Chapter 17: Problem 74
If \(y=c_{1} e^{2 x}+c_{2} e^{x}+c_{3} e^{-x}\) satisfies the differential equation \(\frac{d^{3} y}{d x^{3}}+a \frac{d^{2} y}{d x^{2}}+b \frac{d y}{d x}+c y=0\), then \(\frac{a^{3}+b^{3}+c^{3}}{a b c}\) is equal to (A) \(\frac{1}{4}\) (B) \(-\frac{1}{4}\) (C) \(\frac{1}{2}\) (D) \(-\frac{1}{2}\)
Short Answer
Step by step solution
Compute the First Derivative
Compute the Second Derivative
Compute the Third Derivative
Substitute into the Differential Equation
Collect Terms and Solve for Coefficients
Calculate \(\frac{a^3 + b^3 + c^3}{abc}\)
Consider Constant Solutions
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
First Derivative
The derivative of \( e^{2x} \) is \( 2e^{2x} \), hence the term \( c_1 e^{2x} \) becomes \( 2c_1 e^{2x} \). Similarly, the derivative of \( e^{x} \) is \( e^{x} \), making \( c_2 e^{x} \) into \( c_2 e^{x} \). Finally, \( e^{-x} \) differentiates to \( -e^{-x} \), turning the last term into \( -c_3 e^{-x} \).
Now, putting all these together, the first derivative is:
\[ \frac{d y}{d x} = 2c_{1}e^{2x} + c_{2}e^{x} - c_{3}e^{-x} \] This expression allows us to understand the rate at which the original function changes with respect to \( x \).
Second Derivative
Building on the first derivative, the second derivative is found by differentiating the first derivative. For \( y = c_1 e^{2x} + c_2 e^{x} + c_3 e^{-x} \), once we have \( \frac{dy}{dx} = 2c_{1}e^{2x} + c_{2}e^{x} - c_{3}e^{-x} \), differentiate once more:
- \( 2c_1 e^{2x} \) becomes \( 4c_1 e^{2x} \)
- \( c_2 e^{x} \) remains \( c_2 e^{x} \)
- \( -c_3 e^{-x} \) becomes \( -c_{3} e^{-x} \).
Thus, the second derivative is:
\[ \frac{d^2 y}{d x^2} = 4c_{1}e^{2x} + c_{2}e^{x} + c_{3}e^{-x} \]
This indicates the acceleration of change over \( x \).
Third Derivative
Continuing with our function, we now differentiate the second derivative \( \frac{d^2 y}{d x^2} = 4c_{1}e^{2x} + c_{2}e^{x} + c_{3}e^{-x} \) again:
- \( 4c_1 e^{2x} \) becomes \( 8c_1 e^{2x} \)
- \( c_2 e^{x} \) stays \( c_2 e^{x} \)
- \( c_3 e^{-x} \) changes to \( -c_3 e^{-x} \).
This calculation yields the third derivative:
\[ \frac{d^3 y}{d x^3} = 8c_{1}e^{2x} + c_{2}e^{x} - c_{3}e^{-x} \]
The third derivative is crucial for identifying inflection points and understanding dynamic systems described by differential equations.
Homogeneous Equation
The term "homogeneous" implies equal balance; this means that no external inputs or driving forces are contributing to the system—everything is natural growth or decay.
In the context of solving such equations, homogeneous solutions lead us to find various constants or coefficients that satisfy the zero outcome. For instance here, by substituting the derivatives back into the equation, it tests whether an expression like \( 8c_{1}e^{2x} + c_{2}e^{x} - c_{3}e^{-x} \), etc., can settle into zero with certain values for \( a, b, \) and \( c \).
Students encounter homogeneous equations frequently, as they succinctly represent core system dynamics without extra influences.
Eigenvalues
An eigenvalue provides early insights about the solution before details are fully worked out. For example, in a setup like here, where expressions are reduced via coefficients \( a, b, \), and \( c \), these act like constraints, playing into a consistent and functional result, often guiding system behavior.
Obtaining eigenvalues can involve setting the determinant of a matrix to zero and solving for the roots of that characteristic equation. Algebraically, eigenvalues hint at whether solutions grow, stagnate, or shrink over time. This maintains simplicity in complex dynamic systems equations, leading to expressions like the symbolic representation of eigen balancings closer to known limits—like the symbolic answer of \(-1/4\) in such systems.
This comprehensive understanding helps predict how systems react to different conditions and letting students comprehend the role of eigenvalues simplifies the modeling of real-world phenomena.