Question

Solve the following initial value problems. $$u^{\prime}(x)=2 u+6, u(1)=6$$

Short answer

The solution is \(u(x) = 3 + 3e^2 e^{-2x}\).

Step-by-step solution

Identify the type of differential equation

We can see that the given differential equation is a first-order linear ordinary differential equation (ODE) in the form: $$u^{\prime}(x)=2 u+6$$ with an initial condition \(u(1)=6\).

Find the integrating factor

To solve this ODE, we first need to find the integrating factor. For a first-order linear ODE of the form $$u^{\prime}(x)+p(x)u(x)=q(x),$$ the integrating factor \(I(x)=e^{\int p(x) dx}\). In our case, we have \(p(x)=2\). Therefore, the integrating factor is $$I(x)=e^{\int 2 dx}=e^{2x}.$$

Multiply the ODE by the integrating factor

Now, multiply the entire ODE by the integrating factor: $$e^{2x}u^{\prime}(x)+2e^{2x}u(x)=6e^{2x}.$$

Integrate to find the solution

Notice that the left-hand side is the derivative of a product: $$\frac{d}{dx}\left(u(x)e^{2x}\right) = 6e^{2x}.$$ Now we can integrate both sides with respect to \(x\): $$\int \frac{d}{dx}\left(u(x)e^{2x}\right) dx = \int 6e^{2x} dx.$$ This gives: $$u(x)e^{2x} = 3e^{2x} + C,$$ where \(C\) is the constant of integration.

Solve for \(u(x)\)

Divide both sides by \(e^{2x}\) to obtain the general solution: $$u(x) = 3 + Ce^{-2x}.$$

Apply the initial condition

We use the initial condition \(u(1)=6\) to find the specific constant \(C\): $$6 = 3 + Ce^{-2(1)}.$$ Solving for \(C\), we get: $$C = 3e^2.$$

Write down the final solution

Substitute the value of \(C\) into the general solution to get the final solution: $$u(x) = 3 + 3e^2 e^{-2x}.$$ Thus, the solution to the initial value problem is: $$u(x) = 3 + 3e^2 e^{-2x}.$$

Key concepts

First-Order Linear Ordinary Differential Equation

Understanding a first-order linear ordinary differential equation (ODE) is crucial for solving various problems in mathematics and the sciences. This type of equation is characterized by a derivative of the first order that involves the function and its first derivative only, typically following the standard form \( u'^{(1)}(x) + p(x)u(x) = q(x) \). Here, \( p(x) \) and \( q(x) \) are functions of \( x \) which can be constants as well. The linearity implies that both \( u(x) \) and its derivative \( u'(x) \) appear to the first power and are not multiplied together.

For the given problem, \( u'^{(1)}(x) = 2u + 6 \), the equation is linear because the unknown function \( u(x) \) and its derivative are both to the first power and not multiplied together. There are no quadratic or higher-order terms, and the equation does not include nonlinear functions like \( u(x)^2 \) or \( \text{sin}(u(x)) \) which would make solving much more complex.

Integrating Factor Method

The integrating factor method is a powerful technique for solving first-order linear ODEs. It is based on the idea of multiplying the entire equation by an appropriately chosen function, called an 'integrating factor', that facilitates the integration process. For the equation \( u'^{(1)}(x) + p(x)u(x) = q(x) \), the integrating factor is often denoted by \( I(x) \) and is calculated as \( I(x) = e^{\int p(x) dx} \).

By multiplying the original ODE by this integrating factor, the left side becomes the derivative of the product \( u(x)I(x) \), which can be integrated easily. This transformation is crucial because it turns the problem into a simpler one that can be solved with a straightforward integration. For our example, the integrating factor is \( e^{2x} \) which simplifies our original equation into a form where the product rule of derivatives can be observed and utilized.

Exponential Functions

Exponential functions play a central role in differential equations and various fields of mathematics. An exponential function is of the form \( f(x) = a^{x} \), where \( a \) is a positive real number, often referred to as the base. The most important and frequently used exponential function in calculus and differential equations is \( e^{x} \), where \( e \) is Euler's number, approximately equal to 2.71828.

In our context, the integrating factor involves an exponential function \( e^{2x} \), due to the constant coefficient \( 2 \) in our ODE. Exponential functions have unique properties such as the derivative of \( e^{x} \) being itself, which enable us to solve equations involving growth and decay processes. In solving ODEs, exponential functions generally appear in the solution either as part of the integrating factor or the general solution itself.

Integration of Functions

Integration is one of the fundamental operations in calculus, alongside differentiation. It is essentially the reverse process of finding a derivative and is used to calculate the area under curves, among other things. In the context of differential equations, integration is used to find the antiderivative or indefinite integral of functions. This process involves finding a function whose derivative equals the original function being integrated.

After applying the integrating factor to the ODE, the next step is to integrate both sides of the equation, which provides us with a general solution that includes a constant of integration (\( C \)). Determining this constant requires an additional piece of information, which is typically an initial condition. In the worked example, integrating the function \( 6e^{2x} \) leads to the general solution \( u(x)e^{2x} = 3e^{2x} + C \) before we apply the initial condition.

Applying Initial Conditions

Initial conditions allow us to find a unique, specific solution to a differential equation from a family of potential solutions. When we integrate a differential equation, we usually obtain a general solution that includes a constant of integration. This constant represents an infinite number of possible functions that satisfy the equation. However, the initial condition provides a specific value for the function at a particular point, which we use to solve for the constant.

In the step-by-step example provided, the initial condition is \( u(1)=6\). We use this information to find the particular value of \( C \) by plugging the initial condition into the general solution. By doing so, we narrow down to one unique solution that not only satisfies the ODE but also the given initial value. This final step is what allows us to move from a general understanding of the problem to a concrete, specific solution.