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Chapter 6: Problem 2

Verify that the function \(y\) satisfies the given differential equation. $$ \begin{array}{l} y=e^{5 x}-4 e^{x}+1 \\ y^{\prime \prime}-6 y^{\prime}+5 y=5 \end{array} $$

Short Answer

Expert verified
The function satisfies the differential equation; substitution leads to true equality.

Step by step solution

01

Differentiate Function y

First, we need to find the first derivative \(y'\) of the function \(y = e^{5x} - 4e^x + 1\). Using the power rule and chain rule, we get:\[y' = \frac{d}{dx}(e^{5x}) - 4\frac{d}{dx}(e^x) + \frac{d}{dx}(1) = 5e^{5x} - 4e^x.\]
02

Differentiate Again to Find y''

Differentiate the first derivative \(y' = 5e^{5x} - 4e^x\) to find \(y''\):\[y'' = \frac{d}{dx}(5e^{5x}) - 4\frac{d}{dx}(e^x) = 25e^{5x} - 4e^x.\]
03

Substitute y, y', y'' into Differential Equation

Substitute \(y = e^{5x} - 4e^x + 1\), \(y' = 5e^{5x} - 4e^x\), and \(y'' = 25e^{5x} - 4e^x\) into the differential equation \(y'' - 6y' + 5y = 5\):\[(25e^{5x} - 4e^x) - 6(5e^{5x} - 4e^x) + 5(e^{5x} - 4e^x + 1) = 5.\]
04

Simplify the Left Side of Equation

Simplify the expression from Step 3:\[25e^{5x} - 4e^x - 30e^{5x} + 24e^x + 5e^{5x} - 20e^x + 5.\]Combine like terms:\[(25e^{5x} - 30e^{5x} + 5e^{5x}) + (-4e^x + 24e^x - 20e^x) + 5 = 5.\]This simplifies to:\[0e^{5x} + 0e^x + 5 = 5.\]
05

Verify the Equation Holds True

The simplified equation from Step 4 is:\[5 = 5,\]which is a true statement. Thus, the function \(y = e^{5x} - 4e^x + 1\) satisfies the differential equation \(y'' - 6y' + 5y = 5\).

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Key Concepts

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

First Derivative
The first step in solving differential equations often involves finding the first derivative of a given function. To find the first derivative, you need to differentiate the function with respect to its variable. For example, if you have a function given by \( y = e^{5x} - 4e^x + 1 \), you apply rules of differentiation.
  • The derivative of \( e^{5x} \) involves using the chain rule, giving us \( 5e^{5x} \).
  • Next, the derivative of \( -4e^x \) is straightforward, as it results in \( -4e^x \).
  • The derivative of the constant \( 1 \) is zero, since constants vanish upon differentiation.
Thus, the first derivative of the function is \( y' = 5e^{5x} - 4e^x \). Calculating the first derivative correctly is crucial, as it sets the pathway for further differentiation.
Second Derivative
Once you have the first derivative, the next step is often to find the second derivative, especially when dealing with second-order differential equations.Differentiating the first derivative yields the second derivative. Take the example of the function's first derivative \( y' = 5e^{5x} - 4e^x \). To find the second derivative \( y'' \), you differentiate again:
  • The derivative of \( 5e^{5x} \) is \( 25e^{5x} \), accomplished using the chain rule once more.
  • The derivative of \( -4e^x \) remains \( -4e^x \).
Combining these, the second derivative is \( y'' = 25e^{5x} - 4e^x \). The accuracy of this computation affects the verification of the differential equation.
Chain Rule
The chain rule is a vital tool in calculus, particularly when differentiating composite functions.In the context of exponential functions, such as \( e^{5x} \), the chain rule tells us to first differentiate the outer function, \( e^u \), and then multiply by the derivative of the inner function, in this example, \( 5x \). Therefore, the chain rule implies:
  • Differentiate \( e^{5x} \) to get \( e^{5x} \), then multiply by the derivative of \( 5x \) which is 5.
This rule facilitates correct differentiation in steps leading to derivatives like \( 5e^{5x} \) and subsequently \( 25e^{5x} \). Being familiar with the chain rule allows you to tackle complex derivative problems efficiently.
Verifying Solutions
Verification is the final, essential step in confirming the function as a solution to a differential equation.In this process, you replace \( y \), \( y' \) and \( y'' \) with their calculated expressions in the given differential equation. Considering our equation \( y'' - 6y' + 5y = 5 \), you would:
  • Plug \( y = e^{5x} - 4e^x + 1 \) into the equation.
  • Substitute \( y' = 5e^{5x} - 4e^x \).
  • Use \( y'' = 25e^{5x} - 4e^x \).
You simplify the expression, combining like terms, and check if the equation holds true, e.g. resulting in \( 5 = 5 \). If true, the function satisfies the differential equation, confirming your solution is correct.

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Most popular questions from this chapter

Evaluate each definite integral using integration by parts. (Leave answers in exact form.) $$ \int_{0}^{2} x e^{x} d x $$

Integration by parts often involves finding integrals like the following when integrating \(d v\) to find \(v\). Find the following integrals without using integration by parts (using formulas 1 through 7 on the inside back cover). Be ready to find similar integrals during the integration by parts procedure. $$ \int \sqrt{x} d x $$

For each differential equation and initial condition: a. Use SLOPEFLD or a similar program to graph the slope field for the differential equation on the window \([-5,5]\) by \([-5,5]\). b. Sketch the slope field on a piece of paper and draw a solution curve that follows the slopes and that passes through the point \((0,2)\). c. Solve the differential equation and initial condition. d. Use SLOPEFLD or a similar program to graph the slope field and the solution that you found in part (c). How good was the sketch that you made in part (b) compared with the solution graphed in part (d)? $$ \left\\{\begin{array}{l} \frac{d y}{d x}=\frac{6 x^{2}}{y^{4}} \\ y(0)=2 \end{array}\right. $$

Find in two different ways and check that your answers agree. \(\int x(x-2)^{5} d x\) a. Use integration by parts. b. Use the substitution \(u=x-2 \quad\) (so \(x\) is replaced by \(u+2\) ) and then multiply out the integrand.

Suppose that you now have $$\$ 6000$$, you expect to save an additional $$\$ 3000$$ during each year, and all of this is deposited in a bank paying \(10 \%\) interest compounded continuously. Let \(y(t)\) be your bank balance (in thousands of dollars) \(t\) years from now. a. Write a differential equation that expresses the fact that your balance will grow by 3 (thousand dollars) and also by \(10 \%\) of itself. [Hint: See Example 7.] b. Write an initial condition to say that at time zero the balance is 6 (thousand dollars). c. Solve your differential equation and initial condition. d. Use your solution to find your bank balance \(t=25\) years from now.
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