Question

Find the solution of the given initial value problem. Sketch the graph of the solution and describe its behavior as \(t\) increases. $$ 6 y^{\prime \prime}-5 y^{\prime}+y=0, \quad y(0)=4, \quad y^{\prime}(0)=0 $$

Short answer

Answer: The specific solution is \(y(t) = 2e^{(\frac{1}{2})t} + 2e^{(\frac{1}{3})t}\).

Step-by-step solution

Find the characteristic equation

Given the differential equation: $$6y^{\prime\prime} - 5y^{\prime} + y = 0$$ In order to find the corresponding characteristic equation, we replace \(y^{\prime\prime}\) with \(r^2\), \(y^{\prime}\) with \(r\), and \(y\) with \(1\). This results in a quadratic equation: $$6r^2 - 5r + 1 = 0$$

Solve the quadratic equation

Now, we will solve the quadratic equation \(6r^2 - 5r + 1 = 0\) to find the roots. Using the quadratic formula, where \(a = 6\), \(b = -5\), and \(c = 1\): $$r = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(6)(1)}}{2(6)}$$ $$r = \frac{5 \pm \sqrt{25 - 24}}{12}$$ $$r = \frac{5 \pm 1}{12}$$ There are two roots: \(r_1 = \frac{6}{12} = \frac{1}{2}\) and \(r_2 = \frac{4}{12} = \frac{1}{3}\).

Form the general solution

Since we have two distinct real roots, the general solution to the differential equation is given as: $$y(t) = C_1 e^{(\frac{1}{2})t} + C_2 e^{(\frac{1}{3})t}$$

Apply the initial conditions and find the specific solution

We have the initial conditions \(y(0) = 4\) and \(y^{\prime}(0) = 0\). We will apply these initial conditions to find the constants \(C_1\) and \(C_2\). For \(y(0) = 4\): $$4 = C_1 e^{(\frac{1}{2})(0)} + C_2 e^{(\frac{1}{3})(0)}$$ $$4 = C_1 + C_2$$ For \(y^{\prime}(0) = 0\), first find the derivative of the general solution: $$y^{\prime}(t) = \frac{1}{2}C_1 e^{(\frac{1}{2})t} + \frac{1}{3}C_2 e^{(\frac{1}{3})t}$$ Now, apply the initial condition \(y^{\prime}(0) = 0\): $$0 = \frac{1}{2}C_1 e^{(\frac{1}{2})(0)} + \frac{1}{3}C_2 e^{(\frac{1}{3})(0)}$$ $$0 = \frac{1}{2}C_1 + \frac{1}{3}C_2$$ Solve this system of linear equations: $$ \begin{cases} C_1 + C_2 = 4 \\ \frac{1}{2}C_1 + \frac{1}{3}C_2 = 0 \\ \end{cases} $$ After solving this system, we obtain \(C_1 = 2\) and \(C_2 = 2\). The specific solution is therefore: $$y(t) = 2e^{(\frac{1}{2})t} + 2e^{(\frac{1}{3})t}$$

Graph the solution and describe its behavior

Plot the specific solution function for \(t \ge 0\). The graph should show that as \(t\) increases, the function \(y(t)\) keeps decreasing, converging to zero without touching the x-axis. This shows that the solution approaches zero as time goes to infinity.

Key concepts

Characteristic Equation

The characteristic equation plays a critical role in solving linear homogeneous differential equations. It's essentially a bridge between differential equations and algebra. When we have a second-order differential equation like
\(6y^{\textprime\textprime} - 5y^{\textprime} + y = 0\),
we can convert it into its characteristic form by replacing
\(y^{\textprime\textprime}\) with \(r^2\),
\(y^{\textprime}\) with \(r\), and \(y\) with 1. The equation then transforms from one involving calculus to a much more manageable quadratic equation
\(6r^2 - 5r + 1 = 0\).
This process simplifies the problem because we have a multitude of techniques at our disposal for solving quadratic equations, which lead us to the roots that determine the general solution of the differential equation.

Differential Equations

Differential equations are mathematical equations that relate a function with its derivatives. In the realm of calculus, they serve as formulas that describe various physical phenomena, such as motion, heat, and growth. The equation given in the exercise,
\(6y^{\textprime\textprime} - 5y^{\textprime} + y = 0\),
is a homogenous linear differential equation with constant coefficients, indicating that the rate of change of a quantity (and its second derivative) is directly proportional to the quantity itself. It is 'homogeneous' because it is set equal to zero, and 'linear' because the variable \(y\) and its derivatives appear to the first power and are not multiplied together. Solving these equations often requires finding the general solution that includes all possible solutions, usually involving constants that are determined by initial conditions.

Solving Quadratic Equations

To solve quadratic equations like the characteristic equation we derived from our differential equation, we often use the quadratic formula, which provides the solutions to any quadratic equation of the form
\(ax^2 + bx + c = 0\).
The formula
\(x = \frac{-b \textpm \textsqrt{b^2 - 4ac}}{2a}\)
applies to our characteristic equation, allowing us to find the roots \(r_1\) and \(r_2\). These roots are crucial because they determine the nature of the solution to the differential equation: real and distinct, real and repeated, or complex conjugates. Each scenario requires a different approach for writing the general solution. In our example, the roots \(\frac{1}{2}\) and \(\frac{1}{3}\) are real and distinct, leading to a combination of exponential functions in the general solution.

Exponential Functions

Exponential functions are pervasive in equations that model growth and decay processes. In our specific case, the general solution
\(y(t) = C_1 e^{\frac{1}{2}t} + C_2 e^{\frac{1}{3}t}\),
involves two exponential terms where \(C_1\) and \(C_2\) are constants determined by the initial conditions. Because the base of the exponentials, \(e\), is a mathematical constant approximately equal to 2.71828, the behavior of the solution as \(t\) increases depends on the powers of \(e\). When the initial conditions are applied, they provide unique values for \(C_1\) and \(C_2\), which help to sketch the specific function's graph and understand the solution's behavior over time. The positive exponents lead to the function increasing at first but eventually approaching zero, as the negative signs in the exponents in the general solution imply decay.