Problem 3
Question
Answer Problems \(1-8\) without referring back to the text. Fill in the blank or answer true/false. A constant multiple of a solution of a linear differential equation is also a solution.
Step-by-Step Solution
Verified Answer
False.
1Step 1: Understanding the Problem
We need to determine whether a constant multiple of a solution of a linear differential equation is also a solution. This involves understanding the properties of linear differential equations.
2Step 2: Recall the Form of a Linear Differential Equation
A linear differential equation can generally be expressed as \( L(y) = f(x) \), where \( L \) is a linear operator acting on a function \( y \). If \( y = y_1(x) \) is a solution, then \( L(y_1(x)) = f(x) \).
3Step 3: Consider the Constant Multiple
Let \( y_2(x) = c \, y_1(x) \), where \( c \) is a constant. We need to check if \( y_2(x) \) is a solution, i.e., if \( L(y_2(x)) = f(x) \).
4Step 4: Apply the Linear Operator to the Constant Multiple
Since the operator \( L \) is linear, we know \( L(c \, y_1(x)) = c \, L(y_1(x)) \). Given that \( L(y_1(x)) = f(x) \), it follows that \( c \, L(y_1(x)) = c \, f(x) \).
5Step 5: Analyze the Result
For the equation to remain valid, we would need \( c \, f(x) = f(x) \). However, this equality only holds if \( c = 1 \), which is generally not true for any constant \( c eq 1 \).
6Step 6: Conclusion
Since \( c \, f(x) = f(x) \) is not true for arbitrary constants, a constant multiple of a solution generally does not satisfy the original differential equation unless certain conditions are met, like \( f(x) = 0 \). Therefore, the statement is false.
Key Concepts
Solution PropertiesMathematical OperatorsDifferential Equation Solutions
Solution Properties
In the realm of linear differential equations, understanding solution properties is vital. A linear differential equation has the fascinating property where each of its solutions can be manipulated while still remaining a valid solution under specific circumstances. For example, if you have a function, say, a solution to the equation, multiplying it by a constant usually modifies the solution, unless specific conditions are met.
Why are solution properties important? Knowing how solutions behave allows us to transform known solutions into new ones, potentially simplifying complex problems. However, a constant multiple of a solution may not always be a valid solution itself. This introduces the importance of understanding each equation's unique properties, such as its homogeneity or non-homogeneity. A homogeneous equation, where the function equals zero ( L(y) = 0), permits any constant multiple of a solution to also be a solution. This property does not hold for non-homogeneous equations unless further conditions are applied.
Why are solution properties important? Knowing how solutions behave allows us to transform known solutions into new ones, potentially simplifying complex problems. However, a constant multiple of a solution may not always be a valid solution itself. This introduces the importance of understanding each equation's unique properties, such as its homogeneity or non-homogeneity. A homogeneous equation, where the function equals zero ( L(y) = 0), permits any constant multiple of a solution to also be a solution. This property does not hold for non-homogeneous equations unless further conditions are applied.
Mathematical Operators
In mathematics, operators are symbols or functions that signify the process of taking some action on certain mathematical objects. In the context of linear differential equations, mathematical operators play a crucial role.
A common operator used for these types of equations is the differential operator, represented here as \( L \). This operator acts on a function \( y \) and involves derivatives. Linear operators, in particular, have the property that they work well with addition and scalar multiplication of functions. This means that \( L(y_1 + y_2) = L(y_1) + L(y_2) \) and \( L(c \, y) = c \, L(y) \), where \( c \) is a constant.
Understanding how operators function helps us predict how they affect solutions. Since the differential operator follows the linearity principle, manipulating one solution doesn't always guarantee another when external parts of the equation (like \( f(x) \)) remain unaffected by this constant multiplication.
A common operator used for these types of equations is the differential operator, represented here as \( L \). This operator acts on a function \( y \) and involves derivatives. Linear operators, in particular, have the property that they work well with addition and scalar multiplication of functions. This means that \( L(y_1 + y_2) = L(y_1) + L(y_2) \) and \( L(c \, y) = c \, L(y) \), where \( c \) is a constant.
Understanding how operators function helps us predict how they affect solutions. Since the differential operator follows the linearity principle, manipulating one solution doesn't always guarantee another when external parts of the equation (like \( f(x) \)) remain unaffected by this constant multiplication.
Differential Equation Solutions
Differential equations are equations that involve functions and their derivatives. Solutions to these equations can take various forms, but they always must satisfy the original differential equation under given conditions.
There are two main types of solutions for linear differential equations:
There are two main types of solutions for linear differential equations:
- Particular Solution: This is a solution that specifically satisfies a non-homogeneous differential equation. It's not multiplied by a constant since such multiplication would skew the equation's conditions.
- General Solution: This includes all possible solutions, typically expressed as a linear combination of solutions to a homogeneous equation and a particular solution to the non-homogeneous equation.
Other exercises in this chapter
Problem 3
Solve each differential equation by variation of parameters. $$ y^{\prime \prime}+y=\sin x $$
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The given family of functions is the general solution of the differential equation on the indicated interval. Find a member of the family that is a solution of
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In Problems \(1-20\), solve the given system of differential equations by systematic elimination. $$ \begin{aligned} &\frac{d x}{d t}=-y+t \\ &\frac{d y}{d t}=x
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In Problems \(1-4\), the given differential equation is a model of an undamped spring/mass system in which the restoring force \(F(x)\) in (1) is nonlinear. For
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