Question

Find an integrating factor and solve the given equation. $$ y^{\prime}=e^{2 x}+y-1 $$

Short answer

Question: Determine the general solution for the differential equation \(y'(x) - y(x) = e^{2x} - 1.\) Answer: The general solution for the given differential equation is \(y(x) = \frac{1}{3}e^{4x} - e^{2x} + Ke^x\), where \(K\) is an arbitrary constant.

Step-by-step solution

Rewrite the given differential equation

To solve the equation, first rewrite the given equation into the form of a first-order linear differential equation: $$ \frac{dy}{dx} - y = e^{2x} - 1. $$ Now, we see that \(P(x) = -1\) and \(Q(x) = e^{2x} - 1\).

Find an integrating factor

Now, we will find an integrating factor of the form \(e^{\int P(x)dx}\). With \(P(x) = -1\), we have: $$ e^{\int (-1) dx} = e^{-x}. $$

Multiply both sides by the integrating factor

Next, multiply both sides of the linear equation by the integrating factor \(e^{-x}\): $$ e^{-x}\frac{dy}{dx} - ye^{-x} = e^{x}(e^{2x} - 1). $$

Integrate both sides

Notice that the left side of the new equation is a derivative of a product: $$ \frac{d}{dx}(ye^{-x}) = e^{x}(e^{2x} - 1). $$ Integrate both sides of the equation with respect to \(x\) to find the function \(y(x)\): $$ \int d(ye^{-x}) = \int e^{x}(e^{2x} - 1) dx. $$

Solve the integrals

First, we can solve the left-hand side integral: $$ \int d(ye^{-x}) = ye^{-x} + C_1. $$ For the right-hand side integral, we can split the integral into two parts: $$ \int e^{x}(e^{2x} - 1) dx = \int e^{3x} dx - \int e^x dx. $$ Solve the two integrals on the right-hand side separately: $$ \int e^{3x} dx = \frac{1}{3}e^{3x} + C_2, \\ \int e^x dx = e^x + C_3. $$ Combine the results to get the entire right-hand side integral solution: $$ \int e^{x}(e^{2x} - 1) dx = \frac{1}{3}e^{3x} - e^x + C, $$ where \(C = C_2 - C_3\) is an arbitrary constant.

Find the solution for \(y(x)\)

Now combine the results from step 5 to get the following equation: $$ ye^{-x} + C_1 = \frac{1}{3}e^{3x} - e^x + C, $$ Observe that \(C_1\) is also an arbitrary constant, and therefore can be combined with \(C\) into a constant \(K\): $$ ye^{-x} = \frac{1}{3}e^{3x} - e^x + K. $$ Multiply both sides by \(e^x\) to find \(y(x)\): $$ y(x) = \frac{1}{3}e^{4x} - e^{2x} + Ke^x. $$ This is the general solution to the given equation.

Key concepts

First-Order Linear Differential Equation

A first-order linear differential equation takes the form \( \frac{dy}{dx} + P(x)y = Q(x) \). These equations have solutions that can be systematically found using techniques like the integrating factor.

In our exercise, the equation was initially rewritten as \( \frac{dy}{dx} - y = e^{2x} - 1 \). Here, \(P(x) = -1\) and \(Q(x) = e^{2x} - 1\). Recognizing this structure is crucial for applying the integrating factor method effectively.

Integration

Integration is a fundamental tool used to solve differential equations. In our context, after transforming the equation using the integrating factor, we integrate both sides to find the solution.

The left side simplifies to \(d(ye^{-x})\), which is straightforward to integrate. Meanwhile, the right side \( \int e^x(e^{2x} - 1) \, dx \) needs careful handling. It splits into two separate integrals:

• \( \int e^{3x} \, dx = \frac{1}{3} e^{3x} + C_2 \)
• \( \int e^x \, dx = e^x + C_3 \)

Understanding how to evaluate these integrals individually helps in constructing the final solution.

General Solution

The general solution of a first-order linear differential equation includes all possible solutions, represented with an arbitrary constant.

After integrating, we combine constants into a single arbitrary constant, forming the solution as: \( y(x) = \frac{1}{3}e^{4x} - e^{2x} + Ke^x \). This expression represents the complete set of solutions for the original equation. Including constant \(K\) ensures we cover every potential initial condition.

Differential Equation

Differential equations describe relationships involving rates of change. They appear in forms like \( y' = e^{2x} + y - 1 \), which need solving to understand the behavior of the function \(y\) over time.

Solving them helps us find functions that satisfy these changing conditions. By rewriting into linear form, finding integrating factors, and integrating, we can derive solutions.

• Used widely in science and engineering.
• Essential for modeling dynamic systems.

Remember that differential equations connect the function's rate of change directly to the actual values of the function.