Question

In Exercises \(13-17\) solve the initial value problem. $$ 4 y^{\prime \prime}-4 y^{\prime}-3 y=0, \quad y(0)=\frac{13}{12}, \quad y^{\prime}(0)=\frac{23}{24} $$

Short answer

The particular solution is \(y(t) = e^t - \frac{1}{4}e^{-\frac{3}{4}t}\).

Step-by-step solution

Solve the Characteristic Equation

The given differential equation is: $$ 4y'' - 4y' - 3y = 0 $$ Let's find the characteristic equation for this differential equation by replacing \(y''\) with \(r^2\), \(y'\) with \(r\), and \(y\) with \(1\). The characteristic equation is: $$ 4r^2 - 4r - 3 = 0 $$ Now, we need to solve this quadratic equation.

Find the Roots

To find the roots of the characteristic equation, let's use the quadratic formula: $$ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$ Applying the quadratic formula to our characteristic equation, we get: $$ r = \frac{4 \pm \sqrt{(-4)^2 - 4(4)(-3)}}{2(4)} $$ Simplifying, we get: $$ r = \frac{4 \pm \sqrt{64}}{8} $$ The two roots are \(r_1 = 1\) and \(r_2=-\frac{3}{4}\).

Write the General Solution

Since our characteristic equation has two distinct real roots, the general solution for the given differential equation is: $$ y(t) = C_1 e^{r_1t} + C_2 e^{r_2t} $$ Plugging in the roots, we have: $$ y(t) = C_1 e^{t} + C_2 e^{-\frac{3}{4} t} $$

Apply the Initial Conditions

Now we need to apply the initial conditions \(y(0) = \frac{13}{12}\) and \(y^{\prime}(0) = \frac{23}{24}\) to solve for the constants \(C_1\) and \(C_2\). For the first initial condition, substitute \(t = 0\) into the general solution: $$ y(0) = C_1 e^{0} + C_2 e^{-\frac{3}{4} \cdot 0} \Rightarrow \frac{13}{12} = C_1 + C_2 $$ For the second initial condition, first find the derivative of the general solution: $$ y'(t) = C_1 e^{t} - \frac{3}{4} C_2 e^{-\frac{3}{4} t} $$ Then substitute \(t = 0\): $$ y^{\prime}(0) = C_1 e^{0} - \frac{3}{4} C_2 e^{-\frac{3}{4} \cdot 0} \Rightarrow \frac{23}{24} = C_1 - \frac{3}{4} C_2 $$

Solve for \(C_1\) and \(C_2\)

Now we have a system of linear equations: $$ C_1 + C_2 = \frac{13}{12} $$ $$ C_1 - \frac{3}{4} C_2 = \frac{23}{24} $$ Solving this system, we find \(C_1 = 1\) and \(C_2 = -\frac{1}{4}\).

Write the Particular Solution

Now that we have the values of \(C_1\) and \(C_2\), we can write the particular solution of the initial value problem: $$ y(t) = e^{t} - \frac{1}{4} e^{-\frac{3}{4} t} $$ So, the solution to the given initial value problem is \(y(t) = e^{t} - \frac{1}{4} e^{-\frac{3}{4} t}\).

Key concepts

Characteristic Equation

The characteristic equation is a cornerstone concept in solving linear homogenous differential equations. It's derived from the original differential equation by substituting each term with a polynomial expression. For example, take the differential equation from our exercise:

$$4y'' - 4y' - 3y = 0$$
To formulate the characteristic equation, we replace the second derivative, first derivative, and the function itself with powers of an unknown variable, typically denoted by 'r'. So, the equation becomes:

$$4r^2 - 4r - 3 = 0$$
The roots of this equation represent the solution in exponential form of the differential equation. Understanding the characteristic equation is essential, as it sets the stage for determining the general solution of the differential equation.

Differential Equation

A differential equation represents a relationship between a function and its derivatives. In our exercise, we are given a second-order linear homogeneous differential equation with constant coefficients:

$$4y'' - 4y' - 3y = 0$$
The equation describes how a dependent variable, y(t), changes with respect to an independent variable, t. When solving such equations, the goal is to find a function y(t) that satisfies the relationship defined by the differential equation for all t. Once we form the characteristic equation and solve it, we can use its roots to construct 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. If we have a quadratic equation in the form:

$$ax^2 + bx + c = 0$$
then the solutions for x are given by the quadratic formula:

$$x = \frac{-b \

General Solution of Differential Equation

Once the characteristic equation's roots have been identified, they are used to write the general solution of the original differential equation. For second-order linear differential equations with constant coefficients, the general solution can take several forms depending on the nature of the roots. If we have two distinct real roots, as in our exercise, the general solution is written as:

$$y(t) = C_1 e^{r_1t} + C_2 e^{r_2t}$$
where the constants \( C_1 \) and \( C_2 \) will be determined using initial conditions provided in the problem. For our case, with roots \( r_1 = 1 \) and \( r_2 = -\frac{3}{4} \), the general solution before applying initial conditions is:

$$y(t) = C_1 e^{t} + C_2 e^{-\frac{3}{4} t}$$
Indeed, understanding how to derive and apply the general solution is crucial for comprehensively grasping differential equations.