0 Solve the seperable differential equation dy = 2y dx & find the particular solution satisfying the initial condition y(0) = -2 DCO → y(x) = ? >
0 Solve the seperable differential equation dy = 2y dx & find the particular solution satisfying the initial condition y(0) = -2 DCO → y(x) = ? >
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Questions in picture below. Can I get readble handwriting please? :(
![### Differential Equations and Cost Analysis: Problems and Solutions
#### Problem 1:
**Solve the separable differential equation:**
\[ \frac{dy}{dx} = 0.2y \]
\begin{itemize}
\item **Find the particular solution satisfying the initial condition** \( y(0) = -2 \)
\[ y(x) = ? \]
\end{itemize}
#### Problem 2:
**A company's marginal cost function is:**
\[ \frac{120}{x}, \text{ where } x \text{ is the number of units.} \]
\begin{itemize}
\item **Find the total cost of the first 49 units (of increasing production from \( x = 0 \) to \( x = 49 \))**
\[ \text{Total cost} = ? \]
\end{itemize}
These problems involve the application of differential equations and integrals to practical scenarios. For Problem 1, the focus is on solving a basic separable differential equation and finding a particular solution that meets a given initial condition. Problem 2 involves calculating the total cost based on a given marginal cost function, integrating to find the total cost over a specified range.
### Detailed Steps and Solutions
#### Solution to Problem 1:
To solve the separable differential equation:
\[ \frac{dy}{dx} = 0.2y \]
We can rewrite this as:
\[ \frac{1}{y} \frac{dy}{dx} = 0.2 \]
Integrating both sides with respect to \( x \):
\[ \int \frac{1}{y} \, dy = \int 0.2 \, dx \]
This gives us:
\[ \ln|y| = 0.2x + C \]
Solving for \( y \), we get:
\[ y = e^{0.2x + C} = e^C e^{0.2x} \]
Let \( e^C = C_1 \) (a constant), so:
\[ y = C_1 e^{0.2x} \]
To find the particular solution with the initial condition \( y(0) = -2 \):
\[ -2 = C_1 e^{0.2 \cdot 0} \]
\[ -2 = C_1 \]
Therefore, the particular solution is](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F469aef3b-31ef-41c9-aa0d-266ee90c84d9%2Fd5dec7c2-5730-4169-a5d0-9a079ce32371%2Fewyfmbk_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Differential Equations and Cost Analysis: Problems and Solutions
#### Problem 1:
**Solve the separable differential equation:**
\[ \frac{dy}{dx} = 0.2y \]
\begin{itemize}
\item **Find the particular solution satisfying the initial condition** \( y(0) = -2 \)
\[ y(x) = ? \]
\end{itemize}
#### Problem 2:
**A company's marginal cost function is:**
\[ \frac{120}{x}, \text{ where } x \text{ is the number of units.} \]
\begin{itemize}
\item **Find the total cost of the first 49 units (of increasing production from \( x = 0 \) to \( x = 49 \))**
\[ \text{Total cost} = ? \]
\end{itemize}
These problems involve the application of differential equations and integrals to practical scenarios. For Problem 1, the focus is on solving a basic separable differential equation and finding a particular solution that meets a given initial condition. Problem 2 involves calculating the total cost based on a given marginal cost function, integrating to find the total cost over a specified range.
### Detailed Steps and Solutions
#### Solution to Problem 1:
To solve the separable differential equation:
\[ \frac{dy}{dx} = 0.2y \]
We can rewrite this as:
\[ \frac{1}{y} \frac{dy}{dx} = 0.2 \]
Integrating both sides with respect to \( x \):
\[ \int \frac{1}{y} \, dy = \int 0.2 \, dx \]
This gives us:
\[ \ln|y| = 0.2x + C \]
Solving for \( y \), we get:
\[ y = e^{0.2x + C} = e^C e^{0.2x} \]
Let \( e^C = C_1 \) (a constant), so:
\[ y = C_1 e^{0.2x} \]
To find the particular solution with the initial condition \( y(0) = -2 \):
\[ -2 = C_1 e^{0.2 \cdot 0} \]
\[ -2 = C_1 \]
Therefore, the particular solution is
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