Problem 14 - Calculus Early Transcendentals: Pearson New International Edition

Question

First find the general solution (involving a constant C) for the given differential equation. Then find the particular solution that satisfies the indicated condition. (See Example 2.) \(\frac{d y}{d x}=-y^{2} x\left(x^{2}+2\right)^{4} ; y=1\) at \(x=0\)

Step-by-Step Solution

Answer

The particular solution is \( y(x) = -\frac{1}{\frac{(x^2 + 2)^5}{10} - 4.2} \).

Step 1: Identify the Type of Differential Equation

The given differential equation is \( \frac{d y}{d x}=-y^{2} x (x^{2}+2)^{4} \). Notice that the equation can be rearranged as \( \frac{d y}{d x}=-y^{2} f(x) \), which is a separable differential equation.

Step 2: Separate the Variables

We can separate the variables by rewriting the equation: \( \frac{d y}{d x} = -y^{2} x (x^{2}+2)^{4} \). Divide both sides by \( y^{2} \) and multiply by \( dx \) to get: \( \frac{1}{y^{2}} dy = -x (x^{2}+2)^{4} dx \).

Step 3: Integrate Both Sides

Integrate the left-hand side with respect to \( y \) and the right-hand side with respect to \( x \):\[ \int \frac{1}{y^{2}} dy = \int -x (x^{2}+2)^{4} dx \]The left-hand side integrates to \( \frac{-1}{y} \). The right-hand side requires integration by substitution.

Step 4: Integrate the Right-Hand Side

To integrate \( \int x (x^{2}+2)^{4} dx \), let \( u = x^2 + 2 \). Then, \( du = 2x dx \), or \( x dx = \frac{1}{2} du \). The integral becomes:\[ \int x (x^{2}+2)^{4} dx = \frac{1}{2} \int u^{4} du \]This evaluates to \( \frac{1}{2} \cdot \frac{u^5}{5} + C \), or \( \frac{(x^2 + 2)^5}{10} + C \).

Step 5: Form the General Solution

Equating the integrated forms gives:\[ \frac{-1}{y} = \frac{(x^2 + 2)^5}{10} + C \]Solve for \( y \) by taking reciprocals:\[ y = -\frac{1}{\frac{(x^2 + 2)^5}{10} + C} \]This is the general solution.

Step 6: Apply the Initial Condition

Use the condition \( y = 1 \) at \( x = 0 \) to find the value of \( C \). Substitute into the general solution:\[-1 = \frac{(0^2 + 2)^5}{10} + C \]\[-1 = \frac{32}{10} + C \]\[-1 = 3.2 + C \]Solving for \( C \) gives \( C = -4.2 \).

Step 7: Write the Particular Solution

With the value of \( C \), the particular solution is given by:\[ y(x) = -\frac{1}{\frac{(x^2 + 2)^5}{10} - 4.2} \]

Key Concepts

Separable Differential Equations

In the world of differential equations, a separable differential equation is one that allows us to isolate each variable on different sides of the equation. This makes solving them much simpler because we can solve each variable individually. The equation \( \frac{dy}{dx} = -y^2 x (x^2 + 2)^4 \) is an example of a separable differential equation. Initially, it might look complex, but notice how we can manipulate it so that all the \( y \) terms are on one side and all the \( x \) terms on another. This separation is achieved by dividing both sides by \( y^2 \) and multiplying by \( dx \), yielding \( \frac{1}{y^2} dy = -x (x^2 + 2)^4 dx \).

Separable differential equations are significant because once the variables are separated, what follows is classic calculus: integration. Integration then provides us with a general solution, characterized by an integration constant, usually denoted as \( C \). This constant arises because integration is an anti-derivative process, introducing an additional unknown that represents any constant function.

Initial Conditions

Initial conditions play a critical role in solving differential equations. They allow us to find a specific solution, known as the particular solution, from the general solution of an equation. This specific solution fits a given point, reflecting real-world scenarios where conditions are known at certain moments.

In our exercise, the initial condition provided is \( y = 1 \) when \( x = 0 \). Essentially, this means that the particular solution is the curve that passes through the point \( (0, 1) \).

With the general solution derived as \[ y(x) = -\frac{1}{\frac{(x^2 + 2)^5}{10} + C} \], we use this initial condition to solve for \( C \). Substitute \( x = 0 \) and \( y = 1 \) into this equation to find \( C \). This transforms the solution from a family of curves, all defined by the different values of \( C \), to a single curve that precisely meets the desired initial condition.

Integration By Substitution

Integration by substitution is a powerful calculus technique used to simplify the integration process by changing variables. It's especially useful when the integrand is a composite function. In our differential equation exercise, the term \( \int x (x^2 + 2)^4 dx \) requires this method.

By setting \( u = x^2 + 2 \), we change the variable from \( x \) to \( u \). Thus, the derivative \( du = 2x dx \) is used to express \( x dx \) as \( \frac{1}{2} du \). The original integral \( \int x (x^2 + 2)^4 dx \) then becomes \( \frac{1}{2} \int u^4 du \), which is simpler to evaluate. This integral resolves to \( \frac{1}{2} \cdot \frac{u^5}{5} + C \), or \( \frac{(x^2 + 2)^5}{10} + C \).

Substitution is an essential skill. It operates like reverse engineering: you reinvent a complex expression into a simpler, more solvable one. By practicing this technique, students build a deeper understanding of how integrals can be maneuvered through clever variable manipulation.