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Determine by inspection at least one solution of the given differential equation. That is, use your knowledge of derivatives to make an intelligent guess. Then test your hypothesis. $$ y^{\prime}=y $$

Short Answer

Expert verified
\( y(x) = e^x \) is a solution to the differential equation.

Step by step solution

01

Understanding the Differential Equation

We have the differential equation \( y' = y \). This means the derivative of \( y \) with respect to \( x \) is equal to \( y \) itself. Our goal is to find a function \( y(x) \) that satisfies this condition.
02

Making an Intelligent Guess

One well-known function whose derivative is itself is the exponential function. Thus, we hypothesize that \( y(x) = e^x \) may be a solution to the differential equation.
03

Differentiating the Hypothesized Solution

To test our hypothesis, we differentiate \( y(x) = e^x \) with respect to \( x \). The derivative of \( e^x \) is also \( e^x \). Thus, \( y' = e^x \).
04

Verifying the Hypothesis

We compare the derivative \( y' = e^x \) with the original differential equation \( y = e^x \). Since both sides are equal, \( y(x) = e^x \) is indeed a solution to the differential equation \( y' = y \).

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Key Concepts

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

Exponential Function
An exponential function is a type of mathematical function where the variable appears as the exponent of a constant base. In the context of differential equations, the exponential function often appears due to its unique properties. For instance, in our exercise, we are looking at the differential equation \( y' = y \). One of the most common functions that meets this criterion is the exponential function of the form \( y(x) = e^{kx} \), where \( e \) is the base of the natural logarithm, approximately equal to 2.718, and \( k \) is a constant multiplier.

This function is significant because its derivative retains a similar form, meaning \( \frac{d}{dx} e^{kx} = ke^{kx} \). Consequently, when \( k = 1 \), the derivative of \( e^x \) is \( e^x \) itself, making it a straightforward solution to the differential equation \( y' = y \). This unique characteristic makes exponential functions particularly important in the context of exponential growth or decay processes, commonly modeled in fields such as physics, biology, and finance.
Derivative
The derivative represents a fundamental concept in calculus that measures the rate of change of a function. For a given function \( y(x) \), the derivative \( y'(x) \) is calculated as the limit of the change in \( y \) over the change in \( x \) as those changes approach zero. In simpler terms, it tells you how steep or flat the function is at any given point, indicating how quickly the function's value is rising or falling.

In the provided exercise, understanding the derivative is crucial. The equation \( y' = y \) tells us that the rate of change of \( y \) is directly proportional to its current value. This implies that if you plot the function and its derivative, they overlap, highlighting the unique nature of solutions like the exponential function. This understanding allows us to identify plausible solutions to differential equations, even by inspection. The notion of derivatives helps us test hypotheses about these solutions by comparing the calculated derivative to the original equation.
Solution Verification
Solution verification involves checking whether a proposed function truly satisfies a given differential equation. This is a critical step to ensure that the guessed solution is not just plausible, but also correct.

For our differential equation \( y' = y \), we guessed that \( y(x) = e^x \) might be a solution. To verify this, we differentiate \( y(x) = e^x \) to obtain \( y' = e^x \). By comparing this result to the original equation \( y = e^x \), we observe that both expressions on either side of the equation are equal. This confirms that our hypothesis was correct.

Verification is essential because it confirms the validity of the function across any value within its domain. It ensures that the solution doesn’t just meet the equation under specific conditions, but comprehensively satisfies it for all points of interest. This step underscores the critical importance of both analytical skills and understanding of differential components in mathematics.

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Most popular questions from this chapter

Early one morning it began to snow at a constant rate. At 7 A.M. a snowplow set off to clear a road. By 8 A.M. it had traveled 2 miles, but it took two more hours (until 10 A.M.) for the snowplow to go an additional 2 miles. (a) Let \(t=0\) when it began to snow, and let \(x\) denote the distance traveled by the snowplow at time \(t\). Assuming that the snowplow clears snow from the road at a constant rate (in cubic feet per hour, say), show that $$ k \frac{d x}{d t}=\frac{1}{t} $$ where \(k\) is a constant. (b) What time did it start snowing? (Answer: 6 A.M.)

A water tank has the shape obtained by revolving the curve \(y=x^{4 / 3}\) around the \(y\) -axis. A plug at the bottom is removed at 12 noon, when the depth of water in the tank is \(12 \mathrm{ft}\). At 1 P.M. the depth of the water is \(6 \mathrm{ft}\). When will the tank be empty?

A river \(100 \mathrm{ft}\) wide is flowing north at \(w\) feet per second. A dog starts at \((100,0)\) and swims at \(v_{0}=4 \mathrm{ft} / \mathrm{s}\), always heading toward a tree at \((0,0)\) on the west bank directly across from the dog's starting point. (a) If \(w=2 \mathrm{ft} / \mathrm{s}\), show that the dog reaches the tree. (b) If \(w=4 \mathrm{ft} / \mathrm{s}\) show that the dog reaches instead the point on the west bank \(50 \mathrm{ft}\) north of the tree. (c) If \(w=6 \mathrm{ft} / \mathrm{s}\), show that the dog never reaches the west bank.

(a) Show that \(y(x)=C x^{4}\) defines a one-parameter family of differentiable solutions of the differential equation \(x y^{\prime}=4 y\) (Fig. 1.1.9). (b) Show that $$ y(x)=\left\\{\begin{aligned} -x^{4} & \text { if } x<0, \\ x^{4} & \text { if } x \geqq 0 \end{aligned}\right. $$ defines a differentiable solution of \(x y^{\prime}=4 y\) for all \(x\), but is not of the form \(y(x)=C x^{4}\). (c) Given any two real numbers \(a\) and \(b\), explain why -in contrast to the situation in part (c) of Problem 47 - there exist infinitely many differentiable solutions of \(x y^{\prime}=4 y\) that all satisfy the condition \(y(a)=b\)

Find a general solution of each reducible second-order differential equation. Assume \(x, y\) and/or \(y^{\prime}\) positive where helpful (as in Example I1). $$ y^{\prime \prime}+4 y=0 $$

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