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Chapter 2: Problem 2

Find the solution \(\phi\) of the initial-value problem $$ y^{\prime \prime \prime}+y=0, \quad y(0)=0, \quad y^{\prime}(0)=1, \quad y^{\prime \prime}(0)=0 . $$

Short Answer

Expert verified
The solution is \(\phi(x) = \frac{3}{4} e^{-x} - \frac{3}{4} e^{\frac{1}{2}x}\cos\left(\frac{\sqrt{3}}{2}x\right) + \frac{2}{\sqrt{3}} e^{\frac{1}{2}x}\sin\left(\frac{\sqrt{3}}{2}x\right)\).

Step by step solution

01

Solve the Characteristic Equation

Convert the differential equation \( y''' + y = 0 \) into its characteristic equation by assuming a solution of the form \( y = e^{rx} \). This results in the characteristic equation \( r^3 + 1 = 0 \).
02

Find the Roots of the Characteristic Equation

Solve \( r^3 + 1 = 0 \) by factoring it as \( (r+1)(r^2 - r + 1) = 0 \). The roots are \( r = -1 \) and the complex roots from \( r^2 - r + 1 = 0 \) using the quadratic formula are \( r = \frac{1 \pm \sqrt{3}i}{2} \).
03

Write the General Solution

The general solution is a combination of the real and complex roots: \[ y(x) = C_1 e^{-x} + C_2 e^{\frac{1}{2}x} \cos\left(\frac{\sqrt{3}}{2}x\right) + C_3 e^{\frac{1}{2}x} \sin\left(\frac{\sqrt{3}}{2}x\right). \]
04

Apply Initial Conditions to Find Constants

Use the initial conditions to find the constants \( C_1, C_2, \text{and } C_3 \). Substituting \( y(0) = 0 \) gives \( C_1 + C_2 = 0 \), \( y'(0) = 1 \) implies \( -C_1 + \frac{1}{2}C_3 \cdot \sqrt{3} = 1 \), and \( y''(0) = 0 \) leads to \( C_1 - \frac{3}{4}C_2 = 0 \).
05

Solve the System of Equations for Constants

Solve the system:1. \( C_1 + C_2 = 0 \)2. \( -C_1 + \frac{1}{2} C_3 \sqrt{3} = 1 \)3. \( C_1 - \frac{3}{4}C_2 = 0 \)From these, we find that \( C_1 = \frac{3}{4}, C_2 = -\frac{3}{4}, C_3 = \frac{2}{\sqrt{3}} \).
06

Write the Specific Solution

Substitute the constants back into the general solution:\[\phi(x) = \frac{3}{4} e^{-x} - \frac{3}{4} e^{\frac{1}{2}x}\cos\left(\frac{\sqrt{3}}{2}x\right) + \frac{2}{\sqrt{3}} e^{\frac{1}{2}x}\sin\left(\frac{\sqrt{3}}{2}x\right).\]

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Key Concepts

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

Characteristic Equation
A characteristic equation is a key concept when solving linear differential equations. It helps convert a complex differential problem into an algebraic one.
To find this equation, we typically assume a solution of the form \( y = e^{rx} \). For our differential equation \( y''' + y = 0 \), substituting this form gives us \( r^3e^{rx} + e^{rx} = 0 \).
Since \( e^{rx} \) never equals zero, we simplify this to \( r^3 + 1 = 0 \) by factoring out \( e^{rx} \). Solving the characteristic equation is crucial, as it provides the basis for finding the general solution.
Initial-Value Problem
An initial-value problem (IVP) consists of a differential equation along with specified values at the beginning of the interval. These values are called initial conditions.
For our equation, we have the following initial conditions: \( y(0) = 0 \), \( y'(0) = 1 \), and \( y''(0) = 0 \). These conditions help us determine specific constants in the general solution.
Applying the initial conditions ensures that the solution we find matches the specific scenario given in the problem. Without them, we'd have a family of solutions rather than one unique solution.
Complex Roots
When solving the characteristic equation, we may encounter complex roots, which lead to trigonometric functions in the solution.
For \( r^3 + 1 = 0 \), we factor it to get \( (r+1)(r^2 - r + 1) = 0 \). The root \( r = -1 \) is real, and the equation \( r^2 - r + 1 = 0 \) gives two complex roots. Using the quadratic formula, we find these roots as \( r = \frac{1 \pm \sqrt{3}i}{2} \).
These complex roots contribute cosine and sine terms in the general solution, leading to oscillating behavior in the function鈥檚 graph.
General Solution
The general solution of a differential equation combines solutions that satisfy the characteristic equation.
In our case, we combine solutions from both real and complex roots: \( y(x) = C_1 e^{-x} + C_2 e^{\frac{1}{2}x} \cos\left(\frac{\sqrt{3}}{2}x\right) + C_3 e^{\frac{1}{2}x} \sin\left(\frac{\sqrt{3}}{2}x\right) \).
This expression represents the most general form of the solution without considering specific conditions. Each term corresponds to a root from the characteristic equation, and constants \( C_1, C_2, \) and \( C_3 \) will be adjusted using the initial conditions to find the specific solution unique to the problem.

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Most popular questions from this chapter

Show that every solution of the constant coefficient equation $$ y^{\prime \prime}+a_{1} y^{\prime}+a_{2} y=0 $$ tends to zero as \(x \rightarrow \infty\) if, and only if, the real parts of the roots of the characteristic polynomial are negative. (Note: In this case the solutions are often called transients.)

Show that \(\phi_{n}(x)=\sin n x\) satisfies the boundary value problem $$ y^{\prime \prime}+n^{2} y=0, \quad y(0)=0, \quad y(\pi)=0 $$ where \(n=1,2, \cdots\). (b) Using (a) show that $$ \int_{0}^{\pi} \sin n x \sin m x d x=0 $$ if \(n \neq m\). (Hint: See,Ex. 5 (a).) (c) Prove that for any positive integer \(n, \phi_{1}, \cdots, \phi_{n}\) are linearly independent on \(0 \leqq x \leqq \pi\). (Hint: Suppose \(a_{1} \phi_{1}+\cdots+a_{n} \phi_{n}=0 .\) Multiply both sides of this equality by \(\phi_{k}(k\) fixed between 1 and \(n\) ) and integrate from 0 to \(\pi\). Use (b).)

Let \(b\) be a continuous function on an interval \(I\), and let \(x_{0}\) be a fixed point in \(I\). Show that the \(\phi\) given by $$ \phi(x)=e^{a x} \int_{x_{0}}^{x} \frac{(x-t)^{k-1}}{(k-1) !} e^{-a t} b(t) d t $$ satisfies $$ (D-a)^{k}(\phi)=b $$ and $$ \phi\left(x_{0}\right)=\phi^{\prime}\left(x_{0}\right)=\cdots=\phi^{(k-1)}\left(x_{0}\right)=0 $$ Here \(a\) is a constant. (Hint: From Ex. 1 (b), $$ (D-a)^{k}(\phi)=e^{a x} D^{k}\left(e^{-a x} \phi\right) $$ To differentiate a function \(F\) of the form $$ F(x)=\int_{\alpha(x)}^{\beta(x)} f(x, t) d t $$ where \(\alpha, \beta, f\) are "nice" functions, use the formula $$ F^{\prime}(x)=f(x, \beta(x)) \beta^{\prime}(x)-f(x, \alpha(x)) \alpha^{\prime}(x)+\int_{\alpha(x)}^{\beta(x)} \frac{\partial f}{\partial x}(x, t) d t $$ A proof of this formula for appropriate \(\alpha, \beta, f\) may be found in most texts on advanced calculus.)

Suppose \(\phi_{1}, \phi_{2}\) are linearly independent solutions of the constant coefficient equation $$ y^{\prime \prime}+a_{1} y^{\prime}+a_{2} y=0, $$ and let \(W\left(\phi_{1}, \phi_{2}\right)\) be abbreviated to \(W\). Show that \(W\) is a constant if and only if \(a_{1}=0\). (Hint: Compute \(W^{\prime}\).)

Suppose the constants \(a_{1}, \cdots, a_{n}\) in $$ L(y)=y^{(n)}+a_{1} y^{(n-1)}+\cdots+a_{n} y $$ are all real. (a) Show that if \(\phi\) is a solution of \(L(y)=0\) then so are $$ \phi_{1}=\operatorname{Re} \phi \text { and } \phi_{2}=\operatorname{Im} \phi $$ (b) If \(\alpha+i \beta(\alpha, \beta\) real) is a root of the characteristic polynomial of \(L\) show that the functions $$ e^{\alpha x} \cos \beta x, \quad e^{\alpha x} \sin \beta x $$ are solutions of \(L(y)=0\).
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