Problem 28 - A First Course in Differential Equations with Modeling Applications

Question

Solve the given initial-value problem. $$2 y^{\prime \prime}+3 y^{\prime}-2 y=14 x^{2}-4 x-11, \quad y(0)=0, y^{\prime}(0)=0$$

Step-by-Step Solution

Answer

\( y = -\frac{247}{12} e^{x/2} + \frac{37}{12} e^{-2x} + 7x^2 - \frac{4}{3}x + \frac{45}{2} \).

Step 1: Identify the type of differential equation

The given differential equation is a second-order linear non-homogeneous differential equation: \(2 y'' + 3 y' - 2 y = 14x^2 - 4x - 11\). The left side is the linear differential operator, and the right side represents the non-homogeneous part.

Step 2: Solve the homogeneous equation

First, solve the homogeneous equation \(2 y'' + 3 y' - 2 y = 0\). The characteristic equation is formed as \(2r^2 + 3r - 2 = 0\). Solve this quadratic equation using the quadratic formula \(r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 2\), \(b = 3\), \(c = -2\).

Step 3: Compute roots of the characteristic equation

Plug into the quadratic formula: \(r = \frac{-3 \pm \sqrt{3^2 - 4(2)(-2)}}{2(2)} = \frac{-3 \pm \sqrt{9 + 16}}{4} = \frac{-3 \pm 5}{4}\). We find the roots are \(r_1 = \frac{1}{2}\) and \(r_2 = -2\).

Step 4: Form the general solution of the homogeneous equation

The general solution of the homogeneous equation is \(y_h = C_1 e^{r_1 x} + C_2 e^{r_2 x} = C_1 e^{x/2} + C_2 e^{-2x}\).

Step 5: Solve the particular solution using undetermined coefficients

Assume a particular solution of the form \(y_p = Ax^2 + Bx + C\). Differentiate: \(y_p' = 2Ax + B\) and \(y_p'' = 2A\). Substitute \(y_p\), \(y_p'\), and \(y_p''\) into \(2 y'' + 3 y' - 2 y = 14 x^2 - 4 x - 11\) and match coefficients.

Step 6: Substitute and equate coefficients

Substitute: \(2(2A) + 3(2Ax + B) - 2(Ax^2 + Bx + C) = 14x^2 - 4x - 11\). Simplify and equate coefficients of like powers of \(x\): \(x^2 (0) + x (4A - 2A) + (4A + 3B - 2B) + (3A + 2B - 2C) = 14x^2 - 4x -11\). This yields \(2A = 14\), \(3B - 2B = -4\), and \(4A + 3B - 2C = -11\).

Step 7: Solve for A, B, C

From the equations: \(2A = 14\) gives \(A = 7\). Substituting \(A = 7\) into the second equation \(3B = -4\) gives \(B = -\frac{4}{3}\). Now substitute \(A = 7\), \(B = -\frac{4}{3}\) into \(4(7) + 3(-\frac{4}{3}) - 2C = -11\) to solve for \(C\). Simplifying gives \(28 - 4 - 2C = -11\) which simplifies to \(-2C = -11 - 24\). Solving gives \(C = \frac{45}{2}\).

Step 8: Write particular solution

Thus, the particular solution becomes \( y_p = 7x^2 - \frac{4}{3}x + \frac{45}{2} \).

Step 9: Combine to form the general solution

Combine the homogeneous and particular solutions: \( y = y_h + y_p = C_1 e^{x/2} + C_2 e^{-2x} + 7x^2 - \frac{4}{3}x + \frac{45}{2} \).

Step 10: Apply initial conditions

Use the initial conditions \(y(0) = 0\) and \(y'(0) = 0\) to solve for \(C_1\) and \(C_2\). Plug into the general solution: \( y(0) = C_1 + C_2 + \frac{45}{2} = 0 \) and \( y'(0) = \frac{C_1}{2} - 2C_2 - \frac{4}{3} = 0\).

Step 11: Solve for constants

From the first equation, \( C_1 + C_2 = -\frac{45}{2} \). From the second equation, \( \frac{C_1}{2} - 2C_2 - \frac{4}{3} = 0 \). Solve this system of equations to find \( C_1 = -\frac{247}{12} \) and \( C_2 = \frac{37}{12} \).

Step 12: Write the particular solution with constants

Substitute \( C_1 \) and \( C_2 \) back into the general solution: \( y = -\frac{247}{12} e^{x/2} + \frac{37}{12} e^{-2x} + 7x^2 - \frac{4}{3}x + \frac{45}{2} \).

Key Concepts

Initial-value problem

In mathematics, an initial-value problem (IVP) refers to a scenario where a differential equation must be solved, considering a set of initial conditions. These conditions specify the value of the unknown function (and possibly its derivatives) at a particular point, giving the solution a starting point from which to progress. In the given problem, the differential equation is a second-order non-homogeneous equation, and the initial conditions are specified as \( y(0) = 0 \) and \( y'(0) = 0 \).
These conditions help determine the specific solution of the differential equation that fits the problem at hand. Without initial conditions, one would end up with a family of solutions.

• The initial condition \( y(0) = 0 \) means that the solution must pass through the origin at time \( t = 0 \).
• The condition \( y'(0) = 0 \) implies that the rate of change of the solution at time \( t = 0 \) is zero.

By applying these conditions, you tailor the general solution to specifically address the given problem.

Non-homogeneous differential equation

A non-homogeneous differential equation includes a term that is not a function of the unknown variable alone, which often involves external forces or influences acting on the system. Homogeneous equations, by contrast, have zero on their right-hand side.
In this particular exercise, the equation \( 2 y'' + 3 y' - 2 y = 14x^2 - 4x - 11 \) is non-homogeneous due to the presence of the polynomial \( 14x^2 - 4x - 11 \) on the right side. This polynomial represents external forces acting upon the system, requiring consideration while solving for the particular solution.

• To work with a non-homogeneous differential equation, one typically finds the general solution, which combines the complementary (homogeneous) solution \( y_h \) with the particular solution \( y_p \).
• The goal is to consider both the natural behavior of the system and the effects of external forces.

Solving such an equation involves first addressing the homogeneous part before determining and adding the particular solution.

Quadratic characteristic equation

The quadratic characteristic equation arises when solving linear differential equations with constant coefficients. This "characteristic equation" is derived from the homogeneous part of the differential equation.
For the equation \( 2 y'' + 3 y' - 2 y = 0 \), we form the characteristic equation \( 2r^2 + 3r - 2 = 0 \) by substituting \( y = e^{rx} \) in the differential equation. Solving this quadratic for \( r \) gives the roots, which helps determine the general solution of the homogeneous differential equation.

• The quadratic characteristic equation can often be solved using the quadratic formula: \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
• In this problem, substituting into the formula gives roots \( r_1 = \frac{1}{2} \) and \( r_2 = -2 \).
• These roots shape the complementary solution, taking the form \( y_h = C_1 e^{r_1 x} + C_2 e^{r_2 x} \), where \( C_1 \) and \( C_2 \) are constants.

This general form reflects the intrinsic behavior of the system without external forces.

Undetermined coefficients

Undetermined coefficients is a method for finding a particular solution to a non-homogeneous differential equation. This approach works well when the non-homogeneous term on the right-hand side of the equation is a polynomial, exponential, or sine/cosine function. It involves guessing a form for the particular solution and then determining the coefficients by substitution.
For the equation \( 2 y'' + 3 y' - 2 y = 14x^2 - 4x - 11 \), we assume a particular solution in the form \( y_p = Ax^2 + Bx + C \).
This form reflects the polynomial nature of the non-homogeneous term. To find the unknown coefficients \( A \), \( B \), and \( C \):

• Differentiate the assumed \( y_p \) to find its derivatives \( y_p' \) and \( y_p'' \).
• Substitute \( y_p \), \( y_p' \), and \( y_p'' \) back into the left-hand side of the original differential equation.
• Match coefficients of corresponding powers of \( x \) with the right-hand side to form a system of equations to solve for \( A \), \( B \), and \( C \).

This technique simplifies the process of finding the particular form tailored to the specific non-homogeneous influences.