Problem 44 - Advanced Engineering Mathematics

Question

Separating variables we have \(d y /(\sqrt{y}+y)=d x /(\sqrt{x}+x) .\) To integrate \(\int d x /(\sqrt{x}+x)\) we substitute \(u^{2}=x\) and get \\[\int \frac{2 u}{u+u^{2}} d u=\int \frac{2}{1+u} d u=2 \ln |1+u|+c=2 \ln (1+\sqrt{x})+c\\]. Integrating the separated differential equation we have\\[2 \ln (1+\sqrt{y})=2 \ln (1+\sqrt{x})+c \text { or } \ln (1+\sqrt{y})=\ln (1+\sqrt{x})+\ln c_{1}.\\] Solving for \(y\) we get \(y=\left[c_{1}(1+\sqrt{x})-1\right]^{2}\).

Step-by-Step Solution

Answer

The solution is \( y = \left[c_{1}(1+\sqrt{x})-1\right]^{2} \).

Step 1: Separate the Variables

We start with the differential equation \( \frac{d y}{\sqrt{y}+y} = \frac{d x}{\sqrt{x}+x} \). The equation is already separated as two integrals, one for \( y \) and one for \( x \).

Step 2: Integrate with Respect to x

To solve the integral \( \int \frac{d x}{\sqrt{x}+x} \), we use the substitution \( u^2 = x \), leading to \( 2u \, du = dx \). The integral becomes \( \int \frac{2u}{u + u^2} \, du = \int \frac{2}{1+u} \, du \).

Step 3: Solve the Substituted Integral

The integral \( \int \frac{2}{1+u} \, du \) is a standard logarithmic form, yielding \( 2 \ln |1+u| + C \). Substituting back \( x = u^2 \), we have \( 2 \ln (1+\sqrt{x}) + C \).

Step 4: Integrate with Respect to y

Similarly, we perform the integration on the \( y \) side of the equation, resulting in a form parallel to the \( x \) integration: \( 2 \ln (1+\sqrt{y}) + C \).

Step 5: Formulate the General Solution

Equating the expressions obtained from both integrations gives: \( 2 \ln (1+\sqrt{y}) = 2 \ln (1+\sqrt{x}) + C \). Divide by 2 to simplify: \( \ln (1+\sqrt{y}) = \ln (1+\sqrt{x}) + \ln c_1 \), where \( c_1 = e^{C/2} \).

Step 6: Solve for y

Exponentiate both sides to eliminate the logarithms: \( 1+\sqrt{y} = c_1(1+\sqrt{x}) \). Solve for \( y \) by isolating \( \sqrt{y} \): \( \sqrt{y} = c_1(1+\sqrt{x}) - 1 \). Square both sides to solve for \( y \): \( y = \left[c_{1}(1+\sqrt{x})-1\right]^{2} \).

Key Concepts

Variable Separation

Variable separation is a method used in solving a certain type of differential equations. The main idea is to rearrange the equation so that each variable, along with its differential, is on one side of the equation. This allows for separate integrations with respect to each variable.

Let's consider the differential equation from our example: \( \frac{d y}{\sqrt{y}+y} = \frac{d x}{\sqrt{x}+x} \). The key is to separate this into two parts: one that deals with \( y \) and another that deals with \( x \).

Once separated, you treat each side as an independent integral:

• The left side becomes \( \int \frac{1}{\sqrt{y}+y} \, d y \)
• The right side becomes \( \int \frac{1}{\sqrt{x}+x} \, d x \)

By converting the differential equation into an equality of two integrals, variable separation simplifies the process of finding a solution.

Definite Integrals

Definite integrals allow us to calculate the area under a curve within a specific interval. However, in the context of solving differential equations like the one we're working with, we typically focus on indefinite integration first. This step is essential before applying limits of definite integration.

The focus here is on finding an antiderivative which can be further evaluated if limits are specified. For instance, \( \int \frac{d x}{\sqrt{x}+x} \) in our example, becomes necessary to solve using techniques such as substitution. The goal is to simplify before making any evaluations of definite values.

Substitution Method

The substitution method is useful to simplify integrals into a more recognizable form. In our exercise, the integral \( \int \frac{d x}{\sqrt{x}+x} \) is tackled by setting \( u^2 = x \).

This changes the differential \( dx \) to \( 2u \, du \), modifying the integral to an easier form: \( \int \frac{2u}{u+u^2} \, du \).

• Choose a substitution that simplifies the equation; here \( u = \sqrt{x} \) translates the \( \sqrt{x} \) terms neatly.
• Reformulate both the integrand and the differential \( dx \). Work through the derivative to replace \( dx \).

Substitution transforms challenging integrals into forms that are more straightforward to integrate.

Logarithmic Integration

Logarithmic integration deals with integrals that result in expressions containing logarithms. These are particularly common when the integral has the form \( \int \frac{1}{x} \, dx \), leading directly to a logarithmic function.

In our case, after substitution, we reduced the integral into \( \int \frac{2}{1+u} \, du \), using integration rules to solve it as \( 2 \ln |1 + u| + C \).

• Logarithmic integration is vital in solving complex expressions typically seen in differential equations.
• Always remember to include the constant of integration \( C \), as it reflects the family of possible solutions.

Recovering the initial variable after solving in terms of the substitution variable is essential to form your final solution. Ultimately, this leads to progress in solving the differential equation, as demonstrated by replacing back \( u = \sqrt{x} \) in our case.