DE-BOOK Initial Conditions & Particular Solutions

Section 2.5 Initial Conditions & Particular Solutions

Solving a differential equation typically starts with finding the general solution that defines all the possible solutions (family of solutions). For example, later on in this text, we will show that the differential equation

\begin{equation} y'' - y' - 12y = 0\tag{5} \end{equation}

has the general solution

\begin{equation} y(t) = c_1e^{-3t} + c_2e^{4t}.\tag{6} \end{equation}

While the general solution gives us a family of possible solutions, in practical scenarios, we are usually interested in the specific solution where the constants are known. This is where initial conditions come into play—they allow us to determine the particular solution that applies to the problem at hand.

An initial condition is a starting value of the solution or its derivatives. For example, consider the free fall of an object due to gravity where \(s(t)\) is the position (distance from the ground) of this object at time \(t\text{.}\) Assuming the ball is dropped from the top of a 100 ft building, what initial information do you know? Let \(t=0\) be the time when the ball is dropped, then the initial conditions are given in the following table.

The following table summarizes the initial conditions for our example, highlighting the key information required to determine the particular solution:

Information at \(t=0\)	Initial Condition
The ball is at the top of a 100 ft building.	\(s(0) = 100 \text{ ft}\)
The ball is dropped from rest (initial velocity is zero).	\(s'(0) = 0 \text{ ft/s}\)

Without these initial conditions, the solution is incomplete and not applicable to any specific scenario. Assuming the solution must be consistent with these conditions forces the constants in the general solution to take specific values, giving us a particular solution.

Example 26.

\(\ \ \)Solve the initial value problem

\begin{equation*} y'' - y' - 12y = 0, \quad y(0) = 4, \quad y'(0) = -2\text{.} \end{equation*}

Solution.

Since we already know the general solution to (5) is given by

\begin{equation*} y = c_1e^{-3t} + c_2e^{4t}\text{.} \end{equation*}

If you are interested in the verification steps, see the solution below.

Solution. Verification as a solution

First find \(y, y', \text{ and } y''\text{:}\)

\begin{align*} y =\amp\ c_1e^{-3t} + c_2e^{4t} \\ y' =\amp\ -3c_1e^{-3t} + 4c_2e^{4t} \\ y'' =\amp\ 9c_1e^{-3t} + 16c_2e^{4t} \end{align*}

and substitute these into \(y'' - y' - 12y\) and simplify:

\begin{align*} y'' - y' - 12y =\amp\ \left(9c_1e^{-3t} + 16c_2e^{4t}\right) \\ \amp\ \hspace{1em} - \left( -3c_1e^{-3t} + 4c_2e^{4t} \right) -12\left( c_1e^{-3t} + c_2e^{4t} \right)\\ =\amp\ \ul{\ul{9c_1e^{-3t}}} + \ul{16c_2e^{4t}} \\ \amp\ \hspace{1em} + \ul{\ul{3c_1e^{-3t}}} - \ul{4c_2e^{4t}} - \ul{\ul{12c_1e^{-3t}}} - \ul{12c_2e^{4t}}\\ =\amp\ 0 \end{align*}

Which shows \(0 = 0\) and so \(y = c_1e^{-3t} + c_2e^{4t}\) is a solution.

We would like determine the spcific values of \(c_1\) and \(c_2\) that correspond to the initial conditions \(y(0) = 4\) and \(y'(0) = -2\text{.}\)

Since one initial condition involves the derivative, we will need to compute this before applying this condition.

\begin{equation*} y' = -3c_1e^{-3t} + 4c_2e^{4t}. \end{equation*}

Now let’s apply the initial conditions:

\begin{align*} y(0) =\amp\ 4 \amp\Rightarrow \hspace{0.2cm} 4 =\amp\ c_1e^{-3\cdot 0} + c_2e^{4\cdot 0}\\ \amp \amp 4 =\amp\ c_1 + c_2 \\ \amp \amp \amp \\ y'(0) =\amp\ -5 \amp\Rightarrow \hspace{0.2cm} -5 =\amp\ -3c_1e^{-3\cdot} + 4c_2e^{4\cdot} \\ \amp \amp -5 =\amp\ -3c_1 + 4c_2 \end{align*}

We now have two equations and two unknowns,

\begin{align*} c_1 + c_2 =\amp\ 4 \\ -3c_1 + 4c_2 =\amp\ -5 \end{align*}

There are multiple ways to solve this system of equations, and the reader is encouraged to verify that \(c_1 = 3\) and \(c_2 = 1\text{.}\) Here is one

Solution. Solving for \(c_1\) and \(c_2\)

We will solve for \(c_2\) in the first equation and then substitute into the second.

\begin{align*} 4 =\amp\ c_1 + c_2 \\ \Rightarrow c_2 =\amp\ 4 - c_1 \\ \amp \\ -5 =\amp\ -3c_1 + 4c_2 \\ \Rightarrow -5 =\amp\ -3c_1 + 4(4 - c_1) \\ -5 =\amp\ -3c_1 + 16 - 4c_1 \\ -21 =\amp\ -7c_1 \\ 3 =\amp\ c_1 \\ \amp \\ c_2 =\amp\ 4 - c_1 \\ =\amp\ 4 - 3 \\ =\amp\ 1 \end{align*}

Thus, the solution we that satisfies the DE and both of the provided initial conditions is

\begin{equation*} y = 3e^{-3t} + e^{4t}. \end{equation*}

Example 27.

\(\ \ \)Solve the initial value problem

\begin{equation} \frac{dy}{dx} = 2xy - 6x, \quad y(0) = 2\tag{7} \end{equation}

Given that the general solution is \(\ds\ y = ce^{x^2} + 3\text{.}\)

Solution.

The initial condition, \(y(0) = 2\text{,}\) tells us that “\(y\) must equal 2 when \(x\) is 0”. When we substitute \(y=2\) and \(x=0\) into the general solution, we see that we can solve for \(c\text{.}\)

\begin{align*} y = ce^{x^2} + 3 \quad \overset{y\ =\ 2,\ x\ =\ 0}{\Rightarrow} \quad 2 =\amp\ ce^{0^2} + 3 \\ 2 =\amp\ c + 3 \\ -1 =\amp\ c \end{align*}

Therefore, \(y(0) = 2\) forces \(c = -1\) and we get the particular solution

\begin{equation*} y = -e^{x^2} + 3\text{.} \end{equation*}

Initial conditions can also be visualized. Figure 28 the initial condition, \(y(0) = 2\text{,}\) is shown as the point \((0,2)\) on the \(y\)-axis. The blue curve is the particular solution we found above and is the only solution in the family that passes through this point.

In the graph, the point on the \(y\)-axis is the initial condition. You can change the initial condition by dragging the point up and down along the \(y\)-axis.

As you drag the point, make the following observations:

Each new value of \(y(0)\) gives a new value of \(c\text{,}\) which corresponds to a new particular solution (blue curve).
Only one particular solution from the family of solutions (green curves) passes through the initial condition.

Figure 28. The family of solutions (green curves) and a particular solution (blue curve) passing through \(y(0)\text{.}\)

In summary, the process of solving a differential equation typically begins with finding a general solution. However, to apply this solution to real-world problems, we rely on initial conditions to determine a particular solution that accurately represents the system’s specific circumstances.

Reading Questions Check your Understanding

1. What is an initial condition?

What is an initial condition?

The starting point of the independent variable.
Incorrect. The initial condition is related to the solution or its derivatives at a specific point.
The starting value of the solution or its derivatives.
Correct! An initial condition specifies the value of the solution or its derivative at a particular point.
The final value of the solution.
Incorrect. The initial condition is at the starting point, not the end.
The point where the solution intersects the x-axis.
Incorrect. The initial condition is about the value at a specific point, not necessarily where it intersects the x-axis.

2. The differential equation, \(\ds\frac{dy}{dx} = 2xy - 6x\text{,}\) is an initial value problem.

The differential equation, \(\ds\frac{dy}{dx} = 2xy - 6x\text{,}\) is an initial value problem

True
Incorrect. An initial value problem includes specific initial conditions, which are not provided in this equation.
False
Correct! The equation is a differential equation without initial conditions, so it is not an initial value problem.

3. The solution to the initial value problem, \(\ds\frac{dy}{dx} = 2xy - 6x, \hspace{0.25cm} y(5) = -10\text{,}\) is a function whose curve passes through the point .

The solution to the initial value problem, \(\ds\frac{dy}{dx} = 2xy - 6x, \hspace{0.25cm} y(5) = -10\text{,}\) is a function whose curve passes through the point

\((5, 0)\)
Incorrect. The solution curve does not pass through the point (5, 0).
\((0, 5)\)
Correct! The solution curve passes through the point (0, 5).
\((0, 0)\)
Incorrect. The solution curve does not pass through the point (0, 0).
\((2, 6)\)
Incorrect. The solution curve does not pass through the point (2, 6).

4. In which case would you need to find a particular solution rather than just a general solution?

In which case would you need to find a particular solution rather than just a general solution?

When the differential equation is linear.
Incorrect. Linearity doesn’t determine whether you need a particular solution.
When initial conditions are provided.
Correct! A particular solution is found when you need to satisfy initial conditions.
When the differential equation is has a general solution.
Incorrect. The existence of a general solution doesn’t determine whether you need a particular solution.
When the function depends on more than one variable.
Incorrect. Multivariable functions are not the reason for finding a particular solution.

5. What role do initial conditions play in solving differential equations?

Select all that apply.

They determine the general form of the solution.
Incorrect. Initial conditions are not used to find the general solution.
They used to determine the constants in the general solution.
Correct! Initial conditions are used to find specific values for constants in the general solution.
They are used to find the particular solution.
Correct! Initial conditions are used to find the specific solution that applies to a particular scenario.
They are not needed if the general solution is already known.
Incorrect. If provided, initial conditions are always needed to get the particular solution from the general solution.

6. What happens if you do not apply initial conditions to the general solution of a differential equation?

What happens if you do not apply initial conditions to the general solution of a differential equation?

You get a family of solutions without identifying the particular solution.
Correct! Without initial conditions, you only have the general solution, which represents a family of possible solutions.
You cannot find any solution to the differential equation.
Incorrect. You can find the general solution without initial conditions, but not the specific one.
The solution to the differential equation does not exist.
Incorrect. The solution exists, but you need initial conditions to find the specific one.
The solution to the differential equation is unique.
Incorrect. The general solution represents a family of possible solutions, not a unique one.

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