Question

In each of Problems 1 through 10 find the general solution of the given differential equation. \(4 y^{\prime \prime}-4 y^{\prime}-3 y=0\)

Short answer

The general solution for the given differential equation is: \(y(x) = C_1 e^{\frac{1}{2}x} + C_2 e^{-\frac{3}{2}x}\), where \(C_1\) and \(C_2\) are constants determined by the initial conditions.

Step-by-step solution

Write down the given differential equation

The given differential equation is: \(4y'' - 4y' - 3y = 0\).

Find the characteristic equation

The characteristic equation for a second-order linear homogeneous differential equation of the form \(ay'' + by' + cy = 0\) is given by: \(r^2 - \frac{b}{a} r + \frac{c}{a} = 0\). For the given differential equation, we have \(a = 4, b=-4\), and \(c = -3\). Plugging these values into the formula, we get: \(r^2 + r - \frac{3}{4} = 0\).

Solve the characteristic equation

Now, we'll solve the quadratic equation \(r^2 + r - \frac{3}{4} = 0\). Using the quadratic formula, we have \(r = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}\). Plugging A=1, B=1, and C=-\(\frac{3}{4}\) into the formula, we get the two roots: \(r_1 = \frac{-1 + \sqrt{1^2 - 4(1)(-\frac{3}{4})}}{2(1)} = \frac{-1 + \sqrt{4}}{2} = \frac{1}{2}\). \(r_2 = \frac{-1 - \sqrt{1^2 - 4(1)(-\frac{3}{4})}}{2(1)} = \frac{-1 - \sqrt{4}}{2} = -\frac{3}{2}\).

Write the general solution

Since the roots of the characteristic equation are distinct, the general solution of the given differential equation is: \(y(x) = C_1 e^{r_1 x} + C_2 e^{r_2 x}\). Substituting our roots \(r_1 = \frac{1}{2}\) and \(r_2 = -\frac{3}{2}\), we get the general solution: \(y(x) = C_1 e^{\frac{1}{2}x} + C_2 e^{-\frac{3}{2}x}\). The general solution for the given differential equation \(4y'' - 4y' - 3y = 0\) is: \(y(x) = C_1 e^{\frac{1}{2}x} + C_2 e^{-\frac{3}{2}x}\).

Key concepts

Characteristic Equation

The characteristic equation is a fundamental component for solving second-order linear homogeneous differential equations. It casts the differential equation into a quadratic equation which is easier to solve and reveals crucial information about the behavior of the solution.

In our example, we start by identifying the coefficients from the given differential equation, which is in the standard form of \( ay'' + by' + cy = 0 \). Here, \( a = 4 \), \( b = -4 \), and \( c = -3 \). We create the characteristic equation by translating the differential equation's coefficients into a quadratic equation, which results in \( r^2 + r - \frac{3}{4} = 0 \).

Effectively, solving the characteristic equation gives us the roots that dictate the form of the general solution. If the roots are real and distinct, the solution will be a sum of exponential terms. If they are complex, it will involve sines and cosines, and if they are repeated, it will have an additional polynomial factor.

Second-Order Linear Homogeneous Differential Equation

A second-order linear homogeneous differential equation has a standard form of \( ay'' + by' + cy = 0 \) where \( y \) is the function of the independent variable (often time or space), and the apostrophes denote its derivatives. The key to solving this type of equation is understanding that 'homogeneous' refers to the absence of a function of the independent variable on the right side of the equation.

The solutions to these differential equations are important in various fields including engineering, physics, and economics due to their ability to describe systems with constant coefficients, oscillating systems, and growth or decay processes. Our textbook example is one such equation where the solution, \( y(x) \), describes the behavior of the system modeled by the differential equation. The general strategy involves calculating the roots of the characteristic equation, which then inform the solution's form.

Quadratic Formula

The quadratic formula, \( r = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \), is used to find the roots of quadratic equations, which are equations of the form \( Ax^2 + Bx + C = 0 \). In our context, A, B, and C correspond to the coefficients of the characteristic equation derived from the differential equation.

Applying this formula to solve the characteristic equation provides the roots, which are the possible values of \( r \) that satisfy the equation. In our exercise, by substituting the values \( A = 1 \), \( B = 1 \), and \( C = -\frac{3}{4} \) into the quadratic formula, we obtained the distinct real roots \( r_1 = \frac{1}{2} \) and \( r_2 = -\frac{3}{2} \). These roots are pivotal in forming the general solution, which involves exponential functions based on these roots.