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Chapter 8: Problem 25

Find a particular solution satisfying the given conditions. . \(3 x^{2} y d x+x^{3} d y=0, \quad y=2\) when \(x=1\).

Short Answer

Expert verified
The particular solution is \(y = 2x^{-3}\).

Step by step solution

01

- Separate Variables

First, rewrite the given differential equation in a separable form. The given equation is: \[3x^2 y \, dx + x^3 \, dy = 0\] Rewrite it as: \[3x^2 y \, dx = -x^3 \, dy\] Then separate variables: \[\frac{dy}{y} = -\frac{3x^2}{x^3} \, dx = -\frac{3}{x} \, dx\]
02

- Integrate Both Sides

Integrate both sides of the equation to find the general solution. The separated equation is: \[\int \frac{1}{y} \, dy = -3 \int \frac{1}{x} \, dx\]Integrating both sides gives: \[\ln|y| = -3 \ln|x| + C\] where \(C\) is the constant of integration.
03

- Exponentiate Both Sides

Exponentiate both sides to solve for \(y\). The equation is: \[\ln|y| = -3 \ln|x| + C\]Exponentiating gives: \[|y| = e^C \cdot |x|^{-3}\] Letting \(A = e^C\), we get: \[|y| = A \cdot x^{-3}\].
04

- Determine the Particular Solution

Use the given initial condition \(y = 2\) when \(x = 1\) to find the value of \(A\). The equation is: \[2 = A \cdot 1^{-3} \Rightarrow 2 = A\] Thus, \(A = 2\). The particular solution is: \[y = 2x^{-3}\].
05

- Write the Final Solution

The particular solution that satisfies the given conditions is: \[y = 2x^{-3}\].

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Key Concepts

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

Separable Differential Equation
A **separable differential equation** is one that can be written so that all terms involving one variable are on one side of the equation, and all terms involving the other variable are on the other side. For example, in our given differential equation 3x^2 y \, dx + x^3 \, dy = 0, ajust a moment we'll separate the variables. Firstly, rewrite it open so that each term involving y on one side:

3x^2 y \, dx = -x^3 \, dy.
Now, we'll divide both sides by y and by x^3, so we have the terms involving y and dy on one side, and terms involving x and dx on the other:

\( \frac{dy}{y} = -\frac{3x^2}{x^3} \, dx = -\frac{3}{x} \, dx \)
This makes it easier to integrate and find the solution.
Integration
Integration involves finding the antiderivative of a function. After separating our differential equation, we have:

\( \int \frac{1}{y} \, dy = -3 \int \frac{1}{x} \, dx \)

Integrating both sides:

\( \ln|y| = -3 \ln|x| + C \)
Here, \( C \) is the constant of integration, representing where the function starts. For the integration process:
  • Separate the variables.
  • Integrate both sides.
  • Combine the constant of integration.

    • The integration essentially 'undoes' the differentiation, allowing us to arrive at the general solution of the differential equation.
Initial Conditions
Initial conditions are used to determine the particular solution to a differential equation from the general solution. They specify the value of the unknown function at a given point. In our problem, the initial conditions are given as between y=2 x=1. This means when x=1 and y=2,

Given this information, we substitute these values into our general solution to determine the constant A:
\( |y| = A \, |x|^{-3} \)
Substituting y = 2 and x = 1, we find:
\( 2 = A \, 1^{-3} \)So, \( 2 = A \)
Thus, A is equal to 2.

Initial conditions are vital in transforming a general solution into the unique, particular solution that fits the given conditions.
Particular Solution
The **particular solution** of a differential equation is the specific solution that satisfies the initial conditions given. Starting with our general solution:
\( |y| = A \, |x|^{-3} \), we already know that A = 2 from the initial conditions.
Therefore, the specific or particular solution becomes:
\( y = 2x^{-3} \)

This particular solution accurately fits our original differential equation and the initial condition provided.
In summary:
  • Find the general solution by integrating.
  • Use initial conditions to solve for the constants.
  • Write out the particular solution that includes these constants.
    Understanding the particular solution is crucial because it ensures the differential equation aligns with the real-world problem it models.

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Most popular questions from this chapter

Find the "general solution" (that is, a solution containing an arbitrary constant) of each of the following differential equations, by separation of variables. Then find a particular solution of each equation satisfying the given boundary conditions. \(y^{\prime} \sin x=y \ln y\) \(y=e\) when \(x=\pi / 3\)

A simple pendulum consists of a point mass \(m\) suspended by a weightless cord of length \(l .\) Find the equation of motion of the pendulum, that is, the differential equation for \(\theta\) as a function of \(t .\) Show that (for small \(\theta\) ) this is approximately a simple harmonic motion equation, and find \(\theta\) if \(\theta=\theta_{0}, d \theta / d t=0\) when \(t=0\).

(a) Consider a light beam traveling downward into the ocean. As the beam progresses, it is partially absorbed and its intensity decreases. The rate at whieh the intensity is decreasing with depth at any point is proportional to the intensity at that depth. The proportionality constant \(\mu\) is called the linear absorption coefficient. Show that if the intensity at the surface is \(I_{0}\), the intensity at a distance \(s\) below the surface is \(l=\) \(I_{0} e^{-\mu s} .\) The linear absorption coefficient for water is of the order of \(10^{-2} \mathrm{ft}^{-1}\) (the exact value depending on the wavelength of the light and the impurities in the water). For this value of \(\mu\), find the intensity as a fraction of the surface intensity at a depth of \(1 \mathrm{ft}, 50 \mathrm{ft}, 500 \mathrm{ft}, 1\) mile. When the intensity of a light beam has been reduced to half. its surface intensity \(\left(I=\frac{1}{2} I_{0}\right)\), the distance the light has penetrated into the absorbing substance is called the half-value thickness of the substance. Find the half-value thickness in terms of \(\mu\). Find the half-value thickness for water for the value of \(\mu\) given above. (b) Note that the differential equation and its solution in this problem are mathematically the same as those in Example I, although the physical problem and the terminology are different. In discussing radioactive decay, we call \(\lambda\) the decay constant, and we define the half-life \(T\) of a radioactive substance as the time when \(N=\frac{1}{2} N_{0}\) (compare half-value thickness). Find the relation between \(\lambda\) and \(T\).

Identify each of the differential equations as to type (for example, separable, linear first order, linear second order, etc.), and then solve it. \(y^{\prime}+x y=x / y\)

Find the "general solution" (that is, a solution containing an arbitrary constant) of each of the following differential equations, by separation of variables. Then find a particular solution of each equation satisfying the given boundary conditions. \(x y^{\prime}=y\), \(y=3\) when \(x=2\).
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