Initial Value Problems: Videos & Practice Problems

Understanding initial value problems involves solving differential equations that relate an unknown function to its derivative. The process begins with integration to find the general solution, which includes a constant of integration. Applying an initial condition allows for determining this constant, leading to the particular solution. For example, integrating a derivative like $dy/dx$ yields the original function, which can then be adjusted using the initial condition to find the specific solution.

Initial Value Problems

In calculus, understanding the relationship between derivatives and integrals is crucial, especially when tackling initial value problems. An initial value problem involves a differential equation, which is an equation that includes an unknown function and its derivative, along with an initial condition that provides a specific point on the function.

To illustrate, consider a differential equation expressed as $dy/dx=f\left(x\right)$. The goal is to find the function $y\left(x\right)$ that satisfies this equation. The process begins by integrating both sides of the equation. For example, if the right side is $3x^2-4$, the integral of the left side $dy/dx$ gives us the original function $y$, while integrating the right side yields $x^3-4x+c$, where $c$ is the constant of integration.

This result is known as the general solution. To find the particular solution, we apply the initial condition, which specifies a value for $y$ at a certain $x$. For instance, if we know that $y\left(1\right)=-1$, we substitute $x$ with 1 in our general solution to solve for $c$. This process leads to a specific value for $c$, allowing us to express the particular solution as $y\left(x\right)=x^3-4x+2$.

In cases involving higher-order derivatives, the same integration process is applied multiple times. For example, if dealing with a second derivative, one would integrate twice to arrive at the original function, often requiring additional initial conditions to fully determine the solution. Mastering these techniques is essential for solving a variety of initial value problems effectively.

Solve the following initial value problem:
$\frac{dy}{dx}=2x-5$; $y\left(0\right)=4$

• A. $y\left(x\right)=2x^2-5x$
• B. $y\left(x\right)=x^2-5x$
• C. $y\left(x\right)=2x^2-5x+4$
• D. $y\left(x\right)=x^2-5x+4$

Answer: (D)

Using the acceleration function below, find the velocity function, if the velocity is v = 5 at time t = 2.
$a\left(t\right)=-20$

• A. $v\left(t\right)=-20t+10$
• B. $v\left(t\right)=-20t+5$
• C. $v\left(t\right)=-20t+45$
• D. $v\left(t\right)=-20t-35$

Answer: (C)

Find the function $f\left(x\right)$ that satisfies the following differential equation.
$f^{\prime\prime}\left(x\right)=3x^2$; $f^{\prime}\left(0\right)=1$; $f\left(1\right)=3$

• A. $f\left(x\right)=3x^4+x-1$
• B. $f\left(x\right)=\frac{x^4}{4}+x+\frac74$
• C. $f\left(x\right)=3x^4+x+\frac74$
• D. $f\left(x\right)=\frac{x^4}{4}+x-1$

Answer: (B)

Initial Value Problems Example 1

In solving an initial value problem involving a second-order differential equation, we start with the equation given by the second derivative of a function \( y \) with respect to \( x \). For example, if we have:

\[\frac{d^2y}{dx^2} = 4x^3 - 10\]

To find the first derivative \( y' \), we integrate the right side of the equation. This process involves applying the power rule for integration:

\[y' = \int (4x^3 - 10) \, dx = x^4 - 10x + C\]

Here, \( C \) is the constant of integration. To determine \( C \), we use the initial condition provided, such as \( y'(1) = -2 \). Substituting \( x = 1 \) into the equation gives:

\[-2 = 1^4 - 10(1) + C\]

Solving this, we find \( C = 7 \). Thus, the first derivative becomes:

\[y' = x^4 - 10x + 7\]

Next, to find the original function \( y \), we integrate \( y' \) again:

\[y = \int (x^4 - 10x + 7) \, dx = \frac{x^5}{5} - 5x^2 + 7x + D\]

Here, \( D \) is another constant of integration. We can find \( D \) using the second initial condition, such as \( y(0) = 3 \). Substituting \( x = 0 \) gives:

\[3 = \frac{0^5}{5} - 5(0^2) + 7(0) + D\]

This simplifies to \( D = 3 \). Therefore, the final solution for \( y \) is:

\[y = \frac{x^5}{5} - 5x^2 + 7x + 3\]

In summary, solving initial value problems with higher-order derivatives involves integrating the function multiple times and applying the given initial conditions to find the constants of integration. This method remains consistent regardless of the number of derivatives involved, allowing for systematic problem-solving in differential equations.

Initial Value Problems Example 2

To find the position function \( s(t) \) of an object given its acceleration function \( a(t) \), initial velocity \( v(0) = 9 \), and initial position \( s(0) = 0 \), we can follow a systematic approach involving integration of the acceleration function.

First, we recognize the relationship between acceleration, velocity, and position. The acceleration function \( a(t) \) is the second derivative of the position function \( s(t) \), while the velocity function \( v(t) \) is the first derivative of \( s(t) \). Therefore, to find the position function, we need to integrate the acceleration function twice.

Given that the acceleration is constant at \( a(t) = -9.8 \, \text{m/s}^2 \), we start by integrating this function to find the velocity:

\[v(t) = \int a(t) \, dt = \int -9.8 \, dt = -9.8t + C\]

Here, \( C \) is the constant of integration. To determine \( C \), we use the initial condition \( v(0) = 9 \):

\[v(0) = -9.8(0) + C = 9 \implies C = 9\]

Thus, the velocity function becomes:

\[v(t) = -9.8t + 9\]

Next, we integrate the velocity function to find the position function:

\[s(t) = \int v(t) \, dt = \int (-9.8t + 9) \, dt = -4.9t^2 + 9t + D\]

Again, \( D \) is a constant of integration. We apply the initial condition \( s(0) = 0 \) to find \( D \):

\[s(0) = -4.9(0)^2 + 9(0) + D = 0 \implies D = 0\]

Therefore, the position function simplifies to:

\[s(t) = -4.9t^2 + 9t\]

This function describes the position of the object over time, taking into account the initial conditions provided. The process illustrates how to work backwards from acceleration to position through integration, while carefully applying initial conditions to find specific solutions.

Here’s what students ask on this topic:

What is an initial value problem in calculus?

An initial value problem in calculus involves solving a differential equation that includes an unknown function and its derivative, along with an initial condition. The initial condition provides a specific value for the function at a particular point, which helps determine the constant of integration. This process allows us to find the particular solution to the differential equation.

How do you solve an initial value problem?

To solve an initial value problem, follow these steps: 1) Integrate the given differential equation to find the general solution, which includes a constant of integration. 2) Apply the initial condition to determine the value of the constant. 3) Substitute the constant back into the general solution to obtain the particular solution. This process ensures that the solution satisfies both the differential equation and the initial condition.

What is the difference between a general solution and a particular solution in initial value problems?

The general solution of a differential equation includes a constant of integration, representing a family of solutions. The particular solution is obtained by applying the initial condition to determine the specific value of the constant. This particular solution satisfies both the differential equation and the initial condition, making it unique to the given problem.

Can initial value problems involve higher-order derivatives?

Yes, initial value problems can involve higher-order derivatives. In such cases, you need to integrate multiple times to find the original function. For example, if you have a second derivative, you would first integrate to find the first derivative, then integrate again to find the original function. Each integration step may require applying initial conditions to determine the constants of integration.

What is a differential equation in the context of initial value problems?

A differential equation in the context of initial value problems is an equation that involves an unknown function and its derivative(s). It describes the relationship between the function and its rate of change. Solving the differential equation involves finding the function that satisfies the equation, often using integration and applying initial conditions to determine specific constants.