Differential Equation Exercise 19

We are asked to solve the following differential equation:

$$ y'' + 4y'+13y = \sin 3x $$

This is a second-order non-homogeneous (complete) linear differential equation.

To solve it, we add the general solution of the corresponding homogeneous equation, \( y_o \), to a particular solution of the full equation, \( y_p \).

So the general solution to the full equation is:

$$ y = y_o + y_p $$

We’ll begin by solving the homogeneous part, then find a particular solution, and finally combine them.

General Solution of the Homogeneous Equation

The associated homogeneous equation is obtained by removing the non-homogeneous term:

$$ y'' + 4y'+13y = 0 $$

We now write the characteristic equation using the auxiliary variable \( t \):

$$ t^2 + 4t +13 = 0 $$

Let's solve for the roots:

$$ t = \frac{-4 \pm \sqrt{16-52}}{2} $$

$$ t = \frac{-4 \pm \sqrt{-36}}{2} $$

$$ t = \frac{-4 \pm \sqrt{36i^2}}{2} $$

Note: The square root of a negative number can be expressed using complex numbers. Since \( -36 = (-1) \cdot 36 \) and \( i^2 = -1 \), we can rewrite the radical as \( i^2 \cdot 36 \), which allows us to compute the square root.

$$ t = \frac{-4 \pm i \sqrt{36}}{2} $$

$$ t = \frac{-4 \pm 6i}{2} $$

$$ t = -2 \pm 3i $$

$$ t = \begin{cases} t_1 = -2-3i \\ \\ t_2 = -2+3i \end{cases} $$

These are two complex roots of the form \( \alpha \pm i\beta \), with \( \alpha = -2 \) and \( \beta = 3 \).

Therefore, the general solution to the homogeneous second-order equation is:

$$ y_o = e^{\alpha x} \cdot [ c_1 \cos(\beta x) + c_2 \sin(\beta x) ] $$

Substituting \( \alpha = -2 \) and \( \beta = 3 \), we get:

$$ y_o = e^{-2x} \cdot [ c_1 \cos(3x) + c_2 \sin(3x) ] $$

This is the general solution of the homogeneous equation.

Particular Solution of the Non-Homogeneous Equation

Next, we need a particular solution to the non-homogeneous equation:

$$ y'' + 4y'+13y = \sin 3x $$

We use the method of undetermined coefficients, since the forcing term \( \sin(3x) \) is an elementary function.

According to this method, when the right-hand side is a sine function, we guess a particular solution of the form:

$$ y_p = A \sin(3x)+B \cos(3x) $$

This has the same functional form as the forcing term, but the constants A and B still need to be determined.

Let's compute the first and second derivatives of \( y_p \):

$$ y'_p = D_x[ A \sin(3x)+B \cos(3x) ] = 3A \cos(3x) - 3B \sin(3x) $$

$$ y''_p = D_x[ 3A \cos(3x) - 3B \sin(3x) ] = -9A \sin(3x) - 9B \cos(3x) $$

Now substitute \( y_p \), \( y'_p \), and \( y''_p \) into the original equation:

$$ y'' + 4y'+13y = \sin (3x) $$

We plug in the expressions step by step:

$$ [-9A \sin(3x) - 9B \cos(3x)] + 4[3A \cos(3x) - 3B \sin(3x)] + 13[A \sin(3x) + B \cos(3x)] = \sin(3x) $$

Group the terms by function:

$$ \sin(3x) \cdot [ -9A - 12B + 13A ] + \cos(3x) \cdot [ -9B + 12A + 13B ] = \sin(3x) $$

$$ \sin(3x) \cdot [ 4A - 12B ] + \cos(3x) \cdot [ 4B + 12A ] = \sin(3x) $$

Now equate the coefficients of \( \sin(3x) \) and \( \cos(3x) \) on both sides:

$$ \begin{cases} 4A - 12B = 1 \\ \\ 4B + 12A = 0 \end{cases} $$

Explanation: On the left-hand side, the coefficient of \( \sin(3x) \) is \( 4A - 12B \), which must equal the coefficient of \( \sin(3x) \) on the right-hand side, i.e., 1. Similarly, the coefficient of \( \cos(3x) \) is \( 4B + 12A \), which must be zero, since there's no cosine term on the right-hand side.

We solve the system by substitution:

$$ \begin{cases} 4A-12B = 1 \\ \\ B= -3A \end{cases} $$

Substitute B into the first equation:

$$ 4A - 12(-3A) = 1 \Rightarrow 40A = 1 \Rightarrow A = \frac{1}{40}, \quad B = -\frac{3}{40} $$

Now plug these values back into the particular solution:

$$ y_p = \frac{1}{40} \sin(3x) - \frac{3}{40} \cos(3x) $$

This is our particular solution to the non-homogeneous equation.

General Solution to the Full Differential Equation

The general solution to the full equation is the sum of the homogeneous and particular solutions:

$$ y = y_o + y_p $$

With \( y_o = e^{-2x} \cdot [ c_1 \cos(3x) + c_2 \sin(3x) ] \) and \( y_p = \frac{1}{40} \sin(3x) - \frac{3}{40} \cos(3x) \), we have:

$$ y = e^{-2x} \cdot [ c_1 \cos(3x) + c_2 \sin(3x) ] + \frac{1}{40} \sin(3x) - \frac{3}{40} \cos(3x) $$

This is the general solution to the given second-order non-homogeneous differential equation.