91Ó°ÊÓ

Non-homogeneous differential equations are a key concept when dealing with differential equations of any order. These equations are termed "non-homogeneous" because they contain a non-zero function on the right-hand side of the equation.
In the given problem, the equation is: \( \frac{\mathrm{d}^{2} i}{\mathrm{~d} t^{2}} + 8 \frac{\mathrm{d} i}{\mathrm{~d} t} + 25 i = 48 \cos 3 t - 16 \sin 3 t \).
Here's how to identify if an equation is non-homogeneous:

• The equation has terms purely dependent on the independent variable (such as \(\cos 3t\) and \(\sin 3t\)).
• If it wasn’t for these terms, the equation would read zero on the right side, making it homogeneous instead.

Non-homogeneous equations often arise in real-world physics problems, where external forces or inputs modify the system's behavior.
Solving such equations involves finding a combination of the homogeneous solution (which solves the associated homogeneous equation) and a particular solution (which solves the non-homogeneous part).
This combination gives you the general solution for the non-homogeneous equation.

The characteristic equation is a critical tool for solving homogeneous parts of second-order linear differential equations. It emerges when you solve the homogeneous version of the differential equation: \( \frac{\mathrm{d}^{2} i}{\mathrm{~d} t^{2}} + 8 \frac{\mathrm{d} i}{\mathrm{~d} t} + 25 i = 0 \).
To find the characteristic equation, start by assuming a solution of the form \(i(t) = e^{rt}\).
Substitute this assumed solution into the homogeneous differential equation, leading to the characteristic polynomial \( r^2 + 8r + 25 = 0 \).
This polynomial is solved using the quadratic formula:

• \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \(a=1\), \(b=8\), and \(c=25\).
• The roots of the characteristic equation, \( r = -4 \pm 3i \), indicate the type of solution form needed for the homogeneous equation.

With these complex roots, the associated homogeneous solution takes the form: \( i_h(t) = e^{-4t}(C_1 \cos 3t + C_2 \sin 3t) \).
The coefficients \(C_1\) and \(C_2\) adjust according to initial conditions or additional system constraints, forming part of the complete solution for the original differential equation.

Finding a particular solution is essential in dealing with non-homogeneous differential equations. The particular solution addresses the non-zero function on the right-hand side of the equation.
In this scenario, we have a non-homogeneous term: \( 48 \cos 3t - 16 \sin 3t \).
Given this, assume a particular solution of the form \( i_p(t) = A \cos 3t + B \sin 3t \).
To find specific coefficients \(A\) and \(B\):

• Differentiate \( i_p(t) \) to obtain the first derivative: \( \frac{\mathrm{d} i_p}{\mathrm{~d} t} = -3A \sin 3t + 3B \cos 3t \).
• Put that into the second derivative: \( \frac{\mathrm{d}^2 i_p}{\mathrm{~d} t^2} = -9A \cos 3t - 9B \sin 3t \).

Substitute both derivatives into the original differential equation and solve:

• Combine like terms which leads to: \( (16A + 24B) \cos 3t + (16B - 24A) \sin 3t = 48 \cos 3t - 16 \sin 3t \).
• From here, solve simultaneously: \( 16A + 24B = 48 \) and \( 16B - 24A = -16 \).
• This gives coefficients \( A = 4 \) and \( B = 2 \).

Your particular solution, \( i_p(t) = 4 \cos 3t + 2 \sin 3t \), combines with the homogeneous solution to form the general solution, addressing all aspects of the differential equation at hand.