Problem 28
Question
Find values of \(m\) so that the function \(y=e^{m x}\) is a solution of the given differential equation. $$ 3 y^{\prime}=4 y $$
Step-by-Step Solution
Verified Answer
\(m = \frac{4}{3}\)
1Step 1: Differentiate the function
The given function is \(y = e^{mx}\). We need to find the derivative \(y' = \frac{dy}{dx}\). Differentiate \(e^{mx}\) using the chain rule: \(y' = me^{mx}\).
2Step 2: Substitute into the differential equation
Substitute \(y' = me^{mx}\) and \(y = e^{mx}\) into the differential equation \(3y' = 4y\). This gives us the equation \(3(me^{mx}) = 4(e^{mx})\).
3Step 3: Simplify the equation
Cancel \(e^{mx}\) from both sides of the equation: \(3m = 4\).
4Step 4: Solve for \(m\)
Divide both sides by 3 to isolate \(m\): \(m = \frac{4}{3}\).
Key Concepts
Exponential FunctionsChain RuleSeparation of VariablesSolving Linear Differential Equations
Exponential Functions
Exponential functions, like the given function \(y = e^{mx}\), form a cornerstone in mathematics, especially in calculus and differential equations. An exponential function is characterized by a constant base raised to an exponent that contains a variable. Here, \(e\) is the natural exponential constant, approximately equal to 2.71828.
This particular function grows or decays exponentially depending on the sign of \(m\). If \(m > 0\), the function grows, and if \(m < 0\), it decays. Exponential functions are unique because they model situations where a quantity grows or declines at a rate proportional to its current value.
In solving differential equations, these functions are often solutions because they simplify many mathematical processes involved in differentiation and integration.
This particular function grows or decays exponentially depending on the sign of \(m\). If \(m > 0\), the function grows, and if \(m < 0\), it decays. Exponential functions are unique because they model situations where a quantity grows or declines at a rate proportional to its current value.
In solving differential equations, these functions are often solutions because they simplify many mathematical processes involved in differentiation and integration.
- They appear in continuous growth processes, like population growth or radioactive decay.
- Their mathematical properties allow for straightforward differentiation and integration.
Chain Rule
The chain rule is a fundamental tool in calculus used for differentiating compositions of functions. When a function is expressed as a composition, such as \(y = e^{mx}\), the chain rule helps us determine its derivative with respect to \(x\).
In our example, \(e^{mx}\) is differentiated with respect to \(x\) by treating \(mx\) as an inner function. The derivative of \(e^u\) (where \(u=mx\)) with respect to \(u\) is simply \(e^u\), and \(u'\), the derivative of \(mx\) with respect to \(x\), is \(m\). Thus, applying the chain rule gives us:
\[ \frac{dy}{dx} = me^{mx} \]
This allows us to express how rapidly the function \(e^{mx}\) changes as \(x\) varies. The chain rule is essential in differentiating both simple and complex functions effectively.
In our example, \(e^{mx}\) is differentiated with respect to \(x\) by treating \(mx\) as an inner function. The derivative of \(e^u\) (where \(u=mx\)) with respect to \(u\) is simply \(e^u\), and \(u'\), the derivative of \(mx\) with respect to \(x\), is \(m\). Thus, applying the chain rule gives us:
\[ \frac{dy}{dx} = me^{mx} \]
This allows us to express how rapidly the function \(e^{mx}\) changes as \(x\) varies. The chain rule is essential in differentiating both simple and complex functions effectively.
- Helps find derivatives of composite functions efficiently.
- Essential for solving higher-level calculus problems.
Separation of Variables
Separation of variables is a method often employed to solve ordinary differential equations, especially when they can be neatly split up into functions of individual variables. In the equation we're working with, \(3y' = 4y\), separating variables might not be directly apparent, but it's a principle of simplifying complex differential equations.
In this method, we rearrange the equation into a form where all terms involving one variable are on one side and all terms involving the other variables are on the other side. This is usually followed by integrating both sides to find a general solution. However, this specific problem uses another technique due to its linear nature, but the principles remain the same.
In this method, we rearrange the equation into a form where all terms involving one variable are on one side and all terms involving the other variables are on the other side. This is usually followed by integrating both sides to find a general solution. However, this specific problem uses another technique due to its linear nature, but the principles remain the same.
- Effective for first-order differential equations.
- Often leads to straightforward integration for solutions.
Solving Linear Differential Equations
Linear differential equations are equations that involve derivatives of a function and the function itself, where each term is either a function of the independent variable or the product of a constant and the derivative of the dependent variable.
In our exercise, the equation \(3y' = 4y\) is linear. Solving such equations often involves finding functions that satisfy the equation for all values of the variable. Here, substituting \(y' = me^{mx}\) and \(y = e^{mx}\) into the equation and simplifying by canceling \(e^{mx}\) lead to \(3m = 4\).
The solution was obtained by solving for \(m\), where \(m = \frac{4}{3}\).
In our exercise, the equation \(3y' = 4y\) is linear. Solving such equations often involves finding functions that satisfy the equation for all values of the variable. Here, substituting \(y' = me^{mx}\) and \(y = e^{mx}\) into the equation and simplifying by canceling \(e^{mx}\) lead to \(3m = 4\).
The solution was obtained by solving for \(m\), where \(m = \frac{4}{3}\).
- Linear differential equations can often be solved using algebraic methods.
- They frequently model real-world linear processes.
Other exercises in this chapter
Problem 27
In Problems 27-30 verify that the indicated expression is an implicit solution of the given differential equation. $$ x \frac{d y}{d x}+y=\frac{1}{y^{2}} ; \qua
View solution Problem 28
$$ 3 y^{\prime}=4 y $$
View solution Problem 29
$$ y^{\prime \prime}-5 y^{\prime}+6 y=0 $$
View solution Problem 29
(a) Byinspection, find a one-parameter family of solutions of the differential equation \(x y^{\prime}=y .\) Verify that each member of the family is a solution
View solution