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Chapter 6: Problem 8

Find the solution of the differential equation that satisfies the given boundary condition(s). $$x^{\prime \prime}+4 x^{\prime}+4 x=0, x(0)=1, x^{\prime}(0)=1$$

Short Answer

Expert verified
The solution is \(x(t) = (1 + 3t)e^{-2t}\).

Step by step solution

01

Identify the Type of Differential Equation

The given equation is a second-order linear homogeneous differential equation with constant coefficients: \( x'' + 4x' + 4x = 0 \).
02

Write the Characteristic Equation

For homogeneous linear differential equations of the form \(a x'' + b x' + c x = 0\), we can write the characteristic equation as \(ar^2 + br + c = 0\). For this equation, it becomes \(r^2 + 4r + 4 = 0\).
03

Solve the Characteristic Equation

The characteristic equation \(r^2 + 4r + 4 = 0\) can be solved using the quadratic formula: \(r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Here, \(a = 1\), \(b = 4\), and \(c = 4\). Substitute these values to get: \[ r = \frac{-4 \pm \sqrt{16 - 16}}{2} = \frac{-4}{2} = -2. \] Thus, we have a repeated root \(r = -2\).
04

Write the General Solution

The general solution for a differential equation with a repeated root \(r\) is given by \(x(t) = (C_1 + C_2 t)e^{rt}\). Since our repeated root is \(r = -2\), the general solution becomes \(x(t) = (C_1 + C_2 t)e^{-2t}\).
05

Apply Initial Conditions

Use the initial conditions to find the constants. We have:1. \( x(0) = 1 \Rightarrow (C_1 + C_2 \cdot 0)e^{0} = 1 \Rightarrow C_1 = 1 \).2. Find \(x'(t)\), the derivative of \((C_1 + C_2 t)e^{-2t}\), which is \(x'(t) = -2(C_1 + C_2 t)e^{-2t} + C_2 e^{-2t} = (-2C_1 - 2C_2 t + C_2)e^{-2t}\). Given the initial condition \(x'(0) = 1\), we substitute to get:\((-2C_1 + C_2)e^{0} = 1\), which simplifies to \(-2(1) + C_2 = 1 \Rightarrow C_2 = 3\).
06

Write the Particular Solution

Substituting \(C_1 = 1\) and \(C_2 = 3\) into the general solution gives the particular solution: \(x(t) = (1 + 3t)e^{-2t}\).

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Key Concepts

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

Second-order Linear Differential Equations
Understandably, second-order linear differential equations might sound a bit daunting. But they're just equations that help us understand how functions and their rates of change behave. They're called 'second-order' because they involve the second derivative, which means how the rate of change of a quantity is itself changing. In our example equation, we have:
  • The second derivative, \(x''\)
  • The first derivative, \(x'\)
  • The original function, \(x\)
These equations often describe natural phenomena, such as motion under gravity. The word 'linear' implies that each term in our equation is either the function itself or its derivative, with no powers or products of these terms. This makes them a bit simpler to handle, as compared to non-linear equations.
Characteristic Equation
The characteristic equation is a handy tool that transforms a differential equation into an algebraic one, which is typically easier to solve. For any second-order linear homogeneous differential equation with constant coefficients, like \[ a x'' + b x' + c x = 0 \] we can create a characteristic equation by replacing each derivative with powers of \(r\) to form: \[ ar^2 + br + c = 0 \]
In our specific case, with \(x'' + 4x' + 4x = 0\), the characteristic equation becomes \(r^2 + 4r + 4 = 0\). Solving this quadratic equation helps us identify the type of solutions we can expect, which, in the case of repeated roots, indicates solutions involve exponential terms multiplied by polynomials.
Initial Conditions
Initial conditions are the cherry on top of solving differential equations. They give us specific values for the function or its derivatives at a certain point, allowing us to find specific solutions rather than a broad set of possibilities. In simple terms, they fine-tune our general solution to fit the particular problem at hand. For the exercise, we had initial conditions \(x(0) = 1\) and \(x'(0) = 1\). This means:
  • When \(t = 0\), the function \(x\) should equal 1.
  • The rate of change of \(x\) (\(x'\)) should also be 1 at \(t = 0\).
Using these, we substitute back into our general solution and tweak the constants to obtain the particular answer \((1 + 3t)e^{-2t}\). These constants, found using initial conditions, ensure the solution is tailor-made for the specific problem scenario.

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Most popular questions from this chapter

Find the solution of the differential equation that satisfies the given boundary condition(s). $$f^{\prime \prime}-2 f^{\prime}+5 f=0, f(0)=1, f(\pi / 4)=0$$

Let \(\\{\mathbf{u}, \mathbf{v}, \mathbf{w}\\}\) be a linearly independent set of vectors in a vector space \(V\) (a) \(\mathrm{Is}\\{\mathbf{u}+\mathbf{v}, \mathbf{v}+\mathbf{w}, \mathbf{u}+\mathbf{w}\\}\) linearly independent? Either prove that it is or give a counterexample to show that it is not. (b) \(\mathrm{Is}\\{\mathbf{u}-\mathbf{v}, \mathbf{v}-\mathbf{w}, \mathbf{u}-\mathbf{w}\\}\) linearly independent? Either prove that it is or give a counterexample to show that it is not.

If \(f\) and \(g\) are in \(\mathscr{C}^{(1)}\), the vector space of all functions with continuous derivatives, then the determinant \\[ W(x)=\left|\begin{array}{ll} f(x) & g(x) \\ f^{\prime}(x) & g^{\prime}(x) \end{array}\right| \\] is called the Wronskian of \(f\) and \(g\) [named after the Polish-French mathematician J贸sef Maria Ho毛n茅Wronski \((1776-1853),\) who worked on the theory of determinants and the philosophy of mathematics] Show that \(f\) and \(g\) are linearly independent if their Wronskian is not identically zero (that is, if there is some \(x\) such that \(W(x) \neq 0\) ).

Determine whether the set \(\mathcal{B}\) is a basis for the vector space \(V\) $$\begin{array}{l} V=M_{22} \\ B=\left\\{\left[\begin{array}{ll} 1 & 2 \\ 2 & 1 \end{array}\right],\left[\begin{array}{ll} 2 & 1 \\ 1 & 2 \end{array}\right],\left[\begin{array}{rr} 1 & 3 \\ -3 & 1 \end{array}\right],\left[\begin{array}{ll} 2 & 3 \\ 3 & 1 \end{array}\right],\left[\begin{array}{ll} 1 & 2 \\ 3 & 2 \end{array}\right]\right\\} \end{array}$$

Let \(T: V \rightarrow W\) be a linear transformation between finite-dimensional vector spaces \(V\) and \(W\). Let \(B\) and \(C\) be bases for \(V\) and \(W,\) respectively, and let \(A=[T]_{c+B}\) If \(V=W\) and \(B=C,\) show that \(T\) is diagonalizable if and only if \(A\) is diagonalizable.
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