Question

In Exercises \(15-19\) find the general solution. $$ y^{\prime \prime}+2 y^{\prime}-3 y=-16 x e^{x} $$

Short answer

Answer: The general solution of the given non-homogeneous differential equation is \(y(x) = c_1 e^{x} + c_2 e^{-3x} + (x-1)e^x\).

Step-by-step solution

Find the complementary function (homogeneous equation)

First, we need to find the complementary function by solving the homogeneous equation: $$ y^{\prime \prime}+2 y^{\prime}-3 y=0 $$ Write down the characteristic equation: $$ r^2 + 2r - 3 = 0 $$ Now, we need to find the roots of the characteristic equation: $$ r_{1,2} = \frac{-b\pm\sqrt{b^2-4ac}}{2a} $$ Substitute with values: $$ r_{1,2} = \frac{-2\pm\sqrt{2^2-4(1)(-3)}}{2(1)} $$ After simplification, we get: $$ r_{1,2} = -1 \pm 2 $$ So, the roots are: $$ r_1 = 1, \quad r_2 = -3 $$ Hence, the complementary function is given by: $$ y_c = c_1 e^{x} + c_2 e^{-3x} $$

Find a particular solution (non-homogeneous equation)

Now, we will find a particular solution for the given non-homogeneous equation: $$ y^{\prime \prime}+2 y^{\prime}-3 y=-16 x e^{x} $$ We will guess a particular solution of the form: $$ y_p = (Ax^2+Bx+C)e^x $$ Let's find the first derivative, \(y_p'\): $$ y_p' = (2Ax+B)xe^x + (Ax^2+Bx+C)e^x $$ And now the second derivative, \(y_p''\): $$ y_p'' = (2A)xe^x + (2Ax+B)e^x + (2Ax+B)xe^x + (Ax^2+Bx+C)e^x $$ Substitute \(y_p\), \(y_p'\), and \(y_p''\) into the non-homogeneous equation: $$ (2A)xe^x + (2Ax+B)e^x + (2Ax+B)xe^x + (Ax^2+Bx+C)e^x + 2(2Ax+B)xe^x + 2(Ax^2+Bx+C)e^x - 3(Ax^2+Bx+C)e^x = -16xe^x $$ Simplify and collect the coefficients for \(e^x\): $$ (2A + 4A - 3A)x^2+(2B + 2B - 3B)x + (2A+2B+2C)x + (Ax^2 + Bx + C)e^x = -16xe^x $$ $$ x^2e^x - 14xe^x + (2A+2B+2C)e^x = -16xe^x $$ Comparing the coefficients, we get the following system of equations: $$ A = 0 \\ -14 = -16 + 2B \\ 2A+2B+2C = 0 $$ Solving the above system gives: $$ A=0, \quad B=1, \quad C=-1 $$ So, our particular solution is: $$ y_p = (x-1)e^x $$

Write the general solution

We have found the complementary function, \(y_c\), and a particular solution, \(y_p\), for the given non-homogeneous equation. The general solution is given by the sum of these two: $$ y(x) = y_c + y_p = (c_1 e^{x} + c_2 e^{-3x}) + (x-1)e^x $$ Therefore, the general solution is: $$ y(x) = c_1 e^{x} + c_2 e^{-3x} + (x-1)e^x $$

Key concepts

Characteristic Equation

When solving homogeneous differential equations, the characteristic equation is a crucial concept. It helps us find solutions that fit the equation. For example, given the differential equation:

• \( y'' + 2y' - 3y = 0 \),

we rewrite this using the characteristic equation:

• \( r^2 + 2r - 3 = 0 \).

This equation is derived by replacing the derivatives with powers of \( r \). Each term that involves a derivative translates to a power of \( r \) in the characteristic equation. This is a type of quadratic equation, which you can solve using the quadratic formula:

• \( r_{1,2} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).

The roots from the equation provide insights into the behavior of solutions to the original differential equation. Here, the roots are real and distinct, leading to solutions involving exponential functions.

Complementary Function

The complementary function, often denoted as \( y_c \), represents solutions derived from the homogeneous aspect of a differential equation. In our example, it comes from the differential equation:

• \( y'' + 2y' - 3y = 0 \).

The roots found from the characteristic equation determine the form of this complementary function. If the roots are real and distinct, as in our example \( r_1 = 1 \) and \( r_2 = -3 \), the complementary function is:

• \( y_c = c_1 e^{x} + c_2 e^{-3x} \).

The constants \( c_1 \) and \( c_2 \) are usually determined by initial conditions, if provided. The complementary function accounts for the natural behavior of the system without external forces or inputs.

Particular Solution

The particular solution addresses the "non-homogeneous" component of the differential equation. Non-homogeneous differential equations have additional terms that change the behavior of the solution, as seen in:

• \( y'' + 2y' - 3y = -16xe^x \).

We guess a solution form that resembles the non-homogeneous part:

• \( y_p = (Ax^2 + Bx + C)e^x \).

By differentiating this guessed form and substituting into the original equation, we adjust the coefficients to satisfy the equation. For our example, solving the system of equations revealed that:

• \( A = 0 \), \( B = 1 \), and \( C = -1 \).

This resulted in a particular solution:

• \( y_p = (x-1)e^x \).

Combining the complementary and particular solutions provides the general solution, incorporating both homogeneous and non-homogeneous aspects.

Non-Homogeneous Equation

Differential equations can either be homogeneous or non-homogeneous. A non-homogeneous equation has terms independent of the function or its derivatives. In our example,

• \( y'' + 2y' - 3y = -16xe^x \),

the term \(-16xe^x\) makes the equation non-homogeneous because it does not involve \( y \), \( y' \), or \( y'' \) directly. Solving non-homogeneous equations involves:

• Finding the complementary function from the associated homogeneous equation.
• Identifying a particular solution that fits the non-homogeneous part.

The general solution results from combining these two parts:

• \( y(x) = y_c + y_p \).

This approach shows how both the natural behavior (complementary) and the response to external inputs (particular) define the system's solution.