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Chapter 9: Problem 11

Find the general solution. $$5 y^{\prime \prime}+\frac{11}{4} y^{\prime}-\frac{3}{4} y=0$$

Short Answer

Expert verified
The general solution of the given second-order linear homogeneous differential equation \(5y'' + \frac{11}{4}y' - \frac{3}{4}y = 0\) is given by: \[y(x) = C_1e^{r_1x} + C_2e^{r_2x}\] where \(r_1\) and \(r_2\) are the roots of the characteristic equation \(5r^2 + \frac{11}{4}r - \frac{3}{4} = 0\), which can be found using the quadratic formula: \[r = \frac{-\frac{11}{4} \pm \sqrt{(\frac{11}{4})^2 - 4(5)(-\frac{3}{4})}}{2(5)}\] and \(C_1\) and \(C_2\) are constants.

Step by step solution

01

Write down the characteristic equation

For the given differential equation: \(5y'' + \frac{11}{4}y' - \frac{3}{4}y = 0\), we can write the corresponding characteristic equation: \[5r^2 + \frac{11}{4}r - \frac{3}{4} = 0\]
02

Solve for the roots of the characteristic equation

We can solve the quadratic equation for the roots \(r_1\) and \(r_2\): \(5r^2 + \frac{11}{4}r - \frac{3}{4} = 0\) Using the quadratic formula: \[r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] In our case, \(a = 5\), \(b = \frac{11}{4}\), and \(c = -\frac{3}{4}\). Substituting these values into the formula: \[r = \frac{-\frac{11}{4} \pm \sqrt{(\frac{11}{4})^2 - 4(5)(-\frac{3}{4})}}{2(5)}\] Simplify to find the roots \(r_1\) and \(r_2\).
03

Construct the general solution

Use the found roots, \(r_1\) and \(r_2\), to construct the general solution of the given differential equation in the form: \[y(x) = C_1e^{r_1x} + C_2e^{r_2x}\] Replace \(r_1\) and \(r_2\) with their respective values and write the general solution for the given differential equation.

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Key Concepts

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

Characteristic Equation
In the context of solving linear differential equations with constant coefficients, the characteristic equation is a bridge that helps us transition from the differential equation to an algebraic one. The purpose of the characteristic equation is to determine the particular form of the solution we seek. For the differential equation given in the exercise:
  • \(5y'' + \frac{11}{4}y' - \frac{3}{4}y = 0\), the typical approach is to assume a solution of the form \(y = e^{rx}\), where \(r\) is a constant.
  • By substituting this assumed solution into the differential equation and simplifying, we end up with a polynomial equation in terms of \(r\). This polynomial is what we refer to as the characteristic equation.
  • In our case, the characteristic equation is: \[5r^2 + \frac{11}{4}r - \frac{3}{4} = 0\]
Finding the roots of this polynomial equation provides us with the constants necessary to construct the general solution for the differential equation. These roots represent the exponential rates of change of the solution in its simplest form.
Quadratic Formula
The quadratic formula is an essential tool for finding the roots of any quadratic equation. When faced with a quadratic equation of the form \(ax^2 + bx + c = 0\), the quadratic formula gives us the solutions for \(x\):
  • \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
  • The expression \(\pm\) indicates that there are generally two solutions鈥攐r roots鈥攆or \(x\).
In our exercise, the characteristic equation derived from the original differential equation is:
  • \(5r^2 + \frac{11}{4}r - \frac{3}{4} = 0\)
  • To apply the quadratic formula here, we substitute \(a = 5\), \(b = \frac{11}{4}\), and \(c = -\frac{3}{4}\) into the formula to find the specific roots \(r_1\) and \(r_2\).
  • This calculation involves evaluating the discriminant \(b^2 - 4ac\), which determines the nature of the roots鈥攚hether they are real, repeated, or complex.
Using this formula and solving step-by-step allows for deriving the necessary roots to build towards the general solution of the differential equation.
General Solution of Linear Differential Equations
The general solution of a linear differential equation is the combination of all possible solutions that satisfy the equation. In the case of second-order linear differential equations with constant coefficients like the one in this exercise, the general solutions can often be expressed through exponentials based on the roots of the characteristic equation.
  • Once the roots \(r_1\) and \(r_2\) of the characteristic equation \(5r^2 + \frac{11}{4}r - \frac{3}{4} = 0\) are found, they are used to express the general solution.
  • When the roots are distinct and real, the solution format is: \[y(x) = C_1 e^{r_1 x} + C_2 e^{r_2 x}\]
  • The constants \(C_1\) and \(C_2\) represent arbitrary constants that account for the initial conditions specific to any particular solution.
This structure allows the general solution to adapt to various initial or boundary conditions, ensuring flexibility to model different scenarios described by the differential equation.

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Most popular questions from this chapter

Solve the initial-value problem. $$y^{\prime}=6 e^{2 x-y}, \quad y(0)=0.$$

Given in reference of the differential equation (9.1.1) $$ y^{\prime}+p(x) y=q(x) $$ with \(p\) and \(q\) continuous on some interval \(l\) Show that if \(y_{1}\) and \(y_{2}\) are solutions of \((9.1 .1),\) then \(y=y_{1}-y_{2}\) is a solution of \(y^{\prime}+p(x) y=0\)

The Gumpertz equation $$ \frac{d P}{d t}=P(a-b \ln P) $$ where \(a\) and \(b\) arc positive constants, is another model of population growth. (a) Find the solution of this differential equation that satisfies the initial condition \(P(0)=P_{0} .\) HINT: Define a new dependent variable \(Q\) by setting \(Q=\ln P\) (b) What happens to \(P(t)\) as \(t \rightarrow \infty ?\) (c) Determine the concavity of the yraph of \(P\). (d) Use a graphing utility to draw the graph of \(P\) in the case where \(a=4, b=2,\) and \(P_{0}=\frac{1}{2} e^{2} .\) Docs the graph confirm your result in part (c)?

(a) Show that the substitution \(y=e^{\alpha x} u\) transforms $$\begin{aligned} &y^{\prime \prime}-2 \alpha y^{\prime}+\alpha^{2} y=0 \quad \text { into }\\\ &u^{\prime \prime}=0 \end{aligned}$$ (b) Show that the substitution \(y=e^{a x} u\) transforms $$y^{\prime \prime}-2 \alpha y^{\prime}+\left(\alpha^{2}+\beta^{2}\right) y=0 \quad \text { into } \quad u^{\prime \prime}+\beta^{2} u=0$$

Find the integral curves. If the curves are the graphs of functions \(y=f(x)\). determine all the functions that satisfy the equation. $$y^{\prime}=y \sin (2 x+3)$$
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