/*! 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 32 From \(4 m^{2}-4 m-3=0\) we obta... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 32

From \(4 m^{2}-4 m-3=0\) we obtain \(m=-1 / 2\) and \(m=3 / 2\) so that \(y=c_{1} e^{-x / 2}+c_{2} e^{3 x / 2} .\) If \(y(0)=1\) and \(y^{\prime}(0)=5\) then \(c_{1}+c_{2}=1,-\frac{1}{2} c_{1}+\frac{3}{2} c_{2}=5,\) so \(c_{1}=-7 / 4, c_{2}=11 / 4,\) and \(y=-\frac{7}{4} e^{-x / 2}+\frac{11}{4} e^{3 x / 2}\).

Short Answer

Expert verified
The specific solution is \(y = -\frac{7}{4} e^{-x/2} + \frac{11}{4} e^{3x/2}\).

Step by step solution

01

Understanding the Given Equation

We start with the quadratic equation in the form of a characteristic equation for differential equations: \(4m^2 - 4m - 3 = 0\). This equation helps find the roots that we will use to construct the general solution for the differential equation.
02

Solving the Quadratic Equation

Using the quadratic formula \(m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) on \(4m^2 - 4m - 3 = 0\), where \(a = 4\), \(b = -4\), and \(c = -3\), we find the roots: \(m = -\frac{1}{2}\) and \(m = \frac{3}{2}\).
03

Forming the General Solution

The general solution given the roots \(m = -\frac{1}{2}\) and \(m = \frac{3}{2}\) is \(y = c_1 e^{-x/2} + c_2 e^{3x/2}\). This is a linear combination of exponential solutions based on the roots.
04

Applying Initial Conditions

We apply the initial conditions \(y(0) = 1\) and \(y'(0) = 5\) to find the constants \(c_1\) and \(c_2\). Evaluating the general solution at \(x=0\), we get \(y(0) = c_1 + c_2 = 1\). Differentiating \(y(x)\) and evaluating at \(x=0\), we find \(y'(0) = -\frac{1}{2}c_1 + \frac{3}{2}c_2 = 5\).
05

Solving the System of Equations for Constants

The system of equations from the initial conditions is \(c_1 + c_2 = 1\) and \(-\frac{1}{2}c_1 + \frac{3}{2}c_2 = 5\). Solving this system gives \(c_1 = -\frac{7}{4}\) and \(c_2 = \frac{11}{4}\).
06

Constructing the Specific Solution

With the values for \(c_1\) and \(c_2\), the specific solution to the differential equation is \(y = -\frac{7}{4} e^{-x/2} + \frac{11}{4} e^{3x/2}\).

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Key Concepts

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

Characteristic Equation
The characteristic equation is key when dealing with linear differential equations with constant coefficients. This type of equation helps us find the nature of the solutions we should prepare for.
The characteristic equation is derived by assuming a solution of the form \(y = e^{mx}\), where \(m\) represents the root.
For our problem, the characteristic equation is given as \(4m^2 - 4m - 3 = 0\), which is a quadratic equation. Solving this allows us to identify the roots \(m = -\frac{1}{2}\) and \(m = \frac{3}{2}\). These roots are crucial as they form the basis of our general solution.
  • A characteristic equation similar to this often occurs when solving second-order linear differential equations with constant coefficients.
  • The roots will help us establish if the solution is real or complex, and dictate the form of the solution.
Using a characteristic equation transforms complex differential equations into simpler algebraic problems.
Quadratic Formula
The quadratic formula is an essential tool when dealing with second-order algebraic equations.
To solve the characteristic equation \(4m^2 - 4m - 3 = 0\), we use the quadratic formula: \[m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]where \(a = 4\), \(b = -4\), and \(c = -3\).
This formula allows us to calculate the roots \(m = -\frac{1}{2}\) and \(m = \frac{3}{2}\), which are essential for forming the general solution.
  • The quadratic formula is versatile and can solve any quadratic equation.
  • It systematically handles cases with varied discriminants (\(b^2 - 4ac\)).
Understanding how to use the quadratic formula simplifies finding roots, especially when a characteristic equation arises in a differential equation context.
Initial Conditions
Initial conditions are the values used to convert a general solution into a specific one. They are vital in making mathematical models relevant to a particular real-world situation.
In our case, the initial conditions are \(y(0) = 1\) and \(y'(0) = 5\). By applying these conditions to the general solution: \[y = c_1 e^{-x/2} + c_2 e^{3x/2}\]we obtain specific values for \(c_1\) and \(c_2\).
This involves solving a system of linear equations: \(c_1 + c_2 = 1\) and \[-\frac{1}{2}c_1 + \frac{3}{2}c_2 = 5\].
  • Initial conditions are typically given at \(x=0\), making calculations straightforward.
  • Solving these equations enables us to identify particular values for constants in the solution.
Setting these conditions ensures the mathematical solution accurately describes the specific problem context.
Exponential Solutions
Exponential solutions are a common outcome when solving second-order linear differential equations with constant coefficients. Recognizing their form can simplify understanding and applying solutions.
In our specific problem, once we find the roots of the characteristic equation, we construct the general solution using a combination of exponential functions: \[y = c_1 e^{-x/2} + c_2 e^{3x/2}\]Here, each root translates into an exponential term where \(m = -\frac{1}{2}\) and \(m = \frac{3}{2}\), reflecting different growth or decay rates.
  • These solutions are characteristic because they inherently solve the original differential equation.
  • In the context of real-world phenomena, they might represent processes with constant proportionality, such as population growth or cooling processes.
Exponential solutions provide significant insights into the behavior of systems modeled by differential equations.

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Most popular questions from this chapter

From \(k_{1}=40\) and \(k_{2}=120\) we compute the effective spring constant \(k=4(40)(120) / 160=120 .\) Now, \(m=20 / 32\) so \(k / m=120(32) / 20=192\) and \(x^{\prime \prime}+192 x=0 . \quad\) Using \(x(0)=0\) and \(x^{\prime}(0)=2\) we obtain \(x(t)=\frac{\sqrt{3}}{12} \sin 8 \sqrt{3} t\).

Solving \(\frac{1}{2} q^{\prime \prime}+20 q^{\prime}+1000 q=0\) we obtain \(q_{c}(t)=e^{-20 t}\left(c_{1} \cos 40 t+c_{2} \sin 40 t\right) .\) The steady- state charge has the form \(q_{p}(t)=A \sin 60 t+B \cos 60 t+C \sin 40 t+D \cos 40 t .\) Substituting into the differential equation we find $$\begin{aligned}(-1600 A-2400 B) & \sin 60 t+(2400 A-1600 B) \cos 60 t \\ +&(400 C-1600 D) \sin 40 t+(1600 C+400 D) \cos 40 t\end{aligned}$$ $$=200 \sin 60 t+400 \cos 40 t$$ Equating coefficients we obtain \(A=-1 / 26, B=-3 / 52, C=4 / 17,\) and \(D=1 / 17 .\) The steady-state charge is \\[q_{p}(t)=-\frac{1}{26} \sin 60 t-\frac{3}{52} \cos 60 t+\frac{4}{17} \sin 40 t+\frac{1}{17} \cos 40 t\\] and the steady-state current is \\[i_{p}(t)=-\frac{30}{13} \cos 60 t+\frac{45}{13} \sin 60 t+\frac{160}{17} \cos 40 t-\frac{40}{17} \sin 40 t\\].

Solving \(\frac{1}{20} q^{\prime \prime}+2 q^{\prime}+100 q=0\) we obtain \(q(t)=e^{-20 t}\left(c_{1} \cos 40 t+c_{2} \sin 40 t\right) .\) The initial conditions \(q(0)=5\) and \(q^{\prime}(0)=0\) imply \(c_{1}=5\) and \(c_{2}=5 / 2 .\) Thus \\[q(t)=e^{-20 t}\left(5 \cos 40 t+\frac{5}{2} \sin 40 t\right)=\sqrt{25+25 / 4} e^{-20 t} \sin (40 t+1.1071)\\] and \(q(0.01) \approx 4.5676\) coulombs. The charge is zero for the first time when \(40 t+1.1071=\pi\) or \(t \approx\) 0.0509 second.

$$\text { since } x=\frac{\sqrt{85}}{4} \sin (4 t-0.219)-\frac{\sqrt{17}}{2} e^{-2 t} \sin (4 t-2.897), \text { the amplitude approaches } \sqrt{85} / 4 \text { as } t \rightarrow \infty$$.

Applying integration by parts twice we have $$\begin{aligned} \int e^{a x} f(x) d x &=\frac{1}{a} e^{a x} f(x)-\frac{1}{a} \int e^{a x} f^{\prime}(x) d x \\ &=\frac{1}{a} e^{a x} f(x)-\frac{1}{a}\left[\frac{1}{a} e^{a x} f^{\prime}(x)-\frac{1}{a} \int e^{a x} f^{\prime \prime}(x) d x\right] \\ &=\frac{1}{a} e^{a x} f(x)-\frac{1}{a^{2}} e^{a x} f^{\prime}(x)+\frac{1}{a^{2}} \int e^{a x} f^{\prime \prime}(x) d x \end{aligned}$$ Collecting the integrals we get $$\int e^{a x}\left(f(x)-\frac{1}{a^{2}} f^{\prime \prime}(x)\right) d x=\frac{1}{a} e^{a x} f(x)-\frac{1}{a^{2}} e^{a x} f^{\prime}(x)$$ In order for the technique to work we need to have $$\begin{aligned} &\int e^{a x}\left(f(x)-\frac{1}{a^{2}} f^{\prime \prime}(x)\right) d x=k \int e^{a x} f(x) d x\\\ &f(x)-\frac{1}{a^{2}} f^{\prime \prime}(x)=k f(x) \end{aligned}$$ where \(k \neq 0 .\) This is the second-order differential equation \(f^{\prime \prime}(x)+a^{2}(k-1) f(x)=0\) If \(k<1, k \neq 0,\) the solution of the differential equation is a pair of exponential functions, in which case the original integrand is an exponential function and does not require integration by parts for its evaluation. Similarly, if \(k=1, f^{\prime \prime}(x)=0\) and \(f(x)\) has the form \(f(x)=a x+b .\) In this case a single application of integration by parts will suffice. Finally, if \(k>1\), the solution of the differential equation is $$f(x)=c_{1} \cos a \sqrt{k-1} x+c_{2} \sin a \sqrt{k-1} x$$ and we see that the technique will work for linear combinations of \(\cos \alpha x\) and \(\sin \alpha x\)
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