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Chapter 31: Problem 19

Solve the given differential equations. $$4 D^{2} y-3 D y=2 y$$

Short Answer

Expert verified
The general solution is \(y = c_1 e^{\frac{3 +  \sqrt{41}}{8}x} + c_2 e^{\frac{3 -  \sqrt{41}}{8}x}\).

Step by step solution

01

Rewrite the Differential Equation

The given differential equation is \(4 D^{2} y - 3 D y = 2 y\). The operator \(D\) indicates differentiation with respect to \(x\). Rewrite the equation as \(4y'' - 3y' = 2y\), where \(y''\) denotes the second derivative of \(y\) and \(y'\) denotes the first derivative.
02

Set Up the Characteristic Equation

To solve the differential equation, we assume a solution of the form \(y = e^{rx}\). Substituting \(y = e^{rx}\), \(y' = re^{rx}\), and \(y'' = r^2 e^{rx}\) into the differential equation gives \(4r^2 e^{rx} - 3r e^{rx} = 2e^{rx}\). Factor out \(e^{rx}\) to obtain the characteristic equation: \(4r^2 - 3r - 2 = 0\).
03

Solve the Characteristic Equation

The characteristic equation is \(4r^2 - 3r - 2 = 0\). Use the quadratic formula \(r = \frac{-b  \sqrt{b^2 - 4ac}}{2a}\) where \(a = 4\), \(b = -3\), and \(c = -2\). Calculate the discriminant \(b^2 - 4ac = 9 + 32 = 41\). The solutions are \(r = \frac{3   \sqrt{41}}{8}\).
04

Write the General Solution

The roots of the characteristic equation are \(r_1 = \frac{3 +  \sqrt{41}}{8}\) and \(r_2 = \frac{3 -  \sqrt{41}}{8}\). Therefore, the general solution of the differential equation is \(y = c_1 e^{r_1 x} + c_2 e^{r_2 x}\), where \(c_1\) and \(c_2\) are arbitrary constants.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Characteristic Equation
The characteristic equation is a crucial step when solving linear differential equations, particularly those with constant coefficients. In the context of a differential equation, it essentially determines the nature of the solutions by converting differential terms into an algebraic equation.

Here's how it works:
  • Assume a trial solution of the form \(y = e^{rx}\). This choice simplifies the process because the exponential function is inherently differentiable.
  • Substitute \(y = e^{rx}\), along with its derivatives, into the original differential equation. This transforms the equation from differential to algebraic.
  • For example, substituting back into \(4y'' - 3y' = 2y\) simplifies to \(4r^2 e^{rx} - 3r e^{rx} = 2e^{rx}\).
By factoring out \(e^{rx}\), which never equals zero, the essence of the equation is captured as the characteristic equation: \(4r^2 - 3r - 2 = 0\). Solving this quadratic equation provides roots that are pivotal in forming the general solution of the differential equation.
Second Derivative
Understanding the second derivative is essential, as it plays a significant role in higher-order differential equations. When you hear about the second derivative, think about how it relates to the "rate of change of the rate of change."

Consider the steps:
  • The first derivative \(y'\) indicates the rate of change or the slope of the function \(y\).
  • The second derivative \(y''\) provides information on how this rate of change itself is changing, offering insights into the concavity of the graph.
In the context of differential equations, transitioning from \(4D^2y\) to \(4y''\) denotes taking the second derivative of \(y\). This step allows the equation to be analyzed for patterns and relationships that are beyond first-order changes. Always remember, evaluating the second derivative helps inform about acceleration and curvature, essential for solving and interpreting differential equations.
Quadratic Formula
The quadratic formula is a dependable tool for solving quadratic equations like \(ax^2 + bx + c = 0\). When dealing with characteristic equations in differential equations, it becomes indispensable. In this exercise, we applied it to find the roots of the equation \(4r^2 - 3r - 2 = 0\).

Here's the breakdown of using the formula:
  • First, ensure the equation is in standard form: \(ax^2 + bx + c = 0\). In our example, \(a = 4\), \(b = -3\), and \(c = -2\).
  • Calculate the discriminant \(b^2 - 4ac\). For our equation, this yields \(9 + 32 = 41\).
  • Substitute these values into the quadratic formula: \(r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
For \(4r^2 - 3r - 2 = 0\), it results in the roots \(r_1 = \frac{3 + \sqrt{41}}{8}\) and \(r_2 = \frac{3 - \sqrt{41}}{8}\). From these roots, the general solution for the differential equation is formed, indicating the versatility and necessity of the quadratic formula in solving complex mathematical problems.

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Most popular questions from this chapter

Find the particular solution of each differential equation for the given conditions. $$\begin{aligned} &3 y^{\prime \prime}-10 y^{\prime}+3 y=x e^{-2 x} ; \quad y^{\prime}=-\frac{9}{35} \text { and } y=-\frac{13}{35} \text { when } &x=0 \end{aligned}$$

Solve the given problems by solving the appropriate differential equation. According to Newton's law of cooling, the rate at which a body cools is proportional to the difference in temperature between it and the surrounding medium. Assuming Newton's law holds, how long will it take a cup of hot water, initially at \(200^{\circ} \mathrm{F},\) to cool to \(100^{\circ} \mathrm{F}\) if the room temperature is \(80.0^{\circ} \mathrm{F},\) if it cools to \(140^{\circ} \mathrm{F}\) in 5.0 min?

Solve the given differential equations. The form of \(y_{p}\) is given. $$\left.y^{\prime \prime}-2 y^{\prime}+y=3+e^{x} \quad \text { (Let } y_{p}=A+B x^{2} e^{x}\right)$$

Solve the given differential equations. The form of \(y_{p}\) is given. $$\left.D^{2} y-y=e^{-x} \quad \text { (Let } y_{p}=A x e^{-x} .\right)$$

Solve the given differential equations by Laplace transforms. The function is subject to the given conditions. $$y^{\prime}+y=0, y(0)=1$$
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