Problem 17 - Technical Mathematics with Calculus

Question

Find the general solution to each differential equation. $$y^{\prime}=2 y+4 e^{2 x}$$

Step-by-Step Solution

Answer

The general solution to the differential equation \(y' = 2y + 4e^{2x}\) is \(y(x) = Ce^{2x} + 2xe^{2x}\), where C is an arbitrary constant.

Step 1: Identify the Types of Terms

Examine the differential equation to identify its elements. It is a first-order linear differential equation with a function of the independent variable x, which in this case is an exponential function, as a non-homogeneous term.

Step 2: Solve the Homogeneous Part

The homogeneous part of the differential equation is given by ignoring the non-homogeneous term: \(y' - 2y = 0\). This can be solved using separation of variables or recognizing it as a first-order linear homogeneous equation with the general solution \(y_h = Ce^{2x}\), where C is an arbitrary constant.

Step 3: Find a Particular Solution

To find a particular solution \(y_p\), we can use the method of undetermined coefficients. Given the non-homogeneous term \(4e^{2x}\), guess a solution of the form \(Ae^{2x}\) because the non-homogeneous term is an exponential function whose exponent matches that of the homogeneous solution. Substitute this into the original equation to find the value of A. After substituting \(y_p = Ae^{2x}\) and its derivative \(y'_p = 2Ae^{2x}\) into the equation, we get \(2Ae^{2x} = 2Ae^{2x} + 4e^{2x}\). Equating coefficients of like terms gives us \(0 = 4\), which implies there is no value of A that will satisfy the equation. Hence, we need to modify our guess to \((Ax+B)e^{2x}\) since the non-homogeneous term is a solution to the homogeneous equation.

Step 4: Determine the Coefficients

Plugging \(y_p = (Ax+B)e^{2x}\) into the original equation we get \(y'_p = (2Ax+2B)e^{2x}\). Substitution gives us \((2Ax+2B)e^{2x} = 2(Ax+B)e^{2x} + 4e^{2x}\). Distributing and collecting like terms, the equation simplifies to \(2Be^{2x} = 4e^{2x}\). Therefore, the coefficient B is equal to 2 and term with A cancels out, implying A can be any real number, which we can set to zero without loss of generality.

Step 5: Combine Homogeneous and Particular Solutions

The general solution to the differential equation is the sum of the homogeneous and particular solutions: \(y(x) = y_h + y_p = Ce^{2x} + 2xe^{2x}\), where C is an arbitrary constant.

Key Concepts

Homogeneous Differential Equation

A homogeneous differential equation is a specific type of differential equation where the solution can be determined without any terms that depend explicitly on the independent variable—in our case, this variable is denoted as 'x'. To solve the homogeneous part of a first-order linear differential equation like the one in our exercise, one often uses the technique of separation of variables or recognizes it as a characteristic equation.

When facing the equation \( y' - 2y = 0 \), the term \(2y\) is the only thing affecting \(y'\) (the derivative of y with respect to x). There are no extra terms that alter the relationship—this is what makes it homogeneous. Solving this, we got the general solution \( y_h = Ce^{2x} \), where \(C\) represents an arbitrary constant. This solution describes a set of functions that all share a common structure, namely an exponential function with a base of \(e\) raised to the power of \(2x\).

Understanding homogeneous equations is crucial, as they form the backbone for finding the general solution to more complex, non-homogeneous equations. The solutions to the homogeneous equation will be a part of the general solution to the entire differential equation.

Particular Solution

In contrast, a particular solution of a differential equation is one that includes all components of the equation, such as non-homogeneous terms. For the exercise at hand, we originally tried to find a particular solution in the form of \(Ae^{2x}\), but we did not arrive at a consistent outcome because the non-homogeneous term \(4e^{2x}\) already appeared as part of the homogeneous solution.

A fundamental insight is that the particular solution expresses the specific response of the system described by the differential equation to the non-homogeneous term — in this case, the term independent of \(y\), which is \(4e^{2x}\). To correctly accommodate this, we refined our approach by considering a solution of the form \( (Ax+B)e^{2x} \), where \(A\) and \(B\) are coefficients to be determined. It turned out that the term with \(A\) canceled itself in the equation, leaving us with \(B\), which we found to be 2 by equating the coefficients of like terms. Therefore, our particular solution is \( y_p = 2xe^{2x} \).

To further aid understanding, it helps to visualize the particular solution as a specific curve that intersects at least one point on the graph of the general solution to the non-homogeneous differential equation. It remains constant for a given non-homogeneous term and is a crucial part of the complete picture—the general solution.

Method of Undetermined Coefficients

The method of undetermined coefficients is a powerful tool for finding particular solutions to linear non-homogeneous differential equations. It's particularly effective when the non-homogeneous term is a simple function like a polynomial, exponential, sine, or cosine.

To use this method, as we did in our exercise, we start by making an educated guess about the form of the particular solution based on the non-homogeneous term. We guessed the form \(Ae^{2x}\) borrowing from the pattern of the non-homogeneous term \(4e^{2x}\). However, this did not work because our chosen form was also a solution to the homogeneous equation. The method requires that the guess be linearly independent from the homogeneous solution.

After adjusting our guess to \( (Ax+B)e^{2x} \), we then plugged this into the differential equation and used the results to solve for the coefficients. This process involved equating the coefficients of like terms to determine specific values for \(A\) and \(B\), which eventually led us to the particular solution that complemented the homogeneous solution. Remember that the method of undetermined coefficients often involves trial and error to find the right form and careful arithmetic to solve for the coefficients once the correct form is established.