6 y=0\)? 
Q2) Is \(y(t) = Ce^{3t}\) also a solution?

Hint

A function \(y(t)\) is a solution if the differential equation is true after 'plugging in' \(y(t)\).

Answer Q1

In order to check if \(y\) is a solution, we also need its derivative \(y' = \frac{d}{dt}y = \frac{d}{dt}e^{rt} = re^{rt}\). In order for \(y\) to be a solution, we need the following to be true:

\[ \notag
\begin{align*}
2y'-6y=0 \quad &\Longleftrightarrow \quad 2(re^{rt}) - 6(e^{rt}) = 0, \\
&\Longleftrightarrow \quad (2r-6)e^{rt} = 0, \\
&\Longleftrightarrow \quad 2r-6 = 0. \quad (\text{Since} \,\, e^{rt} \,\, \text{is never zero}).
\end{align*}
\]

Thus, \(y(t) = e^{rt}\) is a solution to the differential equation only if \(r=3\).

Answer Q2

For \(y(t) = Ce^{3t} \) we have \(y' = \frac{d}{dt}Ce^{3t} = C\frac{d}{dt}e^{3t} = 3Ce^{rt}\). We see that

\[ \notag
\begin{align*}
2y'-6y=0 \quad &\Longleftrightarrow \quad 2(3Ce^{rt}) - 6(Ce^{rt}) = 0, \\
&\Longleftrightarrow \quad (6-6)Ce^{rt} = 0,
\end{align*}
\]

which is true regardless of the value of \(C\).

The answer to Q2 in the previous example shows an interesting feature of linear differential equations: \(Cy\) is a solution whenever \(y\) is. If we want to further narrow down our solutions, we need more information.

Definition \(\PageIndex{8}\): Initial Value Problem (IVP)

An initial value problem is an \(n\)-th order differential equation together with initial conditions on its solutions and their lower-order derivatives. Initial value problems may have no solutions, a unique solution, or infinitely-many solutions.

Examples

Below are examples of IVPs for \(1\)-st order differential equations.

• \(\frac{dv}{dt} = g - \frac{\gamma}{m} v; \quad v(0) = 60\).
• \(2y'-6y=0; \quad y(0)=4\).

Below are examples of IVPs for \(2\)-nd order differential equations.

• \(\frac{d^2x}{dt^2} + 4x = 0; \quad x(0)=1, \quad x'(0)=0\).
• \( y'' + 2y' + 5y = \cos(t); \quad y(0) = 0, \quad y'(0) = 2\).

We will make this more precise in future chapters, but generically an \(n\)-th order IVP with \(n\)-many initial conditions (one for each lower-order derivative) admits a unique solution.

Exercise \(\PageIndex{9}\)

Determine the unique solution of the IVP \(2y' - 6y=0; \quad y(0)=4\).

Answer

From exercise 2 above, we know \(y=C e^{3 t}\) is a solution to \(2 y^{\prime}-6 y=0\).

Plug in initial value to find \(C\) :
If \(y(0)=4\), then \(4=C e^{3(0)}\) implies \(C=4\).
Thus by existence and uniqueness theorem, \(y=4 e^{3 t}\) is the unique solution to IVP: \(2 y^{\prime}+6 y=0, y(0)=4\).

Example \(\PageIndex{10}\)

Suppose salty water with a salt concentration of \(3 \left[\frac{\mathrm{g}}{\mathrm{L}}\right]\) enters and leaves a tank at a rate of \(2 \left[\frac{\mathrm{L}}{\mathrm{min}}\right]\). If the tank contains \(4 [\mathrm{L}]\) of water and initially contains \(5 [\mathrm{g}]\) of salt, find a formula for the amount of salt in the tank after \(t\) minutes.

Solution

Let \(Q(t)\) be the amount of salt in the tank measured in grams \([\mathrm{g}]\). This amount is probably changing with time, so let's try to construct an expression for \(\frac{dQ}{dt}\).  Also since we are tracking something inside the tank, the expression for its rate of change is likely

\[ \notag
\frac{dQ}{dt} = \,\, ``\text{Rate of salt flowing in"} \,\, - \,\, ``\text{Rate of salt flowing out"}
\]

Both sides of this expression must have units \( \left[ \frac{\mathrm{g}}{\mathrm{min}} \right] \). The first two pieces of data are related to something "flowing in", and their product has the right units:

\[ \notag
``\text{Rate of salt flowing in"} \,\, = 3 \left[\frac{\mathrm{g}}{\mathrm{L}}\right] \cdot 2 \left[\frac{\mathrm{L}}{\mathrm{min}}\right]
= 6 \left[\frac{\mathrm{g}}{\mathrm{min}}\right].
\]

The only other "flowing out" data is that the water flow rate is the same. We need a \(\left[\frac{\mathrm{g}}{\mathrm{L}}\right]\) term, which has the type of units of a concentration. We can use \(Q(t)\) and the volume of the tank:

\[ \notag
``\text{Rate of salt flowing out"} \,\, = \frac{Q(t)}{4} \left[\frac{\mathrm{g}}{\mathrm{L}}\right] \cdot 2 \left[\frac{\mathrm{L}}{\mathrm{min}}\right]
= 4Q \left[\frac{\mathrm{g}}{\mathrm{min}}\right].
\]

Putting it all together, and including the initial condition \(Q(0) = 5 [\mathrm{g}]\), we have

\[ \notag
\frac{dQ}{dt} = 6 - \frac{Q}{2}; \quad Q(0) = 5.
\]

This resulting equation is a linear \(1\)-st order IVP. We will learn techniques to solve this differential equation in the following chapter.

(Bump the following to S2.3?)

Using the easier 2.2 (matter of opinion):
\[ \notag
\begin{array}{l}
\frac{d Q}{d t}=6-\frac{Q}{2} \Rightarrow 2 d Q=(12-Q) d t \Rightarrow \frac{2 d Q}{12-Q}=d t \\
\int \frac{2 d Q}{12-Q}=\int d t \Rightarrow-2 \ln |12-Q|=t+C \\
\ln |12-Q|=\frac{-t}{2}+C \Rightarrow|12-Q|=e^{\frac{-t}{2}+C}=e^{\frac{-t}{2}} e^C \\
12-Q= \pm e^C e^{\frac{-t}{2}}=C e^{\frac{-t}{2}} \Rightarrow Q=C e^{\frac{-t}{2}}+12 \\
Q(0)=5: \quad 5=C e^0+12=C+12 . \text { Thus } C=-7\\
\text { Thus } Q(t)=-7 e^{\frac{-t}{2}}+12
\end{array}
\]

We then see that the long term behaviour of the solution \(\lim _{t \rightarrow \infty}-7 e^{\frac{-t}{2}}+12=12\)
Thus the amount of salt in tank approaches \(12 \mathrm{~g}\) as \(t \rightarrow \infty\). Hence the concentration of salt approaches:
\[ \notag
12 \mathrm{~g} / 4 \mathrm{~L}=3 \mathrm{~g} / \mathrm{L} \text { as } t \rightarrow \infty \text {. }
\]