91Ó°ÊÓ

Find the general solutions of the following differential equations: (a) \(4 \frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}-2 \frac{\mathrm{d} x}{\mathrm{~d} t}+7 x=0\) (b) \(\frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}+6 \frac{\mathrm{d} x}{\mathrm{~d} t}-4 x=0\) (c) \(3 \frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}+3 \frac{\mathrm{d} x}{\mathrm{~d} t}+3 x=0\) (d) \(\frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}-8 \frac{\mathrm{d} x}{\mathrm{~d} t}+16 x=0\) (e) \(9 \frac{\mathrm{d}^{3} x}{\mathrm{~d} t^{3}}-9 \frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}-4 \frac{\mathrm{d} x}{\mathrm{~d} t}+4 x=0\) (f) \(\frac{\mathrm{d}^{3} x}{\mathrm{~d} t^{3}}-\frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}+7 \frac{\mathrm{d} x}{\mathrm{~d} t}+9 x=0\) (g) \(\frac{\mathrm{d}^{3} x}{\mathrm{~d} t^{3}}-2 \frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}+3 \frac{\mathrm{d} x}{\mathrm{~d} t}=0\)

Step by step solution

01

Determine the Characteristic Equation for (a)

For the differential equation \(4 \frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}-2 \frac{\mathrm{d} x}{\mathrm{~d} t}+7 x=0\), assume a solution of the form \(x = e^{rt}\). Substitute \(x\) into the differential equation to obtain the characteristic equation. This gives us: \[4r^2 - 2r + 7 = 0\].

02

Solve the Characteristic Equation (a)

The characteristic equation is a quadratic equation: \[4r^2 - 2r + 7 = 0\]. Use the quadratic formula: \[r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]where \(a = 4\), \(b = -2\), and \(c = 7\). Calculate:\[r = \frac{2 \pm \sqrt{(-2)^2 - 4\cdot4\cdot7}}{2\cdot4}\]\[r = \frac{2 \pm \sqrt{4 - 112}}{8}\]\[r = \frac{2 \pm \sqrt{-108}}{8}\]This gives us complex roots \(r = \frac{1}{4} \pm \frac{\sqrt{27}i}{4}\).

03

Write the General Solution for (a)

Given complex roots \( r = \frac{1}{4} \pm \frac{\sqrt{27}i}{4} \), the general solution is: \[x(t) = e^{\frac{1}{4}t}(C_1 \cos(\frac{\sqrt{27}}{4}t) + C_2 \sin(\frac{\sqrt{27}}{4}t))\].

04

Determine the Characteristic Equation for (b)

For the differential equation \(\frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}+6 \frac{\mathrm{d} x}{\mathrm{~d} t}-4 x=0\), assume a solution of the form \(x = e^{rt}\). Substitute \(x\) into the differential equation to get the characteristic equation: \[r^2 + 6r - 4 = 0\].

05

Solve the Characteristic Equation (b)

Use the quadratic formula for \(r^2 + 6r - 4 = 0\) with \(a = 1\), \(b = 6\), and \(c = -4\):\[r = \frac{-6 \pm \sqrt{6^2 - 4\cdot1\cdot(-4)}}{2\cdot1}\]\[r = \frac{-6 \pm \sqrt{36 + 16}}{2}\]\[r = \frac{-6 \pm \sqrt{52}}{2}\]\[r = \frac{-6 \pm 2\sqrt{13}}{2}\]\[r = -3 \pm \sqrt{13}\].

06

Write the General Solution for (b)

Given roots \(r = -3 + \sqrt{13}\) and \(r = -3 - \sqrt{13}\), the general solution is: \[x(t) = C_1 e^{(-3 + \sqrt{13})t} + C_2 e^{(-3 - \sqrt{13})t}\].

07

Repeat Similar Steps for (c), (d)

For each equation, determine the characteristic equation, solve it, and write the general solution. (Detailed steps follow a similar approach to Steps 1-6).

08

Adjust for Higher Order Differential Equations (e), (f) and (g)

For third-order differential equations, assume the solution \(x = e^{rt}\) and find the characteristic equation. Solve for roots using algebraic techniques or factorization if possible, then write the general solution including terms for each real and/or complex root.

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Key Concepts

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

Characteristic Equation

Every second-order linear differential equation, such as those given in the exercise, can be solved by first determining its characteristic equation. The characteristic equation is derived by assuming a solution of the form \(x = e^{rt}\), where \(r\) is a constant to be determined. By substituting \(x = e^{rt}\) into the differential equation, you transform it into a polynomial equation in terms of \(r\). This polynomial equation is the characteristic equation. For example, in equation (a), substituting \(x = e^{rt}\) gives the characteristic equation \(4r^2 - 2r + 7 = 0\). This equation indicates how the differential equation behaves and helps in finding the roots that describe the solution's form.

Complex Roots

When solving the characteristic equation, you can encounter both real and complex roots. Complex roots appear when the discriminant \(b^2 - 4ac\) of the quadratic equation is negative. These roots take the form \(r = \alpha \, \pm \beta i\). In our example (a), the characteristic equation's discriminant is negative, resulting in complex roots \(r = \frac{1}{4} \pm \frac{\sqrt{27}i}{4}\). Complex roots imply oscillatory behavior, which leads to solutions involving sine and cosine functions. The presence of \(i\) (the imaginary unit) converts individual exponential terms into trigonometric functions, capturing the essence of complex dynamics within the system or equation.

General Solution

The general solution of a differential equation provides a family of solutions that covers all possible scenarios defined by the equation. Once the roots of the characteristic equation are found, you can write down the general solution. With complex roots such as those in our example, the solution takes the form:

• \(x(t) = e^{\alpha t}(C_1 \cos(\beta t) + C_2 \sin(\beta t))\)

Here, \(\alpha\) is the real part of the root, and \(\beta\) is the imaginary coefficient. \(C_1\) and \(C_2\) are constants determined by initial conditions. Such solutions reflect the combined effects of growth/decay (from \(\alpha\)) and oscillation (from \(\beta\)). It's crucial to incorporate the effects of both of these parts to accurately represent the behavior of the system described by the differential equation.

Quadratic Formula

The quadratic formula is a critical tool for solving characteristic equations in differential equations. It provides the roots of any quadratic equation \(ax^2 + bx + c = 0\) as:

• \(r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

This formula works by calculating the values of \(r\), and is essential in finding the solutions of the characteristic equations derived from differential equations. In each case, like equation (a), you substitute the coefficients \(a\), \(b\), and \(c\) into this formula to get the roots. Depending on the discriminant value \((b^2 - 4ac)\), you can have two real roots, one real root (repeated), or complex roots, which influence the form of the general solution.