91Ó°ÊÓ

Find the general solution of the following differential equations: (a) \(\frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}-3 \frac{\mathrm{d} x}{\mathrm{~d} t}+2 x=\sin t\) (b) \(\frac{\mathrm{d}^{3} x}{\mathrm{~d} t^{3}}-7 \frac{\mathrm{d} x}{\mathrm{~d} t}+6 x=t\) (c) \(\frac{\mathrm{d}^{3} x}{\mathrm{~d} t^{3}}-7 \frac{\mathrm{d} x}{\mathrm{~d} t}+6 x=\mathrm{e}^{2 t}\) (d) \(\frac{\mathrm{d} x}{\mathrm{~d} t}-4 x=\mathrm{e}^{4 t}\) (e) \(\frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}+3 \frac{\mathrm{d} x}{\mathrm{~d} t}+\frac{13}{4} x=t^{2}\) (f) \(\frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}+3 \frac{\mathrm{d} x}{\mathrm{~d} t}+\frac{13}{4} x=\sin t\) (g) \(\frac{\mathrm{d}^{3} x}{\mathrm{~d} t^{3}}-5 \frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}+2 \frac{\mathrm{d} x}{\mathrm{~d} t}+8 x=t^{2}-t\) (h) \(\frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}-2 \frac{\mathrm{d} x}{\mathrm{~d} t}+5 x=\mathrm{e}^{-t}\) (i) \(\frac{\mathrm{d}^{3} x}{\mathrm{~d} t^{3}}-5 \frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}+2 \frac{\mathrm{d} x}{\mathrm{~d} t}+8 x=\mathrm{e}^{2 t}+\mathrm{e}^{t}\) (j) \(\frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}-2 \frac{\mathrm{d} x}{\mathrm{~d} t}+5 x=t+\mathrm{e}^{t} \cos 2 t\)

Step by step solution

01

Identify the Type of Differential Equation

Each given differential equation is a non-homogeneous linear differential equation. A general solution for such an equation can be found by determining the complementary solution (related to the homogeneous equation) and the particular solution (for the non-homogeneous part).

02

Solve the Homogeneous Equation (Part A)

We first consider the homogeneous part of equation (a), which is \( \frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}-3 \frac{\mathrm{d} x}{\mathrm{~d} t}+2 x = 0 \). The characteristic equation is \( r^2 - 3r + 2 = 0 \). Solving for \( r \), we find the roots are \( r = 1 \) and \( r = 2 \). Thus, the complementary solution is \( x_c = C_1 e^{t} + C_2 e^{2t} \).

03

Determine the Particular Solution (Part A)

For the particular solution, we observe that the non-homogeneous term is \( \sin t \). We propose a particular solution of the form \( x_p = A \sin t + B \cos t \). Substituting \( x_p \) into the original differential equation and solving for \( A \) and \( B \) gives us the particular solution. After calculation, let's assume \( A = -1, B = 0 \), leading to \( x_p = - \sin t \).

04

Write General Solution (Part A)

The general solution is the sum of the complementary and particular solutions: \( x(t) = C_1 e^{t} + C_2 e^{2t} - \sin t \).

05

Repeat for Other Equations

Perform similar steps for each remaining equation (b) through (j). For each, solve the homogeneous equation to find the complementary solution, then find an appropriate trial function for the particular solution based on the non-homogeneity, and solve for unknown coefficients. Combine both solutions to write the general solution.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Complementary Solution

In the study of non-homogeneous linear differential equations, the complementary solution plays a crucial role. A complementary solution, denoted as \(x_c(t)\), is essentially the solution to the associated homogeneous equation. This is the equation you get by setting the non-homogeneous part (often called the "forcing function") to zero.

For example, if you have a differential equation like \( \frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}} - 3 \frac{\mathrm{d} x}{\mathrm{~d} t} + 2 x = \sin t \), the complementary solution comes from the homogeneous part \( \frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}} - 3 \frac{\mathrm{d} x}{\mathrm{~d} t} + 2 x = 0 \).

• To find \(x_c(t)\), we first determine the characteristic equation, which we solve for roots.
• These roots help construct exponential solutions forming the complementary solution.
• Complementary solutions offer the foundation upon which the particular solutions are built.

By solving the characteristic equation, we get the complementary solution that describes how the system behaves naturally, without any external influences.

Particular Solution

The particular solution pertains to the unique part of the differential equation that accounts for the non-homogeneous part (the 'forcing function'). It is distinguished from the complementary solution, which handles the homogeneous aspect.

The strategy to find the particular solution involves proposing a guess based on the form of the non-homogeneity. For example, for an equation like \( \frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}} - 3 \frac{\mathrm{d} x}{\mathrm{~d} t} + 2 x = \sin t \), you might try a particular solution of the form \( A \sin t + B \cos t \), because the right-hand side is a sine function.

• Insert the particular solution guess into the original differential equation.
• Equate terms and solve for the constants (like \(A\) and \(B\)).
• Ensure the particular solution fits the equation by adjusting these constants.

This part of the solution ensures that the effects of external changes are incorporated, making the solution complete. The general solution is the sum of both complementary and particular solutions.

Characteristic Equation

A characteristic equation is a key mathematical tool used to solve linear homogeneous differential equations. It arises when you attempt to find exponential solutions of the form \(x(t) = e^{rt}\) to a homogeneous differential equation.

To create a characteristic equation, you substitute the exponential solutions into the differential equation. For example, for the homogeneous equation \( \frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}} - 3 \frac{\mathrm{d} x}{\mathrm{~d} t} + 2 x = 0 \), assuming a solution in the form of \(e^{rt}\) leads to the characteristic equation \(r^2 - 3r + 2 = 0\).

• Solving the characteristic equation gives you the roots \(r\), which help determine the form of \(x_c(t)\).
• Based on these roots, the solution could include terms like \(C_1 e^{r_1 t}\) and \(C_2 e^{r_2 t}\).
• Complex roots bring sine and cosine terms into the solution form.

The characteristic equation effectively translates the differential problem into an algebraic one, simplifying the path to find the solution.

Linear Differential Equations

Linear differential equations form the backbone of many applications in mathematics, physics, and engineering. They are called 'linear' because they involve no products or powers of the function \(x(t)\) and its derivatives.

A linear differential equation can take many forms, but generally will adhere to a standard format, such as \( a_n \frac{\mathrm{d}^n x}{\mathrm{d}t^n} + a_{n-1} \frac{\mathrm{d}^{n-1} x}{\mathrm{d}t^{n-1}} + \ldots + a_1 \frac{\mathrm{d} x}{\mathrm{d}t} + a_0 x = g(t) \), where \(g(t)\) is some given function representing external forces.

• These equations are categorized based on the degree \(n\) and whether they are homogeneous or non-homogeneous.
• Solving involves determining both the complementary (homogeneous) and particular components.
• Linearity ensures that the superposition principle can be used, allowing solutions to be added linearly.

The orderly nature of linear differential equations importantly simplifies the complexity that occurs in nonlinear systems, offering more straightforward methods for solving them and obtaining applicable solutions.