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Chapter 6: Problem 15

Determine whether the function is a solution of the differential equation \(y^{(4)}-16 y=0\). $$ y=e^{-2 x} $$

Short Answer

Expert verified
Yes, the function \(y=e^{-2x}\) is a solution to the differential equation.

Step by step solution

01

Compute the fourth derivative of y

Calculate the fourth derivative of y, which is given by \(y=e^{-2x}\). Use the chain rule where necessary: The first derivative is \(y'=-2e^{-2x}\), the second derivative is \(y''=4e^{-2x}\), the third derivative is \(y'''=-8e^{-2x}\), and the fourth derivative is \(y^{(4)}=16e^{-2x}\)
02

Substitute y and y' into the differential equation

Substitute \(y=e^{-2x}\) and \(y^{(4)}=16e^{-2x}\) into the equation \(y^{(4)}-16y=0 \). This gives: \(16e^{-2x}-16(e^{-2x}) = 16e^{-2x}-16e^{-2x} = 0\)
03

Verify the equality

As the left side of the equation equals to the right side of the equation, therefore the function \(e^{-2x}\) is indeed a solution for the differential equation \(y^{(4)}-16 y=0\).

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Key Concepts

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

Higher-Order Derivatives
Higher-order derivatives involve differentiating a function more than once. These are crucial in finding solutions to differential equations.
In the given problem, we start with the function \( y = e^{-2x} \) and need to compute its fourth derivative.
Here's how it works:
  • The first derivative \( y' = -2e^{-2x} \) comes from applying the power rule together with the derivative of an exponential function.
  • The subsequent derivatives follow a pattern. The second derivative \( y'' = 4e^{-2x} \), the third derivative \( y''' = -8e^{-2x} \), and finally the fourth derivative is \( y^{(4)} = 16e^{-2x} \).
Each step requires applying derivative rules repeatedly, ensuring that each constant factor is multiplied accurately.
This process is crucial for solving higher-order differential equations as it allows us to determine if the function satisfies the equation after substitution.
Chain Rule
The chain rule is a fundamental technique in calculus, often used to differentiate composite functions. In the case of the function \( y = e^{-2x} \), the chain rule helps differentiate the exponential function which combines an outer function with an inner linear function.
The chain rule states: \( \frac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x) \).
Applying it:
  • When differentiating \( y = e^{-2x} \), consider \( e^u \) where \( u = -2x \).
  • The derivative of \( e^u \) with respect to \( u \) is \( e^u \), while \( u' = -2 \).
  • Therefore, \( y' = -2e^{-2x} \).
By consistently applying the chain rule, you ensure accuracy in obtaining higher-order derivatives.
This underlines the coordination of differentiating nested functions and maintaining correct sign and coefficients throughout the calculations.
Characteristic Equation
The characteristic equation forms the foundation for solving linear differential equations, especially those with constant coefficients.
In our example, the differential equation \( y^{(4)} - 16y = 0 \) involves finding solutions that satisfy this equation.
  • The characteristic approach involves assuming a solution of the form \( y = e^{rx} \).
  • By substituting this in, we discern that the coefficients of the characteristic polynomial guide us.
For the equation \( y^{(4)} - 16y = 0 \), it implies solving \( r^4 - 16 = 0 \).
Solving this gives roots \( r = 2, -2, 2i, -2i \).
These roots relate directly to the general solution of the differential equation, allowing you to construct specific solutions based on initial conditions or boundary values.
Understanding the characteristic equation equips you to solve a wide array of linear differential equations efficiently.

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Most popular questions from this chapter

Solve the first-order differential equation by any appropriate method. $$ 3\left(y-4 x^{2}\right) d x+x d y=0 $$

The table shows the population \(P\) (in millions) of the United States from 1960 to \(2000 .\) (Source: U.S. Census Bureau) $$ \begin{array}{|l|c|c|c|c|c|} \hline \text { Year } & 1960 & 1970 & 1980 & 1990 & 2000 \\ \hline \text { Population, } P & 181 & 205 & 228 & 250 & 282 \\ \hline \end{array} $$ (a) Use the 1960 and 1970 data to find an exponential model \(P_{1}\) for the data. Let \(t=0\) represent \(1960 .\) (b) Use a graphing utility to find an exponential model \(P_{2}\) for the data. Let \(t=0\) represent \(1960 .\) (c) Use a graphing utility to plot the data and graph both models in the same viewing window. Compare the actual data with the predictions. Which model better fits the data? (d) Estimate when the population will be 320 million.

Slope Fields, (a) use a graphing utility to graph the slope field for the differential equation, (b) find the particular solutions of the differential cquation passing through the given points, and (c) use a graphing utility to graph the particular solutions on the slope field. $$ \begin{array}{ll} \underline{\text { Function }} & \underline{\text { Differential Equation }} \\\ \frac{d y}{d x}+4 x^{3} y=x^{3} &\quad\left(0, \frac{7}{2}\right),\left(0,-\frac{1}{2}\right) \end{array} $$

In your own words, describe the relationship between two families of curves that are mutually orthogonal.

Solve the first-order differential equation by any appropriate method. y d x+(3 x+4 y) d y=0
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