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Chapter 8: Problem 8

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. \(y^{\prime}+2 x y^{2}=0\) \(y=1\) when \(x=2\)

Short Answer

Expert verified
The general solution is \( y = \frac{1}{x^2 - C} \). The particular solution is \( y = \frac{1}{x^2 - 3} \).

Step by step solution

01

- Rearrange the Differential Equation

Start by rearranging the given differential equation \( y' + 2xy^2 = 0 \) so that the derivatives are on one side and the other terms on the other side. This gives \( y' = -2xy^2 \).
02

- Separate the Variables

Separate the variables to enable integration. This means writing it as \( \frac{dy}{y^2} = -2x dx \).
03

- Integrate Both Sides

Integrate both sides of the equation. Integrate the left side with respect to \( y \) and the right side with respect to \( x \):\[ \int \frac{1}{y^2} \, dy = \int -2x \, dx \].
04

- Solve the Integrals

Solve the integrals separately. The left side integral is \( \int y^{-2} \, dy = -y^{-1} \). The right side integral is \( \int -2x \, dx = -x^2 \). This leads to \( -\frac{1}{y} = -x^2 + C \), where \( C \) is the integration constant.
05

- Solve for y

Rearrange the equation to solve for \( y \): \( -\frac{1}{y} = -x^2 + C \) becomes \( \frac{1}{y} = x^2 - C \). Then \( y = \frac{1}{x^2 - C} \)
06

- Apply Boundary Condition

Use the boundary condition \( y = 1 \) when \( x = 2 \) to find the particular solution. Substitute these values into the general solution: \( 1 = \frac{1}{4-C} \). Solving for \( C \) gives \( C = 3 \).
07

- Write the Particular Solution

Substitute \( C = 3 \) back into the general solution \( y = \frac{1}{x^2 - C} \). The particular solution is \( y = \frac{1}{x^2 - 3} \).

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Key Concepts

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

Separation of Variables
Separation of variables is a fundamental method used to solve first-order differential equations. It's especially useful when we can arrange the equation such that each variable and its differential are on opposite sides. This way, each side of the equation contains only one variable, which simplifies the problem significantly.
For instance, consider the differential equation in our original exercise:
\text{Step 1: Begin by rearranging the given differential equation:} \( y' + 2xy^2 = 0 \).
Rearrange to isolate the derivative: \( y' = -2xy^2 \).
\text{Step 2: Separate the variables to enable integration:} \( \frac{dy}{y^2} = -2x dx \).
This step makes it clear how separation of variables works. It allows us to prepare each part for integration, making the equation easier to handle.
Boundary Conditions
Boundary conditions are specific values that a solution to a differential equation must satisfy. These conditions often help in finding a particular solution to a differential equation. In the given exercise, we've found the general solution first, which involved an arbitrary constant, and then applied the boundary conditions to find a specific value for that constant.
In our exercise, we had:
  • General solution: \( y = \frac{1}{x^2 - C} \)
  • Boundary condition: \( y = 1 \) when \( x = 2 \)
By substituting the boundary condition into the general solution, we get:
\( 1 = \frac{1}{4 - C} \).
Solving this, we find \( C = 3 \), giving us the particular solution: \( y = \frac{1}{x^2 - 3} \).
Boundary conditions are critical in differential equations because they allow us to hone in on a more precise solution that fits our problem's specific needs.
Integration
Integration is at the heart of solving differential equations, especially when using the separation of variables method. Once the equation is separated, each side is integrated with respect to its corresponding variable.
Let's revisit our exercise. After separating variables, we had:
\( \frac{dy}{y^2} = -2x dx \).
The next step is to integrate both sides:
  • Left side: \( \int y^{-2} \, dy = -y^{-1} \)
  • Right side: \( \int -2x \, dx = -x^2 \)
Combining these, we get: \( -\frac{1}{y} = -x^2 + C \).
Integration often introduces an integration constant, represented here by \( C \), which plays a crucial role in finding both the general and particular solutions.
Solving the integrals carefully while keeping the integration constants in mind is essential for the accuracy of the final solution.

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Most popular questions from this chapter

Find the orthogonal trajectories of each of the following families of curves. In each case sketch several of the given curves and several of their orthogonal trajectories. Be careful to eliminate the constant from \(d y / d x\) for the original curves; this constant takes different values for different curves of the original family and you want an expression for \(d y / d x\) which is valid for all curves of the family crossed by the orthogonal trajectory you are trying to find. \(y=k x^{2}\)

(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. \(y^{\prime \prime}+2 y^{\prime}+2 y=10 e^{x}+6 e^{-x} \cos x\)

Solve the following differential equations. \(y^{\prime \prime}+9 y=0\)

Find the general solution of each of the following differential equations. \(\frac{d y}{d x}=\frac{3 y}{3 y^{2 / 3}-x}\)
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