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In Problems I through 12, verify by substitution that each given function is a solution of the given differential equation. Throughout these problems, primes denote derivatives with \(\mathrm{re}\) spect to \(x\). $$ y^{\prime}=y+2 e^{-x} ; y=e^{x}-e^{-x} $$

Short Answer

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

Step by step solution

01

Differentiate the Function

To verify that the function \( y = e^x - e^{-x} \) is a solution to the differential equation \( y' = y + 2e^{-x} \), we first need to find the derivative \( y' \).The given function is:\[ y = e^x - e^{-x} \]Differentiating with respect to \( x \), we apply the rule that the derivative of \( e^x \) is \( e^x \) and the derivative of \(-e^{-x}\) is \( e^{-x} \).Thus, \[ y' = \frac{d}{dx}(e^x) - \frac{d}{dx}(e^{-x}) = e^x + e^{-x} \].
02

Substitute into the Differential Equation

Substitute \( y \) and \( y' \) back into the differential equation \( y' = y + 2e^{-x} \) to verify the solution.Start with the left side:\[ y' = e^x + e^{-x} \]Now substitute for \( y \) in the equation:\[ y + 2e^{-x} = (e^x - e^{-x}) + 2e^{-x} \]Simplifying the right side:\[ e^x - e^{-x} + 2e^{-x} = e^x + e^{-x} \]
03

Verify Equality of Both Sides

Compare both sides of the differential equation:- From the left side, we have: \( y' = e^x + e^{-x} \)- From the right side, we have: \( y + 2e^{-x} = e^x + e^{-x} \)Since the left side equals the right side:\[ e^x + e^{-x} = e^x + e^{-x} \]This equality confirms that the function \( y = e^x - e^{-x} \) satisfies the differential equation \( y' = y + 2e^{-x} \).

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Key Concepts

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

Solution Verification
Solution verification is a crucial step in ensuring that a proposed function truly solves a given differential equation. This process involves checking that, after substitution into the differential equation, both sides match equivalently. Here’s a simple breakdown of how it works in our example.

Firstly, we need to identify the function claimed to solve the differential equation. In this case, we’re given that the function \( y = e^x - e^{-x} \) might solve the differential equation \( y' = y + 2e^{-x} \). The task is to verify that this is indeed true.

The method involves substituting both the function \( y \) and its derivative \( y' \) into the original equation. We solve each side separately:
  • The left side becomes the derivative \( y' \).
  • The right side involves substituting the entire original function and adding any extra components from the differential equation.
If after simplification, both sides equal each other, we confirm the function is indeed a solution. This rigorous verification is key in affirming that no computational errors were made in reasoning.
Derivatives
The concept of derivatives is fundamental in calculus and crucial for solving and verifying differential equations. A derivative represents how a function changes as its input changes.

In our problem, we start with the function \( y = e^x - e^{-x} \), and we need to find its derivative \( y' \). This requires understanding basic derivative rules:
  • The derivative of \( e^x \) with respect to \( x \) is \( e^x \).
  • The derivative of \( -e^{-x} \) is calculated using the chain rule, which gives \( e^{-x} \).
Applying these rules helps us compute \( y' \) as:\[ y' = e^x + e^{-x} \]Finding the derivative is critical for substitution and verifying the solution against the given differential equation. It helps us check dynamic aspects of how changes in \( x \) affect the function \( y \).
Substitution Method
The substitution method in solving differential equations involves replacing variables or expressions with equivalent counterparts to simplify or solve an equation.

In verification exercises, substitution is used to plug the function and its derivative back into the differential equation. This allows one to test if the function satisfies the equation.

With the given problem, the steps are as follows:
  • Substitute \( y' = e^x + e^{-x} \) into the left side of the equation.
  • Substitute \( y = e^x - e^{-x} \) into the right side, along with \( 2e^{-x} \) as shown in \( y + 2e^{-x} \).
Simplifying, we obtain:\[ y + 2e^{-x} = e^x + e^{-x} \]This method offers a transparent way to verify solutions. By directly substituting into both sides, we make it clear that our function appropriately balances the equation under the operations defined in the differential equation. Through systematic substitution, one builds confidence in their solution's correctness.

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Most popular questions from this chapter

An accident at a nuclear power plant has left the surrounding area polluted with radioactive material that decays naturally. The initial amount of radioactive material present is \(15 \mathrm{su}\) (safe units), and 5 months later it is still \(10 \mathrm{su}\). (a) Write a formula giving the amount \(A(t)\) of radioactive material (in su) remaining after \(t\) months. (b) What amount of radioactive material will remain after 8 months? (c) How long-total number of months or fraction thereof -will it be until \(A=1 \mathrm{su}, \mathrm{so}\) it is safe for people to return to the area?

Find general solutions of the differential equations. Primes denote derivatives with respect to \(x\) throughout. $$ y y^{\prime}+x=\sqrt{x^{2}+y^{2}} $$

Find a general solution of each reducible second-order differential equation . Assume \(x, y\) and/or \(y^{\prime}\) positive where helpful (as in Example 11\() .\) $$ y^{\prime \prime}=\left(x+y^{\prime}\right)^{2} $$

A certain piece of dubious information about phenylethylamine in the drinking water began to spread one day in a city with a population of 100,000 . Within a week, 10,000 people had heard this rumor. Assume that the rate of increase of the number who have heard the rumor is proportional to the number who have not yet heard it. How long will it be until half the population of the city has heard the rumor?

Verify that the given differential equation is exact; then solve it. $$ \left(x+\tan ^{-1} y\right) d x+\frac{x+y}{1+y^{2}} d y=0 $$

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