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For each differential equation, find the particular solution indicated. HINT [See Example 2b.] \(\frac{d y}{d x}=x^{3}-2 x ; y=1\) when \(x=0\).

Short Answer

Expert verified
The particular solution to the differential equation \(\frac{d y}{d x}=x^{3}-2 x\) with initial condition \(y=1\) when \(x=0\) is \(y= \frac{1}{4}x^{4} - x^{2} + 1\).

Step by step solution

01

Integrate the Differential Equation

To find the general solution to the differential equation \(\frac{d y}{d x}=x^{3}-2 x\), we need to integrate both sides of the equation with respect to x. On the left side, integrating \(dy\) with respect to \(x\) gives \(y\). An antiderivative of \(x^{3}-2 x\) is \(\frac{1}{4}x^{4} - x^{2}\). Therefore, the general solution to the differential equation is: \[y= \frac{1}{4}x^{4} - x^{2} + C,\] where \(C\) is the constant of integration, which can be determined using the given initial condition.
02

Apply the Initial Condition

Now you should substitute the given initial condition, \(y=1\) when \(x=0\), into the general solution and solve for \(C\). Plugging \(x=0\) and \(y=1\) into the equation yields: \[1= \frac{1}{4}(0)^{4} - (0)^{2} + C.\] Therefore, \(C=0\).
03

Write Down the Particular Solution

Using the value of the constant \(C\) found in Step 2, the particular solution to the differential equation that satisfies the initial condition is: \[y= \frac{1}{4}x^{4} - x^{2} + 1.\]

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Key Concepts

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

Integrating Differential Equations
Understanding how to integrate differential equations is essential for solving many types of problems in mathematics and applied sciences. A differential equation involves an unknown function and its derivatives, expressing a relationship between the two. The process to solve such an equation usually begins with integration, which is the reverse operation of differentiation.

Consider the given differential equation \(\frac{d y}{d x}=x^{3}-2 x\) which represents the rate of change of a function \(y\) with respect to \(x\). Integration is carried out to find a function \(y\) that describes the accumulated change. Integral calculus provides us with techniques to find an antiderivative, which in this case is \(\frac{1}{4}x^{4} - x^{2}\) plus an arbitrary constant \(C\). This result is called the general solution because it contains the constant \(C\) which needs to be determined for a specific situation.

Integrating differential equations often requires familiarity with a variety of integration techniques, including the power rule, integration by parts, and substitution. It's beneficial for students to practice these methods frequently to gain proficiency and confidence in solving differential equations.
Initial Condition Application
After obtaining the general solution of a differential equation through integration, the next step is to apply the initial conditions to find a specific or 'particular' solution. Initial conditions are values that the solution must satisfy at a particular point, and they are crucial for determining the constant \(C\) in the general solution.

Using the initial condition provided, \(y=1\) when \(x=0\), we can find the specific value of the constant \(C\). Substituting \(x=0\) and \(y=1\) into the general solution, \(y= \frac{1}{4}x^{4} - x^{2} + C\), simplifies the equation to \(1=0+0+C\), leading to \(C=1\).

The application of initial conditions is a fundamental concept in differential equations often used in real-world problems. For instance, when modeling physical phenomena like motion, growth, or decay, the initial state of the system provides these conditions. Understanding and applying initial conditions correctly is vital to ensure accurate and relevant solutions in various scientific fields.
Particular Solution of Differential Equations
A particular solution to a differential equation is a solution that not only satisfies the differential equation but also fulfills a given set of initial conditions. Once the constant of integration, \(C\), is determined through the initial conditions, this constant is plugged back into the general solution to obtain the particular solution.

In our exercise, the particular solution is found by substituting \(C=1\) into the general solution to get \[y= \frac{1}{4}x^{4} - x^{2} + 1\.\] This equation now specifically satisfies the condition that when \(x=0\), \(y\) must equal 1. It's also important to note that, although many different functions could satisfy the original differential equation, the initial condition narrows it down to this unique particular solution.

Particular solutions are incredibly useful because they represent the actual behavior of a system under specific circumstances. In engineering, physics, economics, and other applied fields, finding the particular solution helps predict outcomes and model realistic scenarios. Therefore, fine-tuning the art of determining particular solutions is not just a mathematical exercise but also a practical skill for many fields.

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Most popular questions from this chapter

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