Write an equation for finding initial conditions

So far, we have been finding general solutions to differential equations. However, differential equations are often used to describe physical systems, and the person studying that physical system usually knows something about the state of that system at one or more points in time. For example, if a constant-coefficient differential equation is representing how far a motorcycle shock absorber is compressed, we might know that the rider is sitting still on his motorcycle at the start of a race, time [latex]t=t_0[/latex]. This means the system is at equilibrium, so [latex]y(t_0)=0[/latex], and the compression of the shock absorber is not changing, so [latex]y'(t_0)=0[/latex]. With these two initial conditions and the general solution to the differential equation, we can find the specific solution to the differential equation that satisfies both initial conditions. This process is known as solving an initial-value problem. (Recall that we discussed initial-value problems in Introduction to Differential Equations.) Note that second-order equations have two arbitrary constants in the general solution, and therefore we require two initial conditions to find the solution to the initial-value problem.

Sometimes we know the condition of the system at two different times. For example, we might know [latex]y(t_0)=y_0[/latex] and [latex]y(t_1)=y_1[/latex]. These conditions are called boundary conditions , and finding the solution to the differential equation that satisfies the boundary conditions is called solving a boundary-value problem .

Mathematicians, scientists, and engineers are interested in understanding the conditions under which an initial-value problem or a boundary-value problem has a unique solution. Although a complete treatment of this topic is beyond the scope of this text, it is useful to know that, within the context of constant-coefficient, second-order equations, initial-value problems are guaranteed to have a unique solution as long as two initial conditions are provided. Boundary-value problems, however, are not as well behaved. Even when two boundary conditions are known, we may encounter boundary-value problems with unique solutions, many solutions, or no solution at all.

Example: solving an initial-value problem

Solve the following initial-value problem: [latex]y''+3y'-4y=0[/latex], [latex]y(0)=1[/latex], [latex]y'(0)=-9[/latex].

We already solved this differential equation in Example “Solving Second-Order Equations with Constant Coefficients” part a. and found the general solution to be

When [latex]x=0[/latex], we have [latex]y(0)=c_1+c_2[/latex] and [latex]y^\prime(0)=-4c_1+c_2[/latex]. Applying the initial conditions, we have

Then [latex]c_1=1-c_2[/latex]. Substituting this expression into the second equation, we see that

[latex]\begin -4(1-c_2)+c_2&=-9 \\ -4+4c_2+c_2&=-9 \\ 5c_2&=-5 \\ c_2&=-1 \end[/latex].

So, [latex]c_1=2[/latex] and the solution to the initial-value problem is

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Solve the initial-value problem [latex]y''-3y'-10y=0[/latex], [latex]y(0)=0[/latex], [latex]y'(0)=7[/latex].

Watch the following video to see the worked solution to the above Try It

Example: solving an initial-value problem and graphing the solution

Solve the following initial-value problem and graph the solution:

We already solved this differential equation in Example “Solving Second-Order Equations with Constant Coefficients” part b. and found the general solution to be

When [latex]x=0[/latex], we have [latex]y(0)=c_1[/latex] and [latex]y^\prime(0)=2c_2-3c_1[/latex]. Applying the initial conditions, we obtain

[latex]\begin c_1&=0 \\ -3c_1+2c_2&= 2 \end[/latex].

Therefore, [latex]c_1=0[/latex], [latex]c_2=1[/latex], and the solution to the initial value problem is shown in the following graph.

a graph of the function y = e^−3x sin 2x. The x axis is scaled in increments of tenths. The y axis is scaled in increments of even tenths. The curve passes through the origin and has a horizontal asymptote of the positive x axis." width="723" height="272" />

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Solve the following initial-value problem and graph the solution: [latex]y''-2y'+10y=10=0[/latex], [latex]y(0)=2[/latex], [latex]y'(0)=-1[/latex].

Figure 2. Graph of [latex]y(x)=e^(2\cos3x-\sin3x)[/latex].

Example: initial-value problem representing a spring-mass system

The following initial-value problem models the position of an object with mass attached to a spring. Spring-mass systems are examined in detail in Applications. The solution to the differential equation gives the position of the mass with respect to a neutral (equilibrium) position (in meters) at any given time. (Note that for spring-mass systems of this type, it is customary to define the downward direction as positive.)

Solve the initial-value problem and graph the solution. What is the position of the mass at time [latex]t=2[/latex] sec? How fast is the mass moving at time [latex]t=1[/latex] sec? In what direction?

In Example “Solving Second-Order Equations with Constant Coefficients” part c. we found the general solution to this differential equation to be

When [latex]t=0[/latex], we have [latex]y(0)=c_1[/latex] and [latex]y'(0)=-c_1+c_2[/latex]. Applying the initial conditions, we obtain

Thus, [latex]c_1=1[/latex], [latex]c_2=1[/latex], and the solution to the initial value problem is

This solution is represented in the following graph. At time [latex]t=2[/latex], the mass is at position [latex]y(2)=e^+2e^=3e^\approx0.406[/latex] [latex]m[/latex] below equilibrium.

To calculate the velocity at time [latex]t=1[/latex], we need to find the derivative. We have [latex]y(t)=e^+te^[/latex], so

Then [latex]y^\prime(1)=e^\approx-0.3679[/latex]. At time [latex]t=1[/latex], the mass is moving upward at [latex]0.3679[/latex] m/sec.

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Suppose the following initial-value problem models the position (in feet) of a mass in a spring-mass system at any given time. Solve the initial-value problem and graph the solution. What is the position of the mass at time [latex]t=0.3[/latex] sec? How fast is it moving at time [latex]t=0.1[/latex] sec? In what direction?

At time [latex]t=0.3[/latex], [latex]y(0.3)=0.3^=0.3e^\approx0.0367[/latex]. The mass is [latex]0.0367[/latex] ft below equilibrium. At time [latex]t=0.1[/latex], [latex]y^\prime(0.1)=0.3e^\approx0.1490[/latex]. The mass is moving downward at a speed of [latex]0.1490[/latex] ft/sec.

Example: solving a boundary-value problem

In Example “Solving Second-Order Equations with Constant Coefficients” part f. we solved the differential equation [latex]y''+16y=0[/latex] and found the general solution to be [latex]y(t)=c_1\cos4t+c_2\sin4t[/latex]. If possible, solve the boundary-value problem if the boundary conditions are the following:

1. [latex]y(0)=0[/latex], [latex]y\left(\frac<\pi>4\right)=0[/latex]
2. [latex]y(0)=1[/latex], [latex]y\left(\frac<\pi>8\right)=0[/latex]
3. [latex]y\left(\frac<\pi>8\right)=0[/latex], [latex]y\left(\frac<3\pi>8\right)=0[/latex]

1. Applying the first boundary condition given here, we get [latex]y(0)=c_1=0[/latex]. So the solution is of the form [latex]y(t)=c_2\sin4t[/latex]. When we apply the second boundary condition, though, we get [latex]y\left(\frac<\pi>4\right)=c_2\sin\left(4\left(\frac<\pi>4\right)\right)=c_2\sin\pi=0[/latex] for all values of [latex]c_2[/latex]. The boundary conditions are not sufficient to determine a value for [latex]c_2[/latex], so this boundary-value problem has infinitely many solutions. Thus, [latex]y(t)=c_2\sin4t[/latex] is a solution for any value of [latex]c_2[/latex].
2. Applying the first boundary condition given here, we get [latex]y(0)=c_1=1[/latex]. Applying the second boundary condition gives[latex]y(\frac<\pi>)=c_2=0[/latex], so [latex]c_2=0[/latex]. In this case, we have a unique solution: [latex]y(t)=\cos 4t[/latex].
3. Applying the first boundary condition given here, we get [latex]y\left(\frac<\pi>8\right)=c_2=0[/latex]. However, applying the second boundary condition gives [latex]y\left(\frac<3\pi>8\right)=-c_2=2[/latex], so [latex]c_2=-2[/latex]. We cannot have [latex]c_2=0=-2[/latex] so this boundary value problem has no solution.