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Chapter 4: Problem 25

\(y^{\prime \prime}+4 y^{\prime}+4 y=0, y(0)=1, y^{\prime}(0)=3\)

Short Answer

Expert verified
The solution is \( y(t) = (1 + 5t) e^{-2t} \).

Step by step solution

01

Identify the form of the differential equation

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

Write the characteristic equation

To solve the differential equation, first write the characteristic equation: \[r^2 + 4r + 4 = 0\]
03

Solve the characteristic equation

The characteristic equation can be factored as follows: \[(r + 2)^2 = 0\] This gives a repeated root at \[r = -2\]
04

Write the general solution

Because the characteristic equation has a repeated root, the general solution to the differential equation is: \[ y(t) = (C_1 + C_2 t) e^{-2t}\] where \(C_1\) and \(C_2\) are constants.
05

Apply the initial condition \(y(0) = 1\)

Substitute \(t = 0\) into the general solution and set it equal to 1: \[y(0) = C_1 e^{-2 \times 0} = C_1 = 1\] Thus, \(C_1 = 1\).
06

Apply the initial condition \(y'(0) = 3\)

First, compute the first derivative of the general solution: \[y'(t) = (C_2 e^{-2t} - 2 (C_1 + C_2 t) e^{-2t})\] Substitute \(t = 0\) into this derivative and set it equal to 3: \[y'(0) = C_2 - 2C_1 = 3\] Substituting \(C_1 = 1\), we get \[C_2 - 2 = 3\] Thus, \(C_2 = 5\).
07

Write the particular solution

Using \(C_1 = 1\) and \(C_2 = 5\), the particular solution to the differential equation is: \[ y(t) = (1 + 5t) e^{-2t}\]

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Key Concepts

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

Characteristic Equation
Second-order linear homogeneous differential equations have the form:

\bul y'' + ay' + by = 0\bul

To solve these, we transform the differential equation into a characteristic equation. For the given equation:

\[y'' + 4y' + 4y = 0\]

we replace y'', y', and y with terms involving r. Therefore, the characteristic equation becomes:

\[r^2 + 4r + 4 = 0\]

This transformation makes it easier to solve, as this is now a quadratic equation.
Repeated Roots
Next, we solve the characteristic equation. In our case, the characteristic equation \[ r^2 + 4r + 4 = 0 \]can be factored as \[ (r + 2)^2 = 0\]

This indicates we have a repeated root at: \[ r = -2 \]

When an equation has repeated roots, the solutions take a special form. Knowing how to handle repeated roots is key, because it changes the general solution structure.
General Solution
Given that we have repeated roots, we use the following form for our general solution:

\[ y(t) = (C_1 + C_2 t) e^{-2t} \]

Both constants, \C_1 and \C_2, are determined by initial conditions or boundary values. This form ensures that both parts of the solution remain linearly independent, as per the theory of differential equations.
Initial Conditions
The initial conditions provided are crucial for determining the values of constants in our general solution. In this problem:

\bul y(0) = 1 \bul and y'(0) = 3 \bul

First, we apply the initial condition y(0) = 1: \[ y(0) = C_1 e^{-2 \times 0} = C_1 = 1 \]

This gives C_1 = 1. Next, we apply the initial condition for the first derivative: \[y'(t) = (C_2 e^{-2t} - 2 (C_1 + C_2 t) e^{-2t})\]\[y'(0) = C_2 - 2C_1 = 3\]

Plugging in C_1 = 1, we solve for C_2: \[C_2 - 2 = 3\]\[ C_2 = 5 \]

Thus, the particular solution that satisfies both initial conditions is: \[ y(t) = (1 + 5t) e^{-2t} \].

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Most popular questions from this chapter

Use a computer algebra system to find a general solution of \(y-k^{4} \frac{d^{4} y}{d x^{4}}=0\). Show that the result you obtain is equivalent to $$ y=A e^{-x / k}+B e^{x / k}+C \sin \frac{x}{k}+D \cos \frac{x}{k} . $$ Find conditions on \(A, B, C\), and \(D\) (if possible) so that \(y\) has (a) neither local maxima nor local minima; (b) exactly one local maximum; and (c) exactly one local minimum on the interval \([0, \infty)\).

Laguerre's equation is given by $$ x y^{\prime \prime}+(1-x) y^{\prime}+k y=0, $$ where \(k\) is a constant (usually, it is assumed that \(k>0\) ). (a) Show that \(x=0\) is a regular singular point of Laguerre's equation. (b) Use the Method of Frobenius to determine one solution of Laguerre's equation. (c) Show that if \(k\) is a positive integer, then the solution is a polynomial. This polynomial, denoted \(L_{k}(x)\), is called the Laguerre polynomial of order \(k\).

Solve the initial value problem \(x^{3} y^{\prime \prime \prime}+\) \(9 x^{2} y^{\prime \prime}+44 x y^{\prime}+58 y=0, y(1)=2, y^{\prime}(1)=10\), \(y^{\prime \prime}(1)=-2\). Graph the solution on the interval \([0.2,1.8]\) and approximate all local minima and maxima of the solution on this interval.

A fundamental set of solutions for \(t y^{(4)}+\) \(2 y^{\prime \prime \prime}=0, t>0\), is \(S=\left\\{1, t, t \ln t, t^{2}\right\\}\). Use this information to solve \(t y^{(4)}+2 y^{\prime \prime \prime}=\frac{45}{8} t^{-7 / 2}\), \(y(1)=0, y^{\prime}(1)=0, y^{\prime \prime}(1)=1, y^{\prime \prime \prime}(1)=0\).

The hypergeometric equation is given by \(x(1-x) y^{\prime \prime}+[c-(a+b+1) x] y^{\prime}-a b y=0\), where \(a, b\), and \(c\) are constants. (a) Show that \(x=0\) and \(x=1\) are regular singular points. (b) Show that the roots of the indicial equation for the series \(\sum_{n=0}^{\infty} a_{n} x^{n+r}\) are \(r=0\) and \(r=1-c\). (c) Show that for \(r=0\), the solution obtained with the Method of Frobenius is $$ \begin{aligned} y_{1}=& 1+\frac{a b}{1 ! c} x+\frac{a(a+1) b(b+1)}{2 ! c(c+1)} x^{2} \\ &+\frac{a(a+1)(a+2) b(b+1)(b+2)}{3 ! c(c+1)(c+2)} x^{3} \\ &+\cdots \end{aligned} $$ where \(c \neq 0,-1,-2, \ldots\). This series is called the hypergeometric series. Its sum, denoted \(F(a, b, c ; x)\), is called the hypergeometric function. (d) Show that \(F(1, b, b ; x)\) \(=1 /(1-x)\). (e) Find the solution of the equation that corresponds to \(r=1-c\).
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