Problem 15 - Thomas Calculus in SI Units

Question

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

Step-by-Step Solution

Answer

The solution is \( y = -\ln(2e^{-\sqrt{x}} - C) \).

Step 1: Rearrange the Equation

Start by rearranging the given differential equation: \( \sqrt{x} \frac{dy}{dx} = e^{y+\sqrt{x}} \). Divide both sides by \( e^{y+\sqrt{x}} \) to get \( \frac{1}{e^{y+\sqrt{x}}} \sqrt{x} \frac{dy}{dx} = 1 \).

Step 2: Apply Separation of Variables

Separate the variables by rewriting the equation: \( \frac{1}{e^y} dy = \frac{1}{\sqrt{x} e^{\sqrt{x}}} dx \). This separates the y terms on one side and the x terms on the other.

Step 3: Integrate Both Sides

Integrate both sides with respect to their respective variables. For the left side, integrate \( \int \frac{1}{e^y} \, dy = \int e^{-y} \, dy \) resulting in \(-e^{-y} + C_1\). For the right side, integrate \( \int \frac{1}{\sqrt{x} e^{\sqrt{x}}} \, dx \).

Step 4: Simplify the Right Side Integral

Use substitution \( u = \sqrt{x} \), thus \( du = \frac{1}{2\sqrt{x}} \ dx \). The right side becomes \( 2 \int e^{-u} \, du \), resulting in \( -2e^{-u} + C_2 \), or \( -2e^{-\sqrt{x}} + C_2 \) when substituted back.

Step 5: Combine Integrals and Solve for C

Equate the expressions from both integrals: \(-e^{-y} = -2e^{-\sqrt{x}} + C \). Simplify to find the relationship: \( e^{-y} = 2e^{-\sqrt{x}} - C \). Solve for the constant \( C \) if needed, given any initial conditions.

Step 6: Solve for y

Take the natural logarithm on both sides to isolate y: \( -y = \ln(2e^{-\sqrt{x}} - C) \). Thus, \( y = -\ln(2e^{-\sqrt{x}} - C) \). This is the general solution to the differential equation.

Key Concepts

Separation of Variables

The technique of separation of variables is a common method used to solve differential equations. It involves rearranging the equation such that all terms involving the variable \( y \) and its derivative are on one side, while all terms involving the other variable, in this case \( x \), are on the other side. In the exercise provided, we start with the equation \( \sqrt{x} \frac{dy}{dx} = e^{y+\sqrt{x}} \). By dividing both sides by \( e^{y+\sqrt{x}} \), we can express it as \( \frac{1}{e^{y+\sqrt{x}}} \sqrt{x} \frac{dy}{dx} = 1 \).

• After simplifying, the equation separates into \( \frac{1}{e^y} dy = \frac{1}{\sqrt{x} e^{\sqrt{x}}} dx \).
• This new arrangement indicates that the integral of terms in \( y \) can be handled separately from the integral involving \( x \).

By separating the variables, we prepare the equation for integration, allowing us to proceed with solving the differential equation. This technique is very useful when an equation can be neatly separated into functions of each variable alone.

Integration Techniques

Once the variables have been separated in a differential equation, the next step is to integrate both sides with respect to their respective variables. This process helps to solve for the function \( y(x) \) or \( y \) given \( x \). In the integrated exercise:

• The left side of the equation, \( \int \frac{1}{e^y} \, dy = \int e^{-y} \, dy \), simplifies to \(-e^{-y} + C_1\).
• The right side requires more effort. We integrate \( \int \frac{1}{\sqrt{x} e^{\sqrt{x}}} \, dx \).

Handling these integrals:- The left integral is straightforward by recognizing the exponential relationship,
resulting in the negative exponent's integration.- For the right integral, simplification techniques such as substitution (to be discussed)
aid in the integration process.Integration techniques vary from basic polynomial integration to more advanced methods like partial fractions or specific trigonometric identities, depending on the complexity of the function. Here, understanding exponential integration sets the foundation for more complex manipulations thereafter.

Substitution Method

The substitution method is a powerful technique in integration, especially useful when the integrand is a complex expression. In the differential equation problem discussed, the substitution method simplifies the integration of the right-hand side of the equation.To tackle the integral \( \int \frac{1}{\sqrt{x} e^{\sqrt{x}}} \, dx \), we employ the substitution method:

• Let \( u = \sqrt{x} \), then \( du = \frac{1}{2\sqrt{x}} \, dx \).
• This transforms the integrand into terms of \( u \): \( 2 \int e^{-u} \, du \).

Using substitution simplifies the problem by:- Changing variables from \( x \) to \( u \), allowing the integral to be treated as a straightforward exponential function.- Integration of \( e^{-u} \) results in \(-e^{-u}\), simplified into \(-2e^{-\sqrt{x}} + C_2\) upon re-substitution.The substitution method shines in problems that involve composite functions, exploiting the fact that many complex expressions can be simplified into basic forms through creative variable adjustments.