Question

In Exercises \(1-14\) find a particular solution. $$ y^{\prime \prime}+2 y^{\prime}+y=e^{2 x}\left(-7-15 x+9 x^{2}\right) $$

Short answer

Answer: \(y_p(x) = e^{2x}(\frac32 x^{2} -\frac{11}{2} x - \frac{5}{2})\)

Step-by-step solution

Identify the equation and guess a form for the particular solution

The given equation is a second-order linear inhomogeneous differential equation: $$ y^{\prime\prime} + 2 y^{\prime} + y = e^{2 x}(-7 - 15 x + 9 x^{2}) $$ We guess a form for the particular solution as: $$ y_p(x) = e^{2x}(Ax^{2}+Bx + C) $$

Calculate the first and second derivatives of \(y_p(x)\)

In order to calculate the coefficients, we need to calculate the first and second derivatives of our guessed particular solution, \(y_p(x)\). First derivative: $$ y_p'(x) = e^{2x}(2Ax + B) + 2e^{2x}(Ax^{2}+Bx+C) $$ Second derivative: $$ y_p''(x) = e^{2x}(2A) + 2e^{2x}(2Ax + B) + 4e^{2x}(Ax^{2} + Bx + C) $$

Substitute \(y_p(x)\) and its derivatives into the given equation

Now, we substitute our guessed particular solution and its derivatives into the given differential equation: $$ (e^{2x}(2A) + 2e^{2x}(2Ax + B) + 4e^{2x}(Ax^{2} + Bx + C)) + 2(e^{2x}(2Ax + B) + 2e^{2x}(Ax^{2}+Bx+C)) + e^{2x}(Ax^{2}+Bx + C) = e^{2x}(-7 - 15x + 9x^{2}) $$

Simplify the equation and equate coefficients

Next, we simplify this equation and equate the coefficients of the corresponding terms: $$ (2A+2C) + (4A+4B+2) x + (6A) x^{2} = -7 - 15x + 9x^{2} $$ Equating the coefficients, we get the following system of equations: $$ \begin{cases} 6A = 9 \\ 4A + 4B + 2 = -15 \\ 2A + 2C = -7 \end{cases} $$

Solve the system of equations for the coefficients

Now, we solve this system of equations to get the values for \(A\), \(B\), and \(C\): $$ \begin{cases} A = \frac32 \\ B = -\frac{11}{2} \\ C = -\frac{5}{2} \end{cases} $$

Plug the coefficients back into the particular solution

Finally, we plug the obtained coefficients back into our guessed particular solution, \(y_p(x)\): $$ y_p(x) = e^{2x}(\frac32 x^{2} -\frac{11}{2} x - \frac{5}{2}) $$ This is the particular solution of the given second-order linear inhomogeneous differential equation with constant coefficients.

Key concepts

Second-order Linear Differential Equations

A second-order linear differential equation is a type of equation that involves the second derivative of a function, which essentially describes how the rate of change of the rate of change is behaving. In simpler terms, it's like investigating how something is accelerating, rather than just moving.
Typically, these equations are of the form ax² + bxy + cy² = f(x), where the coefficients 'a', 'b', and 'c' are constants, and the x and y represent variables or functions of a single variable. The key feature here is the presence of the second derivative denoted as \( y'' \).

• The term 'linear' signifies that the dependent variable and all its derivatives appear to the power of one, not squared, cubed, etc.
• In practical uses, second-order equations appear in many physics and engineering problems, such as SHM (simple harmonic motion) and circuits.

In our specific exercise, the equation \( y^{\prime\prime} + 2 y^{\prime} + y = e^{2 x}(-7-15 x+9 x^{2}) \) defines such a second-order linear relationship with a mix of algebraic terms on the right side.

Inhomogeneous Differential Equation

A differential equation is labeled inhomogeneous when it has terms that are not solely dependent on the function and its derivatives. In simpler words, if there's an added function of the independent variable (like \( e^{2x}(-7 - 15x + 9x^{2}) \) in our exercise), it's called inhomogeneous.
This added 'disturbance' makes solving the equation a bit more complex because we must find a solution that satisfies both the homogeneous aspect (with zero external input) and the inhomogeneous part.

• The common technique to address such scenarios involves solving the homogeneous equation first, which would be \( y^{\prime\prime} + 2 y^{\prime} + y = 0 \) in this case.
• Then, a particular solution is found to address the non-homogeneous part, where one guesses a form of \( y_p(x) \) depending on the external terms present.

This blend of solutions provides the complete solution to the inhomogeneous differential equation.

Constant Coefficients in Differential Equations

Constant coefficients in a differential equation imply that the factors accompanying each derivative in the equation do not change—they remain constant at all times. This uniformity simplifies the approach to solving differential equations.
For an equation such as \( y^{\prime\prime} + 2 y^{\prime} + y \), the coefficients before each term (2, and 1, respectively) don't vary with x.

• The advantage here is a consistent strategy for finding solutions, often involving characteristic equations and exponential functions, especially for homogenous solutions.
• Having constant coefficients makes using techniques like the method of undetermined coefficients or variation of parameters feasible for solving the inhomogeneous part as well.

In the exercise we've worked on, these constant coefficients streamline the process of equating terms and finding particular solutions for the given equation.