Question

Finding general solutions Find the general solution of each differential equation. Use \(C, C_{1}, C_{2}, \ldots\) to denote arbitrary constants. $$u^{\prime}(x)=\frac{2(x-1)}{x^{2}+4}$$

Short answer

The general solution of the given first-order differential equation is \(u(x) = x^2 + 4 - 4\ln|x^2 + 4| + C\), where \(C\) is the arbitrary constant.

Step-by-step solution

Integration of the given differential equation

Integrate both sides of the equation: $$\int u'(x) dx = \int \frac{2(x-1)}{x^2 + 4} dx$$

Integration by substitution for the right-hand side

In order to integrate the right-hand side of the equation, we perform a substitution: Let \(v = x^2 + 4\), then \(\frac{dv}{dx} = 2x\). Now, we rewrite the integral: $$\int u'(x) dx = \int \frac{(x-1)\cdot(2x)}{x^2 + 4} dx = \int \frac{(v-4)dv}{v}$$ Note that we multiplied and divided by \(2x\) to match \(v\) substitution.

Integration of both sides

Integrate both sides of the equation: $$u(x) = \int \frac{(v-4)dv}{v} = \int\left(1-\frac{4}{v}\right)dv$$

Calculate the integral

Now, we calculate the integral: $$u(x) = v - 4\ln |v| + C$$

Substitute back the original variable

Substitute \(v = x^2 + 4\) to get the general solution for \(u(x)\): $$u(x) = x^2 + 4 - 4\ln|x^2 + 4| + C$$ Finally, we have the general solution of the given differential equation: $$u(x) = x^2 + 4 - 4\ln|x^2 + 4| + C$$ where \(C\) is the arbitrary constant.

Key concepts

Integration

Integration is a fundamental concept in calculus, closely related to differentiation. If differentiation is about finding rates of change or slopes, integration is about accumulating quantities and finding the total area under curves. In the context of differential equations, integration helps us to solve equations involving unknown functions and their derivatives. By integrating, we can express an unknown function in terms of known functions.

This process often involves indefinite integrals, where the goal is to find antiderivatives. An indefinite integral doesn't produce a specific numerical result but rather a family of functions, represented by adding a constant of integration, typically denoted as \(C\). This represents an infinite number of functions all differing by this constant, which is crucial when solving differential equations.

In our example, we have a differential equation involving a derivative \(u'(x)\). To find \(u(x)\), its original function, we need to integrate the equation. This step fundamentally reverses the process of differentiation, thus leading us from the derivative back to the function.

General Solution

The general solution of a differential equation includes all possible solutions. It accounts for all constants that might arise during integration, reflecting the idea that many different functions could satisfy the same differential equation.

When we integrate a derivative to find the original function, we often introduce an arbitrary constant, \(C\). This constant arises because the derivative of a constant is zero. Therefore, integrating has this inherent ambiguity.

In solving our differential equation, we end up with a general solution: \(u(x) = x^2 + 4 - 4\ln|x^2 + 4| + C\). Each different value of \(C\) represents a different solution, making it a family of solutions. It includes the entire set of functions that satisfy the given differential equation, respecting conditions of existence as the general solution does not imply a unique solution.

Substitution Method

The substitution method is a powerful tool when integrating complex expressions. It simplifies the process by transforming a difficult integral into a more manageable one. By substituting variables, the integral becomes straightforward, often revealing a simple structure.

In the solution we provided, substitution helps us integrate the term \(\frac{2(x-1)}{x^2 + 4}\). Taking \(v = x^2 + 4\) transforms the integral by simplifying the expression into \(\int \frac{(v-4)dv}{v}\).

This process involves initially calculating the derivative of our substitution, specifically \(\frac{dv}{dx} = 2x\), to find a way to replace \(dx\). After substitution, the integral looks more familiar and it becomes much easier to solve without the complication of the original variables. Once solved, substitute back the original variables to return to the initial problem context. The substitution method effectively streamlines integration, showing its utility in differential equations.