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Chapter 8: Problem 7

Verify that the given function \(y\) is a solution of the differential equation that follows it. Assume that \(C\) is an arbitrary constant. $$y(t)=C e^{-5 t} ; y^{\prime}(t)+5 y(t)=0$$

Short Answer

Expert verified
To verify that the given function $$y(t)=Ce^{-5t}$$ is a solution to the differential equation $$y^{\prime}(t)+5y(t)=0$$, we found the first derivative $$y^{\prime}(t) = -5Ce^{-5t}$$. Then, we substituted both $$y(t)$$ and $$y^{\prime}(t)$$ into the differential equation, and the resulting equation simplified to $$0=0$$, which is true. Thus, we can conclude that $$y(t)=Ce^{-5t}$$ is a solution to the given differential equation.

Step by step solution

01

Find the first derivative of the given function

Find the first derivative of the given function $$y(t) = Ce^{-5t}$$. Using the chain rule, we have: $$y^{\prime}(t) = -5Ce^{-5t}$$
02

Substitute the functions into the differential equation

Now, we will substitute the functions $$y(t)$$ and $$y^{\prime}(t)$$ into the given differential equation: $$y^{\prime}(t) + 5y(t) = 0$$ We substitute the expressions: $$(-5Ce^{-5t}) + 5(Ce^{-5t}) = 0$$
03

Check if the resulting equation is true

Let's simplify the equation from Step 2 and see if it holds true: $$-5Ce^{-5t} + 5Ce^{-5t} = 0$$ Since the two terms are equal and have opposite signs, they cancel each other out, resulting in: $$0 = 0$$ Since this equation is true, we can conclude that the given function $$y(t) = Ce^{-5t}$$ is indeed a solution to the differential equation $$y^{\prime}(t) + 5y(t) = 0$$.

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Key Concepts

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

Chain Rule
When tackling derivatives, the chain rule is a powerful tool you will often use. It's a method for finding the derivative of a composite function.
Think of it as a way to "unwrap" functions.
For instance, in our exercise, we had the function \(y(t) = Ce^{-5t}\).
This function is composite, meaning it contains an outer function and an inner function.
  • The outer function here is \(Ce^{u}\), where \(u = -5t\).
  • The inner function is \(-5t\).
To differentiate with the chain rule, you take the derivative of the outer function with respect to the inner function, and then multiply by the derivative of the inner function with respect to \(t\).
In our case:
  • The derivative of the outer function \(Ce^{u}\) with respect to \(u\) is \(Ce^{u}\).
  • The derivative of the inner function \(-5t\) with respect to \(t\) is \(-5\).
Thus, the derivative \(y'(t)\) is \(-5Ce^{-5t}\).
This technique is a key part of many calculus problems, especially those involving differential equations.
First Derivative
Finding the first derivative is a fundamental step in solving differential equations.
The first derivative, often denoted as \(y'(t)\), represents the rate of change of the function with respect to \(t\).
In many contexts, this can be interpreted as the velocity if \(y(t)\) is considered as a position function.
For our function \(y(t) = Ce^{-5t}\), we applied the chain rule to find \(y'(t)\).
This process involved differentiating the composite function, resulting in:
  • \(y'(t) = -5Ce^{-5t}\)
This expression shows how the function \(y(t)\) changes at any point \(t\).
Understanding the incremental changes of functions via their first derivatives forms the basis for analyzing and solving differential equations.
Solution Verification
Once you have calculated the first derivative, the next crucial step is to verify if the given function satisfies the differential equation.
This involves substituting both the original function and its derivative back into the differential equation.
For our exercise, you need to
  • substitute \(y(t) = Ce^{-5t}\) and \(y'(t) = -5Ce^{-5t}\) into the equation: \(y'(t) + 5y(t) = 0\).
  • Simplify: \((-5Ce^{-5t}) + 5(Ce^{-5t}) = 0\).
Notice that when you simplify, both terms \(-5Ce^{-5t}\) and \(5Ce^{-5t}\) cancel each other out perfectly.
The resulting statement \(0 = 0\) confirms that the original function indeed satisfies the differential equation.
Verification like this is essential to ensure that your solution is correct.
It ensures no steps were missed and validates that the function accurately models the differential equation given.
This approach is widely used in calculus and real-world applications to confirm the validity of solutions.

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Most popular questions from this chapter

A physiological model A common assumption in modeling drug assimilation is that the blood volume in a person is a single compartment that behaves like a stirred tank. Suppose that the blood volume is a four-liter tank that initially has a zero concentration of a particular drug. At time \(t=0,\) an intravenous line is inserted into a vein (into the tank) that carries a drug solution with a concentration of \(500 \mathrm{mg} / \mathrm{L} .\) The inflow rate is \(0.06 \mathrm{L} / \mathrm{min}\). Assume that the drug is quickly mixed thoroughly in the blood and that the volume of blood remains constant. a. Write an initial value problem that models the mass of the drug in the blood, for \(t \geq 0\) b. Solve the initial value problem and graph both the mass of the drug and the concentration of the drug. c. What is the steady-state mass of the drug in the blood? d. After how many minutes does the drug mass reach \(90 \%\) of its steady-state level?

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An endowment is an investment account in which the balance ideally remains constant and withdrawals are made on the interest earned by the account. Such an account may be modeled by the initial value problem \(B^{\prime}(t)=r B-m,\) for \(t \geq 0,\) with \(B(0)=B_{0} .\) The constant \(r>0\) reflects the annual interest rate, \(m>0\) is the annual rate of withdrawal, \(B_{0}\) is the initial balance in the account, and \(t\) is measured in years. a. Solve the initial value problem with \(r=0.05, m=\$ 1000 /\) year, and \(B_{0}=\$ 15,000 .\) Does the balance in the account increase or decrease? b. If \(r=0.05\) and \(B_{0}=\$ 50,000,\) what is the annual withdrawal rate \(m\) that ensures a constant balance in the account? What is the constant balance?
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