Question

Solving a First-Order Linear Differential Equation In Exercises \(51-58\) , solve the first-order differential equation by any appropriate method. $$ \frac{d y}{d x}=\frac{e^{2 x+y}}{e^{x-y}} $$

Short answer

The general solution of the differential equation \(\frac{d y}{d x}=\frac{e^{2 x+y}}{e^{x-y}}\) is \(y=-\frac{1}{2} ln(2e^{x} + 2C)\). (Remember that C is the constant of integration, which can be any arbitrary constant)

Step-by-step solution

Simplification

Rewrite the given differential equation in a simplified manner by combining the exponentials. This can be done by subtracting the exponents when the bases are the same. Thus, the equation \(\frac{d y}{d x}=\frac{e^{2 x+y}}{e^{x-y}}\) simplifies to \(\frac{d y}{d x}=e^{x+2y}\)

Identify Differential Equation Type

The equation can be rewritten as \(\frac{d y}{d x}- e^{x+2y}=0\), which is a first order non-linear differential equation.

Solving the Equation

Since this is a non-linear first-order differential equation, a direct analytical solution might not be obtainable. However, an implicit solution can be found by separation of variables and then integrating. By separating variables, the equation becomes \(\int e^{-2y} dy = \int e^{x} dx\). Evaluating these integrals yield \(-\frac{1}{2}e^{-2y}= e^{x} + C\). Hence the (implicit) solution.

Finding the General Solution

To find the general solution, solve for y to make it explicit. Multiply through by -2, and take the natural logarithm of both sides to get \(y=-\frac{1}{2} ln(2e^{x} + 2C)\)

Key concepts

Exponential Functions

Exponential functions are fundamental in the study of first-order differential equations. These functions can be tricky but understanding their properties helps simplify complex problems. An exponential function typically takes the form \( e^x \), where \( e \) is the base of the natural logarithm. When manipulating exponential functions, remember:

• When dividing like bases, subtract the exponents: \( e^{a}/e^{b} = e^{a-b} \).
• Exponential growth functions model continuous growth processes.

In the step-by-step solution, we simplified the given differential equation by combining the exponents of the exponential terms, making the equation easier to handle. The exponentials' properties were key to rewriting the initial expression succinctly.

Separation of Variables

The separation of variables technique is pivotal in solving first-order differential equations, especially non-linear ones. This method involves rearranging the equation so that different variables appear on opposite sides of the equation. That allows us to integrate separately.
Given equation: \( \frac{d y}{d x} = e^{x+2y} \)

• Start by rearranging, placing terms involving \( y \) on one side and terms involving \( x \) on the other.
• We separate the variables to obtain: \( \int e^{-2y} dy = \int e^{x} dx \).

By isolating the terms, integration becomes doable for each side. Separation of variables gives a straightforward path to finding an implicit solution.

Integration Techniques

Integration techniques are essential tools for solving separated differential equations. Once variables are separated, integrating each side provides the path to the solution.

• Exponentials: When integrating exponentials, remember \( \int e^{kx} dx = \frac{1}{k} e^{kx} + C \), where \( C \) is the integration constant.
• Integration by parts: Useful when products of functions are involved, though not required here.

For the equation \( \int e^{-2y} dy = \int e^{x} dx \), the integration results in \( -\frac{1}{2} e^{-2y} = e^{x} + C \). Being proficient in these techniques facilitates finding solutions to differential equations.

Non-Linear Differential Equations

Differential equations that aren't linear, like our example, can be complex to solve because they don't have a direct solution method. Non-linear equations often require more creative methods, like substitution or numerical methods. However, sometimes methods for linear equations, such as separation of variables, can still apply if manipulated correctly.

• Non-linear means terms are not simply proportional, and might involve powers or products of the variable and its derivatives.
• Our example \( \frac{d y}{d x} = e^{x+2y} \) is non-linear because of the multiplicative term of \( y \) in the exponent.

Despite the complexity, non-linear equations can often be solved to give implicit solutions. Finding an explicit solution might not always be possible directly, but it can sometimes be done by further algebraic manipulation.