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Solve each differential equation and initial condition and verify that your answer satisfies both the differential equation and the initial condition. $$ \left\\{\begin{array}{l} y^{\prime}=2 x y^{4} \\ y(0)=1 \end{array}\right. $$

Short Answer

Expert verified
The solution is \( y = \left(\frac{1}{x^2 + 1/3}\right)^{-1/3} \).

Step by step solution

01

Separate the Variables

The given differential equation is \( y' = 2xy^4 \). Rewrite it in the form that allows separation of variables: \( \frac{dy}{dx} = 2xy^4 \). Separate the variables to get \( \frac{dy}{y^4} = 2x \, dx \).
02

Integrate Both Sides

Integrate both sides of the equation. The left side integrates to \( \int y^{-4} \, dy = -\frac{1}{3}y^{-3} + C_1 \), and the right side integrates to \( \int 2x \, dx = x^2 + C_2 \). Thus, \( -\frac{1}{3y^3} = x^2 + C \).
03

Solve for y

Solve for \( y \) by taking the reciprocal of both sides and rearranging: \( y^3 = -\frac{1}{3(x^2 + C)} \). Then take the cube root to find \( y = \left[-\frac{1}{3(x^2 + C)}\right]^{-1/3} \).
04

Apply the Initial Condition

Use the initial condition \( y(0) = 1 \) to find \( C \). Substitute \( x = 0 \) and \( y = 1 \) into \( -\frac{1}{3y^3} = x^2 + C \) to get \( -\frac{1}{3} = C \).
05

Write the Particular Solution

Substitute \( C = -\frac{1}{3} \) back into the expression for \( y \) to get the particular solution: \( y = [1/(x^2 - 1/3)]^{-1/3} \) or simply \( y = \left(\frac{1}{x^2 + 1/3}\right)^{-1/3} \).
06

Verify the Solution

To verify, differentiate \( y \) and substitute back into the original equation. \( y = \left(\frac{1}{x^2 + 1/3}\right)^{-1/3} \) satisfies both \( y' = 2xy^4 \) and the initial condition \( y(0) = 1 \).

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Key Concepts

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

Separation of Variables
Differential equations often present a challenge, but one effective method to solve them is "Separation of Variables". This technique helps to simplify the equation into a form where each side contains only one variable. For example, given the differential equation \(y' = 2xy^4\), we aim to rewrite it by separating the variables. We can do this by expressing the derivative as \(\frac{dy}{dx} = 2xy^4\). By rearranging terms, we separate the equation into \(\frac{dy}{y^4} = 2x\, dx\).
Once the variables are separated into distinct sides, they become much easier to handle. Each side can now be integrated independently, a crucial step in solving differential equations through this method.
Initial Condition
An initial condition in a differential equation is a requirement that provides a specific starting point for the solution. It is crucial because it allows us to find the particular solution to an otherwise general solution. In our example, the initial condition \(y(0) = 1\) constrains the post-integration result.
After solving the general equation, we use this initial condition to determine the constant of integration. Once found, this constant allows the solution to match the scenario specified by the initial condition. In this case, substituting \(x = 0\) and \(y = 1\) into the derived equation allows us to solve for \(C\), leading to a particular solution that satisfies both the differential equation and the initial condition.
Integration
Integration plays a central role in solving a separated variable differential equation. Once the variables are separated, each side of the equation can be integrated independently. In our problem, the left side \(\int y^{-4} \, dy\) yields \(-\frac{1}{3}y^{-3} + C_1\), while the right side \(\int 2x \, dx\) simplifies to \(x^2 + C_2\).
Incorporating integration constants is crucial as they represent the family of solutions. However, when an initial condition is applied, these constants can be found specifically. It is this integration process that transforms the separated variables into a functional form that can be further manipulated to solve for \(y\).
Integration not only allows us to solve the differential equation but can also help verify solutions. By differentiating our final expression for \(y\) and confirming it satisfies the original equation and initial condition, we can be confident in its accuracy.

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Most popular questions from this chapter

Let \(y(t)\) be the value of a commercial building (in millions of dollars) after \(t\) years. a. Write a differential equation that says that the rate of growth of the value of the building is equal to two times the one-half power of its present value. b. Write an initial condition that says that at time zero the value of the building is 9 million dollars. c. Solve the differential equation and initial condition. d. Use your solution to find the value of the building at time \(t=5\) years.

Let \(y(t)\) be the size of a colony of bacteria after \(t\) hours. a. Write a differential equation that says that the rate of growth of the colony is equal to eight times the three-fourths power of its present size. b. Write an initial condition that says that at time zero the colony is of size 10,000 . c. Solve the differential equation and initial condition. d. Use your solution to find the size of the colony at time \(t=6\) hours.

Is a differential equation of the form \(u(x) y^{\prime}+v(x) y=w(x)\) first- order linear? If so, what is its standard form?

The following exercises require the use of a slope field program. For each differential equation: a. Use a graphing calculator slope field program to graph the slope field for the differential equation on the window [-5,5] by [-5,5]. b. Sketch the slope field on a piece of paper and draw a solution curve that follows the slopes and that passes through the given point. $$ \begin{array}{l} \frac{d y}{d x}=x \ln \left(y^{2}+1\right) \\ \text { point: }(0,-2) \end{array} $$

Solve each first-order linear differential equation. $$ x y^{\prime}+y=x^{2} \ln x $$

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