Question

In each of Problems 1 through 12 find the general solution of the given differential equation. $$ y^{\prime \prime}-2 y^{\prime}-3 y=3 e^{2 x} $$

Short answer

Question: Find the general solution to the given non-homogeneous differential equation: \(y'' - 2y' - 3y = 3 e^{2x}\). Answer: The general solution is \(y(x) = C_1 e^{-x} + C_2 e^{3x} - e^{2x}\), where \(C_1\) and \(C_2\) are arbitrary constants.

Step-by-step solution

Solve the homogeneous differential equation

Solve the equation \(y'' - 2y' - 3y = 0\) by finding the characteristic equation. The characteristic equation is given by: $$ r^2 - 2r - 3 = 0 $$ Solve for r: $$ r_{1,2}=\frac{2\pm\sqrt{(-2)^2-4(1)(-3)}}{2}=\frac{2\pm\sqrt{16}}{2}=1\pm2 $$ So the roots are: \(r_1 = -1\) and \(r_2 = 3\). With these roots, the complementary solution is given by: $$ y_c(x) = C_1 e^{-x} + C_2 e^{3x} $$ where \(C_1\) and \(C_2\) are arbitrary constants. ##Step 2: Find a particular solution##

Apply the method of undetermined coefficients

To find a particular solution for the given non-homogeneous differential equation, guess the form of the solution as: $$ y_p(x) = A e^{2x} $$ Now, find the first and second derivatives of \(y_p(x)\): $$ y_p'(x) = 2A e^{2x}, \quad y_p''(x) = 4A e^{2x} $$ Substitute \(y_p(x)\) and its derivatives into the given differential equation: $$ (4A e^{2x}) - 2 (2A e^{2x}) - 3 (A e^{2x}) = 3 e^{2x} $$ simplify the equation: $$ Ae^{2x}(4 - 4 - 3) = 3 e^{2x} $$ Therefore, $$ -3A e^{2x} = 3 e^{2x} $$ Now we can solve for the undetermined coefficient \(A\): $$ -3A = 3 \implies A = -1 $$ So the particular solution is: $$ y_p(x) = - e^{2x} $$ ##Step 3: Find the general solution##

Combine complementary and particular solutions

The general solution to the given non-homogeneous differential equation can be found by combining the complementary solution \(y_c(x)\) and the particular solution \(y_p(x)\): $$ y(x) = y_c(x) + y_p(x) = C_1 e^{-x} + C_2 e^{3x} - e^{2x} $$ So the general solution is: $$ y(x) = C_1 e^{-x} + C_2 e^{3x} - e^{2x} $$

Key concepts

General Solution

When solving differential equations, we often seek the "general solution." This is a solution that accounts for all possible cases by incorporating constants that allow us to generate any particular solution. In the case of the given differential equation \( y^{\prime \prime} - 2y^{\prime} - 3y = 3e^{2x} \), the general solution is found by combining the complementary solution (related to the homogeneous part) and the particular solution (related to the non-homogeneous part).

The general solution, therefore, appears as:

• a part involving arbitrary constants: \( C_1 e^{-x} + C_2 e^{3x} \)
• and a specific function that satisfies the non-homogeneous equation: \( -e^{2x} \).

This results in the total general solution: \( y(x) = C_1 e^{-x} + C_2 e^{3x} - e^{2x} \).
So remember, always combine both parts to form the complete general solution.

Non-Homogeneous Equations

Differential equations are divided into two primary types: homogeneous and non-homogeneous. In a non-homogeneous differential equation, there is an additional function that does not depend on the solution function, as seen with \( 3e^{2x} \) in our example. This term makes the equation non-homogeneous.

When solving such equations, we aim to find solutions that satisfy the entire equation, including this extra non-homogeneous term. To do this, we need to:

• First solve the corresponding homogeneous equation, which sets the non-homogeneous part to zero
• Then find a particular solution for the entire original equation, including the non-homogeneous term.

Both results are combined to develop the complete solution to the differential equation. This approach allows us to tackle complex differential equations effectively.

Method of Undetermined Coefficients

The "method of undetermined coefficients" is a useful strategy for solving non-homogeneous linear differential equations, such as the one in our problem. It involves guessing a form for the particular solution based on the non-homogeneous component.

For instance, with the non-homogeneous term \( 3e^{2x} \), we assume a form \( y_p(x) = Ae^{2x} \). We then find its derivatives and substitute them back into the differential equation. Simplifying gives us a relationship where we can solve for the constant \( A \).

This method works well for equations with polynomial, exponential, or trigonometric terms since these behave predictably when differentiated. Here, our guess yielded \( A = -1 \), leading to the particular solution \( y_p(x) = -e^{2x} \). It’s a systematic way to find a particular solution without guessing blindly.

Characteristic Equation

The characteristic equation is a cornerstone method in solving linear homogeneous differential equations. It transforms a differential equation into an algebraic equation that is easier to solve. For our homogeneous equation \( y'' - 2y' - 3y = 0 \), we assume a solution of the form \( y = e^{rx} \) and derive the characteristic equation:

• \( r^2 - 2r - 3 = 0 \)

By solving this quadratic equation, we find the roots, which guide us in forming the complementary solution.

The roots \( r_1 = -1 \) and \( r_2 = 3 \) lead to two linearly independent solutions \( e^{-x} \) and \( e^{3x} \), which can be combined into \( y_c(x) = C_1 e^{-x} + C_2 e^{3x} \).
Finding these roots is key, as they determine the behaviour of the solution to the differential equation. Without them, we could not accurately describe the system's dynamics.