Question

Find the solution of the given initial value problem. Sketch the graph of the solution and describe its behavior as \(t\) increases. $$ y^{\prime \prime}+5 y^{\prime}+3 y=0, \quad y(0)=1, \quad y^{\prime}(0)=0 $$

Short answer

Answer: As \(t\) increases, the particular solution decays smoothly to 0, approaching the horizontal axis.

Step-by-step solution

Formulate the characteristic equation

From the given ODE, we can create the following characteristic equation: $$ r^2 + 5r + 3 = 0, $$ where \(r\) are the roots of this quadratic equation.

Solve the characteristic equation

We will now find the roots of the characteristic equation using the quadratic formula: $$ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, $$ where \(a=1\), \(b=5\), and \(c=3\). Plugging the values, we get $$ r = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 1 \cdot 3}}{2 \cdot 1} = \frac{-5 \pm \sqrt{13}}{2}. $$ So, we have two roots: \(r_1=\frac{-5+\sqrt{13}}{2}\) and \(r_2=\frac{-5-\sqrt{13}}{2}\).

Write down the general solution of the ODE

Since we have found two distinct real roots, the general solution of the given ODE can be written as: $$ y(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t}, $$ where \(C_1\) and \(C_2\) are constants to be determined using the initial conditions.

Solve for the constants using initial conditions

Using the initial condition \(y(0)=1\), we get: $$ 1 = C_1 e^{r_1 \cdot 0} + C_2 e^{r_2 \cdot 0} = C_1 + C_2. $$ Now, using the other initial condition, \(y^{\prime}(0)=0\), first we find the derivative of \(y(t)\): $$ y^{\prime}(t) = r_1 C_1 e^{r_1 t} + r_2 C_2 e^{r_2 t}. $$ Plugging \(t=0\), we get: $$ 0 = r_1 C_1 e^{r_1 \cdot 0} + r_2 C_2 e^{r_2 \cdot 0} = r_1 C_1 + r_2 C_2. $$ We have a system of two linear equations in terms of \(C_1\) and \(C_2\): 1. \(C_1 + C_2 = 1\) 2. \(r_1 C_1 + r_2 C_2 = 0\) Solve this system to obtain: $$ C_1 = \frac{1}{\sqrt{13}}, \quad C_2 = \frac{-6}{\sqrt{13}}. $$

Write down the particular solution and describe its behavior

We have found the values of the constants \(C_1\) and \(C_2\). Now, we can write down the particular solution of the given ODE: $$ y(t) = \frac{1}{\sqrt{13}} e^{\frac{-5+\sqrt{13}}{2} t} - \frac{6}{\sqrt{13}} e^{\frac{-5-\sqrt{13}}{2} t}. $$ Since both exponential terms have negative exponents, the terms will approach 0 as \(t\) increases, meaning that the graph of the solution will approach the horizontal axis. The overall behavior of the graph is that it starts at the point \((0, 1)\) and decays smoothly to 0 as \(t\) increases.

Key concepts

Characteristic Equation

In solving linear differential equations, the characteristic equation is fundamental. It's essentially a polynomial derived from a linear differential equation which helps identify the roots that determine the nature of the solution. For a second-order linear differential equation like the one given:

• \( y'' + 5y' + 3y = 0 \)

the characteristic equation is formed by replacing the derivatives with powers of \( r \), resulting in \( r^2 + 5r + 3 = 0 \). This transformation allows us to transform the differential equation into an algebraic equation. Solving the characteristic equation gives us the roots \( r \) that are crucial for constructing the general solution of the differential equation.

Quadratic Formula

The quadratic formula is a powerful tool used to find the roots of a quadratic equation. The equation is in the form \( ax^2 + bx + c = 0 \) and the formula for the roots is:

• \[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

For our characteristic equation, \( r^2 + 5r + 3 = 0 \), the coefficients are \( a=1 \), \( b=5 \), and \( c=3 \). Plug these values into the quadratic formula:

• \[ r = \frac{-5 \pm \sqrt{25 - 12}}{2} \]

to obtain the roots \( r_1 = \frac{-5 + \sqrt{13}}{2} \) and \( r_2 = \frac{-5 - \sqrt{13}}{2} \). These roots indicate how the solution behaves over time.

General Solution

The general solution of a differential equation uses the roots from the characteristic equation to form a complete expression involving arbitrary constants. This solution can accommodate any initial conditions provided. For the initial value problem given, the general solution is formed using the roots found:

• \( y(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t} \)

Here, \( C_1 \) and \( C_2 \) are constants that must be determined by applying initial conditions. Given that the roots \( r_1 \) and \( r_2 \) are distinct real numbers, the solution describes the real-valued behavior of the function over time.

System of Linear Equations

In the context of solving differential equations, a system of linear equations often arises when determining unknown constants in the general solution. The initial conditions of the given problem provide these equations:

• \( C_1 + C_2 = 1 \)
• \( r_1 C_1 + r_2 C_2 = 0 \)

Solving this system provides the specific values of \( C_1 \) and \( C_2 \) that satisfy the initial conditions, leading us to the particular solution. For our example, the solution to this system yields \( C_1 = \frac{1}{\sqrt{13}} \) and \( C_2 = \frac{-6}{\sqrt{13}} \). This system ensures that the solution not only solves the differential equation but also meets the initial conditions specified.