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Chapter 9: Problem 1

\(1-10\) Solve the differential equation. $$ \frac{d y}{d x}=3 x^{2} y^{2} $$

Short Answer

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

Step by step solution

01

Separate Variables

To solve the differential equation \( \frac{d y}{d x}=3 x^{2} y^{2} \), start by separating the variables. This means we want to get all \( y \) terms on one side and all \( x \) terms on the other side. Divide both sides by \( y^2 \) and multiply both sides by \( dx \):\[ \frac{1}{y^2} dy = 3x^2dx \]. This separation allows us to integrate both sides with respect to their own variables.
02

Integrate Both Sides

Now, integrate both sides of the equation: \[ \int \frac{1}{y^2} \, dy = \int 3x^2 \, dx \]. Integrating the left side with respect to \( y \), we get \( -\frac{1}{y} + C_1 \). Integrating the right side with respect to \( x \), we get \( x^3 + C_2 \). Thus, the equation becomes \( -\frac{1}{y} = x^3 + C \), where \( C = C_2 - C_1 \) is an arbitrary constant.
03

Solve for y

Rearrange the integrated equation to solve for \( y \). Begin by isolating \( \frac{1}{y} \): \( \frac{1}{y} = -x^3 - C \). Then, take the reciprocal of both sides to obtain \( y = -\frac{1}{x^3 + C} \). This represents the general solution of the differential equation.
04

Verify Solution Form

Check the solution form to ensure it correctly represents possible constants. \( y = -\frac{1}{x^3 + C} \) is the general solution, and you can substitute different values of the constant \( C \) to get specific solutions matching initial conditions, if any are provided.

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Key Concepts

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

Separate Variables
One essential technique in solving differential equations is separating variables. In this method, you rearrange the equation to place all terms containing the dependent variable ( y ) on one side and all terms containing the independent variable ( x ) on the other side. For our equation, \( \frac{dy}{dx} = 3x^2 y^2 \), the goal is to separate y and x .
This step involves simple algebraic manipulation:

  • Divide both sides by y^2 to obtain terms involving y isolated on one side.
  • Multiply both sides by dx to get terms involving x isolated on the other side.
After following these steps, the equation transforms into \( \frac{1}{y^2} dy = 3x^2 dx \), with dy on one side and dx on the other. Now the equation is ready for the next step: integration!
Integration
Once you have separated the variables, the next step is integrating both sides with respect to their respective variables. This process will help you find a function y(x) that satisfies the differential equation.

For the separated equation \( \frac{1}{y^2} dy = 3x^2 dx \):
  • Integrate the left side with respect to y , which gives: \( \int \frac{1}{y^2} \, dy = -\frac{1}{y} + C_1 \).
  • Integrate the right side with respect to x , resulting in: \( \int 3x^2 \, dx = x^3 + C_2 \).
The constants C_1 and C_2 arise from the indefinite integration.
Combining these results gives us the integrated form: \( -\frac{1}{y} = x^3 + C \), where C = C_2 - C_1 \ is a constant representing the difference between the two integration constants.
General Solution
The general solution of a differential equation contains an arbitrary constant and provides a family of solutions valid for a range of conditions or inputs. After integration, we simplify the equation to express y in terms of x .
Starting from \( -\frac{1}{y} = x^3 + C \), we want to isolate y \.:
  • Invert the equation to solve for y : \( y = -\frac{1}{x^3 + C} \).
This expression \( y = -\frac{1}{x^3 + C} \) is the general solution to the original differential equation.

What's important here is the constant C . Different values of C lead to different particular solutions. Essentially, the general solution outlines all possible behaviors the system could exhibit, given diverse initial conditions.
Initial Conditions
Initial conditions are additional pieces of information specifying the value of the unknown function or its derivatives at a certain point. They help in finding the particular solution from the general solution.
Suppose we are given an initial condition such as y(0) = k \.:
  • Substitute x = 0 and y = k into the general solution \( y = -\frac{1}{x^3 + C} \).
  • From this, solve for the constant C \.
Initial conditions narrow down the family of solutions to just the one solution that fits the specified condition.
This focused solution perfectly models the system's behavior under given circumstances.

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Most popular questions from this chapter

Solve the differential equation. $$ \frac{d y}{d x}=x \sqrt{y} $$

Solve the initial-value problem. $$ t^{3} \frac{d y}{d t}+3 t^{2} y=\cos t, \quad y(\pi)=0 $$

A Bernoulli differential equation (named after James Bernoulli) is of the form $$ \frac{d y}{d x}+P(x) y=Q(x) y^{n} $$ Observe that, if \(n=0\) or \(1,\) the Bernoulli equation is linear. For other values of \(n,\) show that the substitution \(u=y^{1-n}\) transforms the Bernoulli equation into the linear equation $$ \frac{d u}{d x}+(1-n) P(x) u=(1-n) Q(x) $$

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For each predator-prey system, determine which of the variables, \(x\) or \(y,\) represents the prey population and which represents the predator population. Is the growth of the prey restricted just by the predators or by other factors as well? Do the predators feed only on the prey or do they have additional food sources? Explain. $$ \begin{array}{l}{\text { (a) } \frac{d x}{d t}=-0.05 x+0.0001 x y} \\ {\quad \frac{d y}{d t}=0.1 y-0.005 x y}\end{array} $$ $$ \text { (b) } \begin{array}{l}{\frac{d x}{d t}=0.2 x-0.0002 x^{2}-0.006 x y} \\\ {\quad \frac{d y}{d t}=-0.015 y+0.00008 x y}\end{array} $$
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