/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 55 Find the general solution of the... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find the general solution of the following differential equations: (a) \(2 \frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}-5 \frac{\mathrm{d} x}{\mathrm{~d} t}+3 x=0\) (b) \(\frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}+2 \frac{\mathrm{d} x}{\mathrm{~d} t}+5 x=0\) (c) \(\frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}+3 \frac{\mathrm{d} x}{\mathrm{~d} t}-4 x=0\) (d) \(\frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}-4 \frac{\mathrm{d} x}{\mathrm{~d} t}+13 x=0\)

Short Answer

Expert verified
The general solutions are: (a) \( x(t) = C_1 e^{t} + C_2 e^{\frac{3}{2}t} \); (b) \( x(t) = e^{-t} (C_1 \cos(2t) + C_2 \sin(2t)) \); (c) \( x(t) = C_1 e^{t} + C_2 e^{-4t} \); (d) \( x(t) = e^{2t} (C_1 \cos(3t) + C_2 \sin(3t)) \).

Step by step solution

01

Identify the Form of the Differential Equation

Each of the given differential equations is a second-order linear homogeneous differential equation of the form: \[ a \frac{\mathrm{d}^2 x}{\mathrm{d} t^2} + b \frac{\mathrm{d} x}{\mathrm{d} t} + c x = 0 \] where \(a\), \(b\), and \(c\) are constants.
02

Write the Characteristic Equation

Convert each differential equation into its corresponding characteristic equation, which is given by: \[ a r^2 + b r + c = 0 \] This characteristic equation is a quadratic that will help in finding the general solutions.
03

Solve the Characteristic Equation for (a)

Equation (a) is \[ 2 \frac{\mathrm{d}^2 x}{\mathrm{~d} t^{2}}-5 \frac{\mathrm{d} x}{\mathrm{~d} t}+3 x=0 \] The characteristic equation is \[ 2r^2 - 5r + 3 = 0 \] Using the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), we find the roots to be \( r_1 = 1 \) and \( r_2 = \frac{3}{2} \). Since both roots are real and distinct, the general solution is \[ x(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t} = C_1 e^{t} + C_2 e^{\frac{3}{2}t} \].
04

Solve the Characteristic Equation for (b)

Equation (b) is \[ \frac{\mathrm{d}^2 x}{\mathrm{~d} t^{2}}+2 \frac{\mathrm{d} x}{\mathrm{~d} t}+5 x=0 \] The characteristic equation is \[ r^2 + 2r + 5 = 0 \] Calculating the discriminant \( b^2 - 4ac = -16 \), which is negative, we find complex roots: \( r = -1 \pm 2i \).
05

General Solution for Complex Roots (b)

For complex roots of the form \(- \alpha \pm \beta i\), the general solution of the differential equation is \[ x(t) = e^{-\alpha t} \left( C_1 \cos(\beta t) + C_2 \sin(\beta t) \right) \] Substituting \( \alpha = 1 \) and \( \beta = 2 \), we get \[ x(t) = e^{-t} \left( C_1 \cos(2t) + C_2 \sin(2t) \right) \].
06

Solve the Characteristic Equation for (c)

Equation (c) is \[ \frac{\mathrm{d}^2 x}{\mathrm{~d} t^{2}}+3 \frac{\mathrm{d} x}{\mathrm{~d} t}-4 x=0 \] The characteristic equation is \[ r^2 + 3r - 4 = 0 \] Using the quadratic formula, the roots are \( r_1 = 1 \) and \( r_2 = -4 \), which are real and distinct.
07

General Solution for Real Distinct Roots (c)

For real and distinct roots, the general solution is \[ x(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t} = C_1 e^{t} + C_2 e^{-4t} \].
08

Solve the Characteristic Equation for (d)

Equation (d) is \[ \frac{\mathrm{d}^2 x}{\mathrm{~d} t^{2}}-4 \frac{\mathrm{d} x}{\mathrm{~d} t}+13 x=0 \] The characteristic equation is \[ r^2 - 4r + 13 = 0 \] The discriminant \( b^2 - 4ac = -36 \) is negative, indicating complex roots: \( r = 2 \pm 3i \).
09

General Solution for Complex Roots (d)

The complex roots result in a general solution \[ x(t) = e^{2t} \left( C_1 \cos(3t) + C_2 \sin(3t) \right) \].

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Characteristic Equation
In the study of second-order linear differential equations, a critical step is forming what we call the characteristic equation. This equation is quite literally derived from the differential equation itself. In a general second-order linear homogeneous differential equation, given by \( a \frac{\mathrm{d}^2 x}{\mathrm{d} t^2} + b \frac{\mathrm{d} x}{\mathrm{d} t} + c x = 0 \), the characteristic equation is obtained by replacing the derivatives with powers of a variable, typically \( r \). Hence, we get a quadratic: \( a r^2 + b r + c = 0 \). This transformation is not just arbitrary; it systematically helps to identify the nature of solutions the differential equation might have, based on the roots of this quadratic. The roots—either real, repeated, or complex—will guide us to the type of solution necessary to satisfy the equation.
Real and Distinct Roots
When solving the characteristic equation, one possibility is finding real and distinct roots. In practical terms, these are simply solutions to the quadratic where each root is real and they are different from each other. Use the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) to find these. If the discriminate \( b^2 - 4ac \) is positive, it indicates two separate real roots. For instance, if you calculate and discover roots \( r_1 \) and \( r_2 \), the general solution of the differential equation will take the form: \( x(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t} \), where \( C_1 \) and \( C_2 \) are arbitrary constants. This result reflects the principle that real and distinct solutions lead to exponential functions that are linearly independent.
Complex Roots
Another scenario in solving the characteristic equation involves complex roots. This occurs when the discriminant \( b^2 - 4ac \) is negative, leading to no real roots. Instead, you find complex conjugates, often expressed as \( r = \alpha \pm \beta i \). Solutions to the differential equations involving complex roots are framed in terms of exponential and trigonometric functions. Specifically, the general solution is \( x(t) = e^{\alpha t} \left( C_1 \cos(\beta t) + C_2 \sin(\beta t) \right) \). This expression captures both the oscillatory nature (due to \( \cos \) and \( \sin \)) and the exponential growth or decay (from \( e^{\alpha t} \)). Complex root cases are rich in applications, especially where systems experience oscillations or waves.
Homogeneous Differential Equations
A homogeneous differential equation, bearing the form \( a \frac{\mathrm{d}^2 x}{\mathrm{d} t^2} + b \frac{\mathrm{d} x}{\mathrm{d} t} + c x = 0 \), signifies that all terms of the equation are proportional to the function or its derivatives. Simply put, there are no constant or function terms added on the right side. Solving these typically involves finding solutions that are linear combinations of exponential components derived from the characteristic equation's roots. Such solutions are notable for not containing arbitrary functions or additions, thus every solution is scale-invariant. Homogeneous equations are foundational in various areas like physics and engineering, particularly when modeling natural systems free from external input.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Denote the Euler-method solution of the initialvalue problem $$ \frac{\mathrm{d} x}{\mathrm{~d} t}=\frac{x t}{t^{2}+2}, \quad x(1)=2 $$ using step size \(h=0.1\) by \(X_{\mathrm{a}}(t)\), and that using \(h=0.05\) by \(X_{\mathrm{b}}(t) .\) Find the values of \(X_{\mathrm{a}}(2)\) and \(X_{\mathrm{b}}(2) .\) Estimate the error in the value of \(X_{\mathrm{b}}(2)\), and suggest a value of step size that would provide a value of \(X(2)\) accurate to \(0.1 \%\). Find the value of \(X(2)\) using this step size. Find the exact solution of the initial-value problem, and determine the actual magnitude of the errors in \(X_{\mathrm{a}}(2), X_{\mathrm{b}}(2)\) and your final value of \(X(2)\).

Find the damping parameters and natural frequencies of the systems governed by the following second-order linear constant-coefficient differential equations: (a) \(\frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}+7 \frac{\mathrm{d} x}{\mathrm{~d} t}+2 x=0\) (b) \(\frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}+p \frac{\mathrm{d} x}{\mathrm{~d} t}+p^{1 / 2} x=0\) (c) \(\frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}+2 a q \frac{\mathrm{d} x}{\mathrm{~d} t}+\frac{1}{2} q x=0\) (d) \(\frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}+14 \frac{\mathrm{d} x}{\mathrm{~d} t}+2 \alpha x=0\)

For what value of \(k\) is the function \(t^{k}\) an integrating factor for the differential equation, $$ \left(t^{2} \cos x t\right) \frac{\mathrm{d} x}{\mathrm{~d} t}+3 \sin x t+x t \cos x t=0 ? $$

Determine which members of the given sets are solutions of the following differential equations. Hence, in each case, write down the general solution of the differential equation. (a) \(\frac{\mathrm{d}^{4} x}{\mathrm{~d} t^{4}}=0 \quad\left\\{1, t, t^{2}, t^{3}, t^{4}, t^{5}, t^{6}\right\\}\) (b) \(\frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}-p^{2} x=0 \quad\left\\{\mathrm{e}^{p t}, \mathrm{e}^{-p t}, \cos p t, \sin p t\right\\}\) (c) \(\frac{\mathrm{d}^{4} x}{\mathrm{~d} t^{4}}-p^{4} x=0\) \(\left\\{\mathrm{e}^{p t}, \mathrm{e}^{-p t}, \cos p t, \sin p t, \cosh p t, \sinh p t\right\\}\) (d) \(\frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}+2 \frac{\mathrm{d} x}{\mathrm{~d} t}=0\) \(\left\\{\cos 2 t, \sin 2 t, \mathrm{e}^{-2 t}, \mathrm{e}^{2 t}, t^{2}, t, 1\right\\}\) (e) \(\frac{\mathrm{d}^{3} x}{\mathrm{~d} t^{3}}+4 \frac{\mathrm{d} x}{\mathrm{~d} t}=0\) \(\left\\{\cos 2 t, \sin 2 t, \mathrm{e}^{-2 t}, \mathrm{e}^{2 t}, t^{2}, t, 1\right\\}\) (f) \(\frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}+2 \frac{\mathrm{d} x}{\mathrm{~d} t}+x=0\) \(\left\\{\mathrm{e}^{t}, \mathrm{e}^{-t}, \mathrm{e}^{2 t}, \mathrm{e}^{-2 t}, t \mathrm{e}^{t}, t \mathrm{e}^{-t}, t \mathrm{e}^{2 t}, t \mathrm{e}^{-2 t}\right\\}\) (g) \(\frac{\mathrm{d}^{3} x}{\mathrm{~d} t^{3}}-\frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}-\frac{\mathrm{d} x}{\mathrm{~d} t}+x=0\) \(\left\\{\mathrm{e}^{t}, \mathrm{e}^{-t}, \mathrm{e}^{2 t}, \mathrm{e}^{-2 t}, t \mathrm{e}^{t}, t \mathrm{e}^{-t}, t \mathrm{e}^{2 t}, t \mathrm{e}^{-2 t}\right\\}\)

Show that by making the substitution $$ v=\frac{\mathrm{d} x}{\mathrm{~d} t} $$ and noting that $$ \frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}=\frac{\mathrm{d} v}{\mathrm{~d} t}=\frac{\mathrm{d} v}{\mathrm{~d} x} \frac{\mathrm{d} x}{\mathrm{~d} t}=v \frac{\mathrm{d} v}{\mathrm{~d} x} $$ the equation $$ \frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}=x \frac{\mathrm{d} x}{\mathrm{~d} t} $$ may be expressed as $$ v \frac{\mathrm{d} v}{\mathrm{~d} x}=x v $$ Show that the solution of this equation is \(v=\frac{1}{2} x^{2}+C\) and hence find \(x(t)\) This technique is a standard method for solving second-order differential equations in which the independent variable does not appear explicitly. Apply the same method to obtain the solutions of the differential equations (a) \(\frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}=p \frac{\mathrm{d} x}{\mathrm{~d} t}\) (b) \(\frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}=\left(\frac{\mathrm{d} x}{\mathrm{~d} t}\right)^{2}\) (c) \(\frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}=\left(\frac{\mathrm{d} x}{\mathrm{~d} t}\right)^{2}\left(2 x-\frac{1}{x}\right)\)

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.