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Chapter 7: Problem 15

Solve the differential equation. $$\sqrt{x} \frac{d y}{d x}=e^{y+\sqrt{x}}, \quad x>0$$

Short Answer

Expert verified
The solution is \( y = -\ln(-2e^{\sqrt{x}} - C) \) for \( x > 0 \).

Step by step solution

01

Separate Variables

The given differential equation is \( \sqrt{x} \frac{dy}{dx} = e^{y+\sqrt{x}} \). We want to separate the variables \( y \) and \( x \). To do this, rewrite the equation as \( \frac{dy}{dx} = \frac{e^{y+\sqrt{x}}}{\sqrt{x}} \). Now, express this as \( e^{-y} dy = e^{\sqrt{x}} \cdot \frac{1}{\sqrt{x}} dx \). This separates the variables \( y \) and \( x \).
02

Integrate Both Sides

Integrate both sides of the separated equation. For the left side, \( \int e^{-y} \, dy = -e^{-y} + C_1 \). For the right side, \( \int \frac{e^{\sqrt{x}}}{\sqrt{x}} \, dx \). Use substitution \( u = \sqrt{x} \), where \( du = \frac{1}{2\sqrt{x}} dx \), resulting in \( \int 2e^{u} \, du = 2e^u + C_2 \). In terms of \( x \), this becomes \( 2e^{\sqrt{x}} \).
03

Solve for Constants

Combine the integrated expressions: \( -e^{-y} = 2e^{\sqrt{x}} + C \), where \( C = C_2 - C_1 \). Solve for \( y \) to express the solution explicitly. Rearrange to find \( e^{-y} = -2e^{\sqrt{x}} - C \). Thus, \( y = -\ln(-2e^{\sqrt{x}} - C) \).
04

Verify Conditions and Simplify

Ensure that \( x > 0 \) in the solution. Simplify and check any initial conditions if provided. The form \( y = -\ln(-2e^{\sqrt{x}} - C) \) is defined given the domain \( x > 0 \), assuming \( -2e^{\sqrt{x}} - C > 0 \). This implies constraints on \( C \).

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Key Concepts

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

Separation of Variables
This is a powerful technique for solving differential equations, especially when the variables can be easily isolated. It involves rearranging the equation to get all terms with the dependent variable on one side, and all terms with the independent variable on the other. In the exercise, the initial equation was rewritten to
  • \(e^{-y} dy = e^{\sqrt{x}} \cdot \frac{1}{\sqrt{x}} dx\)
  • This was achieved by bringing terms involving \(y\) to the left and terms involving \(x\) to the right.
The goal is to find a form that allows you to integrate both sides independently. Separation of variables is not always possible, but when it is, it makes solving the differential equations straightforward.
Integration
Once the differential equation is separated, the next logical step is to integrate both sides. Integration is the process of finding a function given its derivative. In this context:
  • Left : \( \int e^{-y} \, dy = -e^{-y} + C_1 \)
  • Right: \( \int \frac{e^{\sqrt{x}}}{\sqrt{x}} \, dx \)
For the right side, substitution was used (more on that below). Integration ensures that both sides become functions instead of derivatives. This step translates the problem from differential calculus into algebra, allowing us to solve for the unknown function.
Substitution Method
This method is used within the integration process to simplify expressions that are otherwise hard to integrate. In the given exercise, to solve the right-hand side:
  • Set \(u = \sqrt{x}\), making \(du = \frac{1}{2\sqrt{x}} dx\)
  • This transforms the integral into a simpler form: \( \int 2e^{u} \, du \)
The method involves changing the variable of integration, making the integral easier to compute. Once the substitution is complete and the integral is evaluated, you substitute back the original variable. This technique simplifies complexity, making tricky integrals approachable.

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