Problem 83
Question
Solve the initial value problems in Exercises \(71-90\) . $$\frac{d^{2} y}{d x^{2}}=2-6 x ; \quad y^{\prime}(0)=4, \quad y(0)=1$$
Step-by-Step Solution
Verified Answer
The solution is \( y(x) = x^2 - x^3 + 4x + 1. \)
1Step 1: Integrate to find first derivative
Start by integrating the given second-order derivative equation \( \frac{d^2 y}{dx^2} = 2 - 6x \) with respect to \( x \) to find the first derivative, \( y'(x) \). \[ y'(x) = \int (2 - 6x) \, dx = 2x - 3x^2 + C_1,\]where \( C_1 \) is an integration constant.
2Step 2: Use initial condition on first derivative
Use the initial condition \( y'(0) = 4 \) to find \( C_1 \).Substitute \( x = 0 \) and \( y'(0) = 4 \) into the equation:\[ 4 = 2(0) - 3(0)^2 + C_1 \implies C_1 = 4.\]Therefore, \( y'(x) = 2x - 3x^2 + 4. \)
3Step 3: Integrate to find original function
Integrate \( y'(x) = 2x - 3x^2 + 4 \) with respect to \( x \) to find the original function \( y(x) \):\[ y(x) = \int (2x - 3x^2 + 4) \, dx = x^2 - x^3 + 4x + C_2,\]where \( C_2 \) is another integration constant.
4Step 4: Use initial condition on function
Use the initial condition \( y(0) = 1 \) to find \( C_2 \).Substitute \( x = 0 \) and \( y(0) = 1 \) into the equation:\[ 1 = (0)^2 - (0)^3 + 4(0) + C_2 \implies C_2 = 1.\]Thus, \( y(x) = x^2 - x^3 + 4x + 1. \)
Key Concepts
Second-order differential equationInitial value problemIntegrationConstant of integration
Second-order differential equation
A second-order differential equation involves a function and its derivatives up to the second degree. In this context, it shows the rate of change of the rate of change of a function. The given problem is a prime example, expressed as \( \frac{d^{2} y}{d x^{2}} = 2 - 6x \). Here, we're looking for a function \( y(x) \) such that its second derivative equals \( 2 - 6x \). This type of equation is critical in various fields, such as physics and engineering, because it often arises in situations involving acceleration or elasticity.
To solve a second-order differential equation, one generally works to reduce it to a first-order derivative, and then finally to the original function. This is achieved through the process of integration. Each reduction step potentially brings a constant of integration, reflecting the initial conditions influencing the system.
To solve a second-order differential equation, one generally works to reduce it to a first-order derivative, and then finally to the original function. This is achieved through the process of integration. Each reduction step potentially brings a constant of integration, reflecting the initial conditions influencing the system.
Initial value problem
An initial value problem in differential equations is when you're given a differential equation along with some specified values, called initial conditions. These initial conditions ensure a unique solution by setting values at specific points.
- For the first derivative, the initial condition provided is \( y'(0) = 4 \).
- For the function, it is given as \( y(0) = 1 \).
Integration
Integration is a mathematical process used to find a function when its derivative is known. In solving differential equations, it's the reverse operation of differentiation. For both first and second-order differential equations, integration plays a crucial role in addressing parts of the equation back to the original function or its derivatives.
The solution process: Start with integrating the second derivative
- The first integration yields the first derivative: \( y'(x) = \int (2 - 6x) \, dx = 2x - 3x^2 + C_1 \).
Use the initial condition \( y'(0) = 4 \) to find \( C_1 \).
- Then, the second integration gives us the function itself: \( y(x) = \int (2x - 3x^2 + 4) \, dx = x^2 - x^3 + 4x + C_2 \).
Apply the condition \( y(0) = 1 \) to determine \( C_2 \).
Integration can be simple or complex depending on the function, but it remains a fundamental technique in solving differential equations.
The solution process: Start with integrating the second derivative
- The first integration yields the first derivative: \( y'(x) = \int (2 - 6x) \, dx = 2x - 3x^2 + C_1 \).
Use the initial condition \( y'(0) = 4 \) to find \( C_1 \).
- Then, the second integration gives us the function itself: \( y(x) = \int (2x - 3x^2 + 4) \, dx = x^2 - x^3 + 4x + C_2 \).
Apply the condition \( y(0) = 1 \) to determine \( C_2 \).
Integration can be simple or complex depending on the function, but it remains a fundamental technique in solving differential equations.
Constant of integration
When integrating a function, a constant of integration is introduced. This constant represents the family of all possible functions that differ only by a constant addition. It accounts for the indefinite nature of integration.
- When integrating the second derivative to find the first derivative, you get a constant of integration \( C_1 \).
- When integrating the first derivative to find the original function, another constant \( C_2 \) appears.
Other exercises in this chapter
Problem 81
Solve the initial value problems in Exercises \(71-90\) . $$\frac{d v}{d t}=\frac{1}{2} \sec t \tan t, \quad v(0)=1$$
View solution Problem 82
Solve the initial value problems in Exercises \(71-90\) . $$\frac{d v}{d t}=8 t+\csc ^{2} t, \quad v\left(\frac{\pi}{2}\right)=-7$$
View solution Problem 84
Solve the initial value problems in Exercises \(71-90\) . $$\frac{d^{2} y}{d x^{2}}=2-6 x ; \quad y^{\prime}(0)=4, \quad y(0)=1$$
View solution Problem 85
Solve the initial value problems in Exercises \(71-90\) $$\frac{d^{2} r}{d t^{2}}=\frac{2}{t^{3}} ; \quad\left.\frac{d r}{d t}\right|_{t=1}=1, \quad r(1)=1$$
View solution