Question

Consider the initial value problem $$ 5 u^{n}+2 u^{\prime}+7 u=0, \quad u(0)=2, \quad u^{\prime}(0)=1 $$ (a) Find the solution \(u(t)\) of this problem. (b) Find the smallest \(T\) such that \(|u(t)| \leq 0.1\) for all \(t>T\)

Short answer

Answer: The smallest T such that |u(t)| <= 0.1 for all t>T is approximately 19.78.

Step-by-step solution

Solve homogeneous linear differential equation

To solve the homogeneous linear differential equation \(5u^n + 2u' + 7u = 0\), first rewrite it into its standard form: $$ u'' + \frac{2}{5}u' + \frac{7}{5}u = 0. $$ Then we try to find solutions of the form \(u(t) = e^{\lambda t}\) for some \(\lambda\) by substitution. We get the characteristic equation: $$ \lambda^2 + \frac{2}{5}\lambda + \frac{7}{5} = 0. $$ To solve this quadratic equation, we use the quadratic formula: $$ \lambda = \frac{-\frac{2}{5} \pm \sqrt{(\frac{2}{5})^2 - 4(\frac{7}{5})}}{2}. $$

Find general solutions and particular solution with initial conditions

Computing the values of the characteristic equation, we find that the roots are complex conjugate and in the form \(\lambda = a \pm bi\), where \(a = -\frac{1}{5}\) and \(b = \sqrt{\frac{27}{25}}\). So the general solution is: $$ u(t) = e^{-\frac{1}{5}t}(c_1\cos(\sqrt{\frac{27}{25}}t) + c_2\sin(\sqrt{\frac{27}{25}}t)), $$ where \(c_1\) and \(c_2\) are constants that need to be determined using the initial conditions. Initial condition 1: \(u(0) = 2\) $$ 2 = e^0(c_1\cos(0) + c_2\sin(0)) = c_1 \implies c_1 = 2. $$ Initial condition 2: \(u'(0) = 1\) First, compute \(u'(t)\) by differentiating \(u(t)\): $$ u'(t) = e^{-\frac{1}{5}t}(-\frac{1}{5}c_1\cos(\sqrt{\frac{27}{25}}t)-\frac{1}{5}c_2\sin(\sqrt{\frac{27}{25}}t)+c_1\sqrt{\frac{27}{25}}\sin(\sqrt{\frac{27}{25}}t)-c_2\sqrt{\frac{27}{25}}\cos(\sqrt{\frac{27}{25}}t)). $$ Now, substitute the value of \(c_1\), and evaluate \(u'(0) = 1\): $$ 1 = e^0(-\frac{1}{5}(2)\cos(0)-\frac{1}{5}c_2\sin(0)+2\sqrt{\frac{27}{25}}\sin(0)-c_2\sqrt{\frac{27}{25}}\cos(0)) \implies c_2 = -\frac{5}{27}. $$ Thus, the particular solution satisfying the initial conditions is: $$ u(t) = e^{-\frac{1}{5}t}(2\cos(\sqrt{\frac{27}{25}}t) - \frac{5}{27}\sin(\sqrt{\frac{27}{25}}t)). $$

Find the smallest T such that |u(t)| T

We have now obtained the solution: $$ u(t) = e^{-\frac{1}{5}t}(2\cos(\sqrt{\frac{27}{25}}t) - \frac{5}{27}\sin(\sqrt{\frac{27}{25}}t)). $$ To find the smallest \(T\) such that \(|u(t)| \leq 0.1\) for all \(t>T\), we need to find the smallest \(t\) when the amplitude of the oscillation goes below 0.1. Since \(e^{-\frac{1}{5}t}\) is a decreasing function of \(t\), and both cosine and sine terms take on values between -1 and 1, we can derive an inequality for the amplitude by multiplying the exponential term by -1 and 1: $$ -0.1 \leq e^{-\frac{1}{5}t}(2\cos(\sqrt{\frac{27}{25}}t) - \frac{5}{27}\sin(\sqrt{\frac{27}{25}}t)) \leq 0.1. $$ Note that the maximum value for the amplitude is 2 + abs(-\(\frac{5}{27}\)) = \(\frac{61}{27}\), which is attained when t=0 ( i.e., at the initial conditions). Now, to find the smallest \(T\) such that \(|u(t)| \leq 0.1\) for all \(t>T\), we can set the maximum amplitude times the exponential term to be equal to the threshold: $$ \frac{61}{27} e^{-\frac{1}{5}T} = 0.1. $$ Now, we can solve for \(T\): $$ T = -\frac{5}{1}\ln(\frac{27}{610}) \approx 19.78. $$ Hence, the smallest \(T\) such that \(|u(t)| \leq 0.1\) for all \(t>T\) is approximately 19.78.

Key concepts

Understanding Homogeneous Linear Differential Equations

Homogeneous linear differential equations are a core concept in calculus and appear frequently in various fields of science and engineering, like physics, economics, and biology. Essentially, a differential equation is an equation that involves derivatives of a function. A homogeneous linear differential equation of the second order can generally be expressed in the form \[ a u'' + bu' + cu = 0, \] where \(a\), \(b\), and \(c\) are coefficients and can be functions of the independent variable, but not of \(u\) or its derivatives.

The 'homogeneous' term means that the equation is equal to zero. If the equation had an additional term that is not a multiple of \(u(t)\) or its derivatives, it would be considered non-homogeneous. The term 'linear' refers to the property that both \(u\) and its derivatives are to the first power and not multiplied by each other. In solving homogeneous linear differential equations, we typically look for solutions that satisfy certain conditions, which are often described as initial value problems or boundary value problems.

In our exercise, we explored an initial value problem with given conditions at \(t=0\). This setup helps us determine specific constants in the general solution and find a particular solution that fits the real-world scenario being modeled.

The Characteristic Equation Simplified

When dealing with linear homogeneous differential equations with constant coefficients, we employ a powerful technique that involves the characteristic equation. The characteristic equation is a polynomial whose roots are used to construct the general solution to the differential equation. In order to solve for the characteristic equation, we often assume that the solution takes the form \( u(t) = e^{\lambda t} \), where \(\lambda\) represents an unknown constant that will be determined.

By plugging this assumed solution into the differential equation, we end up with a polynomial in terms of \(\lambda\). The solutions to this polynomial, i.e., the values of \(\lambda\), tell us about the behavior of the differential equation's solution. If \(\lambda\) is real and distinct, we will have exponentially growing or decaying solutions. If \(\lambda\) is a repeated root, we need to modify our general solution to account for this, typically by multiplying part of the solution by \(t\). However, if \(\lambda\) is complex, as seen in the exercise, our solutions will involve oscillations, characterized by sine and cosine functions, modulated by an exponential factor that dictates the amplitude's growth or decay over time.

Interpreting Complex Conjugate Roots

When working with characteristic equations, it's not uncommon to encounter complex conjugate roots. These roots come in pairs, \(\lambda = a \pm bi\), where \(i\) is the imaginary unit \(i = \sqrt{-1}\), and \(a\) and \(b\) are real numbers. The presence of complex roots indicates that the solution to the differential equation will involve trigonometric functions.

In the step-by-step solution to our exercise, after finding the complex conjugate roots, we used them to write down the general solution involving \(e^{at}\), \(\cos(bt)\), and \(\sin(bt)\). The role of these functions is quite interesting: while the exponential term represents the envelope of the solution, affecting how the amplitude changes over time, the trigonometric functions reflect the oscillatory nature of the solution.

Hence, the occurrence of complex conjugate roots shifts our solution approach towards exponential-decay-modulated oscillations. Such solutions are essential when dealing with systems that exhibit vibrations or waves, as they precisely describe the behavior of such phenomena over time. In the final answer to our initial value problem, we determined the specific coefficients that define the unique behavior of our system, thus giving us a complete understanding of the solution's dynamics.