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Chapter 4: Problem 14

\(4 y^{\prime \prime}+4 y^{\prime}+y=0\)

Short Answer

Expert verified
The general solution is \( y(t) = (C_1 + C_2 t)e^{-\frac{1}{2}t} \).

Step by step solution

01

- Understanding the Differential Equation

The given differential equation is a second-order linear homogeneous differential equation with constant coefficients: \[ 4y'' + 4y' + y = 0 \]
02

- Characteristic Equation

To solve the differential equation, start by finding the characteristic equation. Replace \(y''\) with \(r^2\), \(y'\) with \(r\), and \(y\) with 1:\[ 4r^2 + 4r + 1 = 0 \]
03

- Solving the Characteristic Equation

Solve the characteristic equation for \(r\):\[ 4r^2 + 4r + 1 = 0 \]Use the quadratic formula \(r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 4\), \(b = 4\), and \(c = 1\):\[ r = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 4 \cdot 1}}{2 \cdot 4} = \frac{-4 \pm \sqrt{16 - 16}}{8} = \frac{-4 \pm 0}{8} = -\frac{1}{2} \]The double root is \(r = -\frac{1}{2}\).
04

- Form the General Solution

Since the characteristic root \(r = -\frac{1}{2}\) is a repeated root, the general solution is:\[ y(t) = (C_1 + C_2 t)e^{-\frac{1}{2}t} \]where \(C_1\) and \(C_2\) are constants.

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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 crucial in solving second-order linear homogeneous differential equations. It translates the differential equation into a polynomial equation. For the given equation:
\[ 4 y^{\prime \prime} + 4 y^{\prime} + y = 0 \], replace derivatives:
  • \( y'' \rightarrow r^2 \)
  • \( y' \rightarrow r \)
  • \( y \rightarrow 1 \)
Yielding the characteristic equation: \[ 4r^2 + 4r + 1 = 0 \]. This equation helps to identify the nature of the roots (real, complex, or repeated).
Quadratic Formula
To solve the characteristic equation, we use the quadratic formula: \[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, coefficients are:
  • \( a = 4 \)
  • \( b = 4 \)
  • \( c = 1 \)
Plugging the values into the formula:
\[ r = \frac{-4 \pm \sqrt{16 - 16}}{8} = \frac{-4 \pm 0}{8} = - \frac{1}{2} \] Results in a repeated root \( r = -\frac{1}{2} \).
Repeated Roots
Repeated roots occur when the discriminant \( b^2 - 4ac \) equals zero. For our equation:
\[ b^2 - 4ac = 4^2 - 4\cdot4\cdot1 = 16 - 16 = 0 \]. Which confirms repeated roots. When roots are repeated, the general solution must account for this by including a term to avoid redundancy. If \( r \) is the repeated root, the solution forms:
\[ y(t) = (C_1 + C_2 t) e^{rt} \]. This ensures both independent solutions are captured.
General Solution
For the differential equation provided, the repeated root is \( r = -\frac{1}{2} \). Employ the general solution format for repeated roots: \[ y(t) = (C_1 + C_2 t) e^{rt} \]. Plugging in our root, we get the particular form:
\[ y(t) = (C_1 + C_2 t) e^{-\frac{1}{2} t} \]. Here, \( C_1 \) and \( C_2 \) are constants determined by initial conditions. This process confirms that our general solution fits the structure needed to solve the original differential equation.

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Most popular questions from this chapter

Use the change of variables \(y=v(x) x^{-1 / 2}\) to transform Bessel's equation \(x^{2} y^{\prime \prime}+x y^{\prime \prime}+\) \(\left(x^{2}-k^{2}\right) y=0\) into the equation \(v^{\prime \prime}+\) \(\left[1+\left(\frac{1}{4}-k^{2}\right) x^{-2}\right] v=0\). By substituting \(k=1 / 2\) into the transformed equation, derive the solution to Bessel's equation with \(k=1 / 2\).

\(4 x^{2} y^{\prime \prime}+y=x^{3}, y(1)=1, y^{\prime}(1)=-1\)

\(y^{(4)}-16 y=1\)

\(x^{3} y^{\prime \prime \prime}+22 x^{2} y^{\prime \prime}+124 x y^{\prime}+140 y=0\)

(a) Show that the transformed equation for the Cauchy-Euler equation \(a x^{2} y^{\prime \prime}+b x y^{\prime}+\) \(c y=0\) is \(a d^{2} y / d t^{2}+(b-a) d y / d t+c y=\) 0 . (b) Show that if the characteristic equation for a transformed Cauchy-Euler equation has a repeated root, then this root is \(r=-(b-a) /(2 a)\). (c) Use reduction of order to show that a second solution to this equation is \(x^{r} \ln x\). (d) Calculate the Wronskian of \(x^{r}\) and \(x^{r} \ln x\) to show that these two solutions are linearly independent.
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