Problem 68 - Calculus Early Transcendentals

Question

Prove that a solution to the initial-value problem $$ h(y) \frac{d y}{d x}=g(x), \quad y\left(x_{0}\right)=y_{0} $$ is defined implicitly by the equation $$ \int_{y_{0}}^{y} h(r) d r=\int_{x_{0}}^{x} g(s) d s $$

Step-by-Step Solution

Answer

The implicit solution is derived by separating variables and integrating both sides with limits using the initial condition.

Step 1: Understand the problem

We have a differential equation of the form \(h(y) \frac{dy}{dx} = g(x)\) with an initial condition \(y(x_0) = y_0\). The goal is to prove that this has an implicit solution \(\int_{y_0}^{y} h(r) \, dr = \int_{x_0}^{x} g(s) \, ds\).

Step 2: Separate variables

Rearrange the differential equation by separating variables, giving \(h(y) \, dy = g(x) \, dx\). This allows us to integrate both sides independently.

Step 3: Integrate both sides

Integrate both sides separately with respect to their respective variables. This yields \(\int h(y) \, dy\) and \(\int g(x) \, dx\).

Step 4: Apply Limits of Integration

Apply the initial condition to set the limits of integration. For the left side, integrate from \(y_0\) to \(y\), giving \(\int_{y_0}^{y} h(r) \, dr\). For the right side, integrate from \(x_0\) to \(x\), giving \(\int_{x_0}^{x} g(s) \, ds\).

Step 5: Conclude the implicit solution

Since both sides involve definite integrals resulting from the separation of variables, equating them gives the implicit solution: \(\int_{y_0}^{y} h(r) \, dr = \int_{x_0}^{x} g(s) \, ds\). This completes the proof.

Key Concepts

Initial Value Problem

An Initial Value Problem (IVP) in differential equations provides us with a differential equation and specific conditions called initial conditions.
These conditions allow us to find a specific solution out of many possible ones.
In simple terms, it's like knowing a starting point from which to solve a road map of paths.For our problem, the IVP is given as:

• Equation: \( h(y) \frac{dy}{dx} = g(x) \)
• Initial condition: \( y(x_0) = y_0 \)

The initial condition tells us the value of \( y \) when \( x = x_0 \), ensuring the uniqueness of the solution we find.
This is integral when solving differential equations as they often have a collection of "family" solutions that satisfy the equation but not necessarily the initial condition.
Thus, an initial value problem specifies which solution to choose from these families.

Implicit Solution

In mathematics, an implicit solution is where a solution to a differential equation is expressed in a form that does not explicitly solve for the dependent variable.
Instead, the relationship is defined within an equation involving both the independent and dependent variables.For the problem at hand, the implicit solution is expressed as:\[\int_{y_{0}}^{y} h(r) \, dr = \int_{x_{0}}^{x} g(s) \, ds\]This implies that instead of getting \( y \) explicitly as a function of \( x \), we define the relationship through integrals.
These integrals, which consider the initial conditions, indirectly resolve the function by setting the changes in \( x \) and \( y \) as equal.
Thus, solving this implicit form often involves finding \( y \) through additional steps or numerical methods, especially when an explicit form is not easily achievable.

Separation of Variables

Separation of Variables is a potent technique to solve differential equations.
This method involves rearranging the equation to isolate each variable and its differential on different sides of the equation.Consider \( h(y) \frac{dy}{dx} = g(x) \).
By separating variables, we reorganize the terms to:
\[ h(y) \, dy = g(x) \, dx \] Each side now contains only one of the variables, either \( y \) or \( x \), and their differentials.
This separation allows for the integration of both sides independently, leading to more manageably solving the equation.Through integration, you obtain:- Left side: \( \int h(y) \, dy \)- Right side: \( \int g(x) \, dx \) By applying the limits of integration based on initial conditions, we can prove that these integrals equate—which is the essence of finding the implicit solution.
This technique is particularly useful because it breaks down complicated differential equations into simpler integral forms that are easier to work with.