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Find values of \(m\) so that the function \(y=e^{m x}\) is a solution of the given differential equation. $$ y^{\prime \prime}-5 y^{\prime}+6 y=0 $$

Short Answer

Expert verified
The values of \( m \) are 3 and 2.

Step by step solution

01

Identify the form of the solution

We are given the function \( y = e^{mx} \). To find if it is a solution of the differential equation, we distinguish the function in terms of derivatives. The form of \( y(x) \) suggests using exponential functions.
02

Compute the first derivative

First, compute \( y' \), the first derivative of \( y \):\( y' = \frac{d}{dx}(e^{mx}) = me^{mx} \).
03

Compute the second derivative

Next, compute \( y'' \), the second derivative of \( y \):\( y'' = \frac{d}{dx}(me^{mx}) = m^2e^{mx} \).
04

Substitute derivatives into the differential equation

Substitute \( y = e^{mx} \), \( y' = me^{mx} \), and \( y'' = m^2e^{mx} \) into the equation \( y'' - 5y' + 6y = 0 \):\[ m^2e^{mx} - 5me^{mx} + 6e^{mx} = 0. \]
05

Factor out the common exponential term

Factor out \( e^{mx} \) as it is common to each term:\[ e^{mx}(m^2 - 5m + 6) = 0. \]Since \( e^{mx} eq 0 \) for any \( m \), we focus on:\[ m^2 - 5m + 6 = 0. \]
06

Solve the quadratic equation

Solve the quadratic equation \( m^2 - 5m + 6 = 0 \) using the quadratic formula:\[ m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Here, \( a = 1 \), \( b = -5 \), \( c = 6 \):\[ m = \frac{-(-5) \pm \sqrt{(-5)^2 - 4 \times 1 \times 6}}{2 \times 1}. \]Calculate the discriminant: \( (-5)^2 - 4 \times 1 \times 6 = 25 - 24 = 1 \).\[ m = \frac{5 \pm 1}{2}. \]The solutions are \( m = 3 \) and \( m = 2 \).

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Key Concepts

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

Quadratic Equations
Quadratic equations are essential in various fields of mathematics and science. They appear frequently when solving problems involving areas, curves, and many other applications. A quadratic equation is typically of the form:\[ ax^2 + bx + c = 0 \]where \( a, b, \) and \( c \) are constants, and \( x \) is the variable. The solution to a quadratic equation can be found using the quadratic formula:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]This formula provides two solutions, often referred to as the roots of the equation. The roots can be either real or complex numbers, depending on the nature of the discriminant \( b^2 - 4ac \).
  • If the discriminant is positive, there are two distinct real roots.
  • If the discriminant is zero, there is exactly one real root (also known as a repeated or double root).
  • If the discriminant is negative, the roots are complex.
The quadratic formula is a powerful tool and a critical step in solving second-order differential equations.
Exponential Functions
Exponential functions are a type of mathematical function where the variable appears in the exponent. They are essential in modeling a wide range of phenomena in science and engineering, such as population growth, radioactive decay, and many more. The general form of an exponential function is:\[ y = e^{mx} \]where \( e \) is the base of the natural logarithm, approximately equal to 2.71828, and \( m \) is a constant. One of the most significant properties of exponential functions is that their derivatives are proportional to the original function:
  • The first derivative of \( e^{mx} \) is \( me^{mx} \).
  • The second derivative of \( e^{mx} \) is \( m^2e^{mx} \).
This property makes exponential functions highly useful for finding solutions to differential equations, especially linear homogeneous equations with constant coefficients. Such functions can transform complex problems into simpler algebraic forms, which is invaluable for analysis.
Second-Order Differential Equations
Second-order differential equations play a fundamental role in modeling and solving real-world problems such as mechanical vibrations, electrical circuits, and many others. They generally take the form:\[ a y'' + b y' + c y = 0 \]These are called linear homogeneous differential equations with constant coefficients. In solving these equations, one common approach is to assume a solution of the form \( y = e^{mx} \), which leads to the characteristic equation:\[ am^2 + bm + c = 0 \]Finding \( m \) often involves solving a quadratic equation. The roots determine the general solution to the differential equation:
  • If the roots are real and distinct, \( y = c_1 e^{m_1 x} + c_2 e^{m_2 x} \).
  • If the roots are real and repeated, \( y = (c_1 + c_2x) e^{m x} \).
  • If the roots are complex, the solution will involve sinusoids and exponentials.
This methodology simplifies the process of identifying solutions by converting the differential equation into an algebraic equation, thus making the problem more tractable.

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Most popular questions from this chapter

Consider the differential equation \(d y / d x=e^{-x^{2}}\). (a) Explain why a solution of the DE must be an increasing function on any interval of the \(x\)-axis. (b) What are \(\lim _{x \rightarrow-\infty} d y / d x\) and \(\lim _{x \rightarrow \infty} d y / d x\) ? What does this suggest about a solution curve as \(x \rightarrow \pm \infty\) ? (c) Determine an interval over which a solution curve is concave down and an interval over which the curve is concave up. (d) Sketch the graph of a solution \(y=\phi(x)\) of the differential equation whose shape is suggested by parts (a)-(c).

\(y=c_{1} \cos 3 x+c_{2} \sin 3 x\) is a two-parameter family of solutions of the second-order DE \(y^{\prime \prime}+9 y=0\). If possible, find a solution of the differential equation that satisfies the given side conditions. The conditions specified at two different points are called boundary conditions. $$ y^{\prime}(\pi / 3)=1, y^{\prime}(\pi)=0 $$

Qualitative information about a solution \(y=\phi(x)\) of a differential equation can often be obtained from the equation itself. Before working Problems \(55-58\), recall the geometric significance of the derivatives \(d y / d x\) and \(d^{2} y / d x^{2}\). Consider the differential equation \(d y / d x=y(a-b y)\), where \(a\) and \(b\) are positive constants. (a) Either by inspection, or by the method suggested in Problems 33-36, find two constant solutions of the DE. (b) Using only the differential equation, find intervals on the \(y\) -axis on which a nonconstant solution \(y=\phi(x)\) is increasing. On which \(y=\phi(x)\) is decreasing. (c) Using only thedifferentialequation, explain why \(y=a / 2 b\) is the \(y\) -coordinate of a point of inflection of the graph of a nonconstant solution \(y=\phi(x)\). (d) On the same coordinate axes, sketch the graphs of the two constant solutions found in part (a). These constant solutions partition the \(x y\) -plane into three regions. In each region, sketch the graph of a nonconstant solution \(y=\phi(x)\) whose shape is suggested by the results in parts (b) and (c).

Use the concept that \(y=c,-\infty

Verify that the indicated function \(y=\phi(x)\) is an explicit solution of the given first-order differential equation. Proceed as in Example 5, by considering \(\phi\) simply as a function, give its domain. Then by considering \(\phi\) as a solution of the differential equation, give at least one interval \(I\) of definition. $$ 2 y^{\prime}=y^{3} \cos x ; \quad y=(1-\sin x)^{-1 / 2} $$

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