Question

Consider the initial value problem $$ 3 u^{\prime \prime}-u^{\prime}+2 u=0, \quad u(0)=2, \quad u^{\prime}(0)=0 $$ (a) Find the solution \(u(t)\) of this problem. (b) Find the first time at which \(|u(t)|=10\).

Short answer

Question: Find the exact solution u(t) for the given initial value problem \(3u'' - u' + 2u = 0\), with \(u(0) = 2\) and \(u'(0) = 0\). Then, determine the first time at which the absolute value of u(t) equals 10. Provide your answer in terms of the exact solution u(t) and the method used to determine the first time at which the absolute value of u(t) equals 10. Answer: The exact solution for u(t) is \(u(t) = 2e^{\frac{1}{6}t}\cos(\frac{\sqrt{23}}{6}t) + \frac{\sqrt{23}}{6}e^{\frac{1}{6}t}\sin(\frac{\sqrt{23}}{6}t)\). To determine the first time at which the absolute value of u(t) equals 10, we need to solve the equation \(|2e^{\frac{1}{6}t}\cos(\frac{\sqrt{23}}{6}t) + \frac{\sqrt{23}}{6}e^{\frac{1}{6}t}\sin(\frac{\sqrt{23}}{6}t)| = 10\) using numerical methods, such as Newton's method or numerical software.

Step-by-step solution

Write the given differential equation in standard form

We have the given initial value problem: $$ 3u'' - u' + 2u = 0, \quad u(0) = 2, \quad u'(0) = 0 $$

Find two linearly independent solutions using the characteristic equation method

For this step, we propose a solution \(u(t)=e^{rt}\) and substitute it into our differential equation. Doing this, we get the characteristic equation: $$ 3r^2-r+2=0 $$ To solve for r, we write down the discriminant and solve the quadratic equation: $$ \Delta = (-1)^2 - 4*3*2 = 1 - 24 = -23 $$ Since the discriminant is negative, we have complex roots for r: $$ r_{1,2}=\frac{-b \pm \sqrt{\Delta}}{2a}=\frac{1 \pm i\sqrt{23}}{6} $$ Thus, two linearly independent solutions are given by: $$ u_1(t)=e^{\frac{1}{6}t} \cos (\frac{\sqrt{23}}{6}t), \quad u_2(t)=e^{\frac{1}{6}t} \sin (\frac{\sqrt{23}}{6}t) $$

Apply the initial values to solve for the constants of the general solution

The general solution is given by: $$ u(t)=C_1e^{\frac{1}{6}t}\cos(\frac{\sqrt{23}}{6}t) + C_2e^{\frac{1}{6}t}\sin(\frac{\sqrt{23}}{6}t) $$ Using the initial conditions, we have: $$ u(0)=C_1=2 $$ Now, finding \(u'(t)\), we have: $$ u'(t)= \frac{1}{6}C_1 e^{\frac{1}{6}t}\cos(\frac{\sqrt{23}}{6}t) - \frac{\sqrt{23}}{6} C_1 e^{\frac{1}{6}t} \sin(\frac{\sqrt{23}}{6}t) + \frac{1}{6} C_2 e^{\frac{1}{6}t}\sin(\frac{\sqrt{23}}{6}t)+\frac{\sqrt{23}}{6} C_2 e^{\frac{1}{6}t}\cos(\frac{\sqrt{23}}{6}t) $$ When applying the initial condition \(u'(0)=0\), we get \(C_2=\frac{\sqrt{23}}{6}\).

Find the exact solution u(t) using the constants found in step 3

With the found values for \(C_1\) and \(C_2\), the solution is given by: $$ u(t)=2e^{\frac{1}{6}t}\cos(\frac{\sqrt{23}}{6}t) + \frac{\sqrt{23}}{6}e^{\frac{1}{6}t}\sin(\frac{\sqrt{23}}{6}t) $$

Determine the first time at which the absolute value of u(t) equals 10.

To determine the first time at which \(|u(t)|=10\), we need to solve for t when: $$ |2e^{\frac{1}{6}t}\cos(\frac{\sqrt{23}}{6}t) + \frac{\sqrt{23}}{6}e^{\frac{1}{6}t}\sin(\frac{\sqrt{23}}{6}t)| = 10 $$ This is a transcendental equation, which cannot be solved analytically. However, it can be solved using numerical methods, such as Newton's method or using numerical software to approximate the solution. This would allow us to find the first time when the absolute value of the solution indeed equals 10.

Key concepts

Differential Equations

Understanding differential equations is essential for various fields such as physics, engineering, and economics because they describe how variables change over time or space. A differential equation is a mathematical equation that relates some function with its derivatives. In simple terms, it represents the rate at which a quantity changes.

For the initial value problem in this exercise, we're asked to solve a second order linear homogeneous differential equation with constant coefficients. The equation has an important form, given by: \( ay'' + by' + cy = 0 \). The equation describes the behavior of a function \( u(t) \) over time, with the second derivative \( u''(t) \) representing acceleration, the first derivative \( u'(t) \) signifying velocity, and the function itself \( u(t) \) denoting position. The initial conditions provide us specific values at time \( t = 0 \), which allows for the precise determination of the solution to the equation.

Characteristic Equation Method

The characteristic equation method is a powerful technique used to solve linear homogeneous differential equations with constant coefficients, like the one presented in our exercise. The characteristic equation is derived by assuming a solution of the form \(u(t) = e^{rt}\), where \( r \) is a constant that needs to be determined. Substituting this form into the differential equation transforms the problem into an algebraic one.

By calculating the characteristic equation \(3r^2 - r + 2 = 0\) and finding its roots, we can determine the nature of the solutions. A key aspect to consider is the discriminant \(\Delta\), which indicates whether the roots are real and distinct, real and repeated, or complex. In the case of complex roots, as we see with \(\Delta = -23\), the solutions involve both exponential and trigonometric functions, reflecting oscillatory behaviors.

Complex Roots

Complex roots arise in the characteristic equation when the discriminant \( \Delta \) is negative, leading to an expression for \( r \) that involves the square root of a negative number. As seen in the exercise, the discriminant of \( \Delta = -23 \) tells us that the characteristic equation does not have real-number solutions. Instead, we obtain a pair of complex conjugates.

When the characteristic equation yields complex roots of the form \( \alpha \pm i\beta \), where \( i \) is the imaginary unit, the general solution to the differential equation involves both exponential and trigonometric functions, specifically sines and cosines.

The Power of Euler's Formula

Euler's formula states that \(e^{i\theta} = \cos(\theta) + i\sin(\theta)\), which explains the appearance of \( \cos \) and \( \sin \) terms in our solutions. These functions represent oscillatory motion, characteristic of systems with complex eigenvalues, such as damped mechanical oscillators or electrical circuits with alternating current.