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Chapter 8: Problem 14

Solve the initial-value problem. $$ y^{\prime}=2 x, y(1)=7 $$

Short Answer

Expert verified
The particular solution is \(y = x^2 + 6\).

Step by step solution

01

Identify the Initial-Value Problem

The initial-value problem provided is: \(y^{\text{'}}=2x\) with initial condition \(y(1)=7\).
02

Integrate the Differential Equation

To find the general solution, integrate both sides of the differential equation: \( y = \int 2x \ dx \). Integrating, we get: \( y = x^2 + C \), where \(C\) is the integration constant.
03

Apply the Initial Condition

Use the initial condition \(y(1)=7\) to find \(C\). Substitute \(x=1\) and \(y=7\) into the general solution: \(7 = 1^2 + C\), solving for \(C\), we get \(C = 6\).
04

Write the Particular Solution

Substitute \(C = 6\) back into the general solution: \(y = x^2 + 6\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Differential Equations
A differential equation is an equation involving a function and its derivatives. In simple terms, it relates a function with its rate of change.
For example, given the differential equation \(y^{'} = 2x\), it shows how the function y changes with respect to x.
Solving differential equations involves finding a function that satisfies the equation.
This is essential in many fields such as physics, engineering, and biology where changes are analyzed.
Better understanding differential equations helps in solving real-life problems like motion, growth, and decay.
Integration
Integration is the reverse process of differentiation. It helps in finding an original function given its derivative.
To solve our differential equation \(y^{'} = 2x\), we integrate both sides with respect to x.
This gives us \(y = \int 2x dx\), which results in \(y = x^2 + C\).
Here, C is the constant of integration which is determined using additional information.
Integration is a fundamental tool in calculus and is widely used to find areas, volumes, and solutions to differential equations.
Initial Conditions
Initial conditions are specific values given to determine the particular solution of a differential equation. They provide additional information that helps find the constant of integration C.
In our problem, the initial condition is \(y(1)=7\). This means when x is 1, y is 7.
By substituting into our general solution \(y = x^2 + C\), we get \(7 = 1^2 + C\), hence C = 6.
Initial conditions are crucial because they allow us to find the unique solution that fits the given situation.
Particular Solution
The particular solution of a differential equation is the specific solution that satisfies both the differential equation and the initial conditions.
After finding C using the initial condition, substitute it back into the general solution.
For our example, substituting C = 6 into \(y = x^2 + C\) gives us \(y = x^2 + 6\).
This is our particular solution, meeting the original differential equation and the initial condition.
Particular solutions are vital because they precisely address the problem at hand, providing a specific answer instead of a general formula.

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Most popular questions from this chapter

Verify that the given function is a solution of the differential equation. $$ \prime^{\prime \prime}-4 y^{\prime}+3 y=0 ; y=2 e^{x}-7 e^{3 x} $$

Solve the differential equation. Be sure to check for possible constant solutions. If necessary, write your answer implicitly. $$ \frac{d y}{d t}=t^{2} e^{t^{3}} y^{3} $$

Solve the differential equation. Be sure to check for possible constant solutions. If necessary, write your answer implicitly. $$ y^{\prime}=x^{2} y^{3} $$

Solve the initial-value problem. State an interval on which the solution exists. $$ t^{2} y^{\prime}+t y=t^{4} ; y(2)=5 $$

Solve the initial-value problem. State an interval on which the solution exists. $$ y^{\prime}+4 t y=t ; y(0)=3 $$
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