Question

find the solution of the given initial value problem. Sketch the graph of the solution and describe its behavior for increasing\(t.\) $$ y^{\prime \prime}+4 y^{\prime}+5 y=0, \quad y(0)=1, \quad y^{\prime}(0)=0 $$

Short answer

Answer: As time (t) increases, the solution oscillates and decays in amplitude, decreasing towards zero.

Step-by-step solution

Form the characteristic equation

In order to find the complementary solution, we must construct the characteristic equation of the given differential equation. We replace \(y''\) with \(m^2\), \(y'\) with \(m\), and \(y\) with \(1\) to form the following equation: $$ m^2 + 4m +5 = 0 $$

Solve the characteristic equation

Now, we need to find the roots of the characteristic equation. Since this is a quadratic equation, we can use the quadratic formula: $$ m = \frac{-b\pm \sqrt{b^2-4ac}}{2a} $$ Plugging in the values \(a=1\), \(b=4\), and \(c=5\): $$ m = \frac{-4\pm \sqrt{4^2-4\cdot 1\cdot 5}}{2\cdot 1} $$ $$ m = \frac{-4\pm \sqrt{16-20}}{2} $$ $$ m = \frac{-4\pm \sqrt{-4}}{2} $$ Since we have a negative value inside the square root, the roots will be complex. Let's rewrite the roots as follows: $$ m_1 = -2 + i,\quad m_2 = -2 - i $$

Write the complementary solution

Since the roots of the characteristic equation are complex, the complementary solution has the following form: $$ y_c(t) = e^{-2t}(A\cos t + B\sin t) $$ where \(A\) and \(B\) are constants that we need to find using the given initial conditions.

Apply the initial conditions and find the constants

We are given the initial conditions \(y(0)=1\) and \(y'(0)=0\). Let's apply these conditions to find the constants \(A\) and \(B\). 1. Apply \(y(0)=1\): $$ y_c(0) = e^{-2\cdot 0}(A\cos 0 + B\sin 0) \\ 1 = (A\cdot 1 + B\cdot 0) \\ A = 1 $$ 2. Apply \(y'(0) = 0\). First, we need to find the derivative of \(y_c\): $$ y_c'(t) = \frac{d}{dt}(e^{-2t}(A\cos t + B\sin t)) \\ y_c'(t) = -2e^{-2t}(A\cos t + B\sin t) + e^{-2t}(-A\sin t + B\cos t) $$ Then, apply the initial condition \(y'(0) = 0\): $$ 0 = -2e^{-2\cdot 0}(A\cos 0 + B\sin 0) + e^{-2\cdot 0}(-A\sin 0 + B\cos 0) \\ 0 = -2(A\cdot 1 + B\cdot 0) + (-A\cdot 0 + B\cdot 1) \\ 0 = -2A + B $$ Since \(A=1\), we can find \(B\): $$ B = 2A \\ B = 2 $$

Write the final solution and describe the behavior

Now that we have the constants \(A\) and \(B\), we can write the final solution of the IVP: $$ y(t) = e^{-2t}(\cos t + 2\sin t) $$ Since the exponential term \(e^{-2t}\) tends to zero as \(t\) approaches infinity, the solution will also tend to zero as \(t\) increases. The oscillatory behavior caused by the \(\sin\) and \(\cos\) terms suggests the solution will oscillate while decaying in amplitude as \(t\) increases.

Sketch the graph of the solution

To sketch the graph of the solution \(y(t)=e^{-2t}(\cos t + 2\sin t)\), use graphing software or plot the function manually, noting the oscillatory and decay behaviors described in step 5. The graph of the solution will show oscillations that decay in amplitude as \(t\) increases. The function starts at an initial value of 1 at \(t=0\), as given by the initial condition.

Key concepts

Characteristic Equation

The characteristic equation plays a pivotal role in solving linear homogeneous differential equations. It is essentially an algebraic equation derived from the ordinary differential equation (ODE) by replacing each term involving the function and its derivatives with a power of a variable, usually denoted as m. For example, if our ODE is of the form y'' + py' + qy = 0, the associated characteristic equation would be m² + pm + q = 0.

In the initial value problem (IVP) y'' + 4y' + 5y = 0, the characteristic equation is thus:
\[m^2 + 4m +5 = 0\].
This forms the foundation for identifying the behavior of the solution, as its roots dictate the form of the complementary solution. If the roots are real and distinct, we get exponential solutions. However, if they are complex, as they are in this case, our solution will have an oscillatory nature, shaped by sine and cosine functions.

Complementary Solution

The complementary solution, often denoted as y_c(t), refers to the part of the general solution to the homogeneous ODE that incorporates the characteristic roots. If the characteristic equation yields real roots, the complementary solution consists of exponential functions. However, when the roots are complex, as they often are in oscillatory systems, the complementary solution typically includes a combination of exponential, sine, and cosine functions.

Specifically, for complex roots of the form m = -a \(pm\) bi, the complementary solution takes the form e^{-at}(Acos(bt) + Bsin(bt)), where A and B are constants to be determined by initial conditions. In our exercise, since the roots are -2 \(pm\) i, the complementary solution is:
\[y_c(t) = e^{-2t}(A\text{cos}(t) + B\text{sin}(t))\].
The complementary solution reflects the homogeneous nature of the ODE and showcases the system's behavior not influenced by external forces or inputs.

Complex Roots

Complex roots arise from the characteristic equation when the discriminant, that is, b² - 4ac from the quadratic formula, is negative. This indicates that the quadratic equation does not intersect the x-axis and thus does not have real-valued roots. Instead, these roots are represented as complex numbers, typically in the form a \(pm\) bi, where i is the imaginary unit, the square root of -1.

Complex roots lead directly to oscillatory solutions when solving second-order linear homogeneous differential equations due to the sine and cosine terms they introduce into the complementary solution. As seen in our related ODE exercise, upon solving the characteristic equation, we get complex roots m1 = -2 + i and m2 = -2 - i. These roots will result in a solution that oscillates because the imaginary part of the complex roots corresponds to the frequency of oscillation in the sine and cosine functions.

Oscillatory Behavior

Oscillatory behavior in the context of differential equations refers to periodic fluctuations of the solution around a central value, often a result of complex characteristic roots. This means the system's state will rise and fall in a regular, repeating pattern. In mechanical systems, oscillatory behavior is akin to the motion of a spring or a pendulum.

In the case of the initial value problem discussed, the presence of a damping coefficient, indicated by the real part of the complex roots, causes the amplitude of the oscillation to decay exponentially over time. The solution y(t) = e^{-2t}(cos(t) + 2sin(t)), derived from these roots, shows such a damped oscillation. As time progresses, the exponential term e^{-2t} diminishes, reducing the impact of the oscillatory components represented by cos(t) and sin(t). This exponential decay indicates that the system will eventually stabilize as t approaches infinity, leading to the cessation of the oscillatory behavior.