Chapter 6: Problem 20
Find the function \(y=f(t)\) passing through the point \((0,10)\) with the given first derivative. Use a graphing utility to graph the solution. $$ \frac{d y}{d t}=\frac{3}{4} y $$
Short Answer
Expert verified
The function that passes through the point (0,10) and has a derivative of \((3/4)y\) is \(y = 10e^{(3/4)t}\)
Step by step solution
01
Identify the differential equation
Given \(dy/dt = (3/4)y\) which is a first order differential equation and the point (0,10). The equation states that the derivative of y with respect to t equals (3/4) times y.
02
Solve the differential equation
Separate the variables by dividing both sides by y and multiplying both sides by dt. This gives: \[1/y dy = (3/4)dt\] Now, integrate both sides with respect to their variables. The integral of 1/y dy is ln|y| and the integral of (3/4)dt is (3/4)t. So by solving the integrals, it gives: \[ln|y| = (3/4)t + C_1\] Where \(C_1\) is the constant of integration.
03
Solve for y
Get y by itself by exponentiating both sides to remove the natural logarithm. This gives: \[y = e^{(3/4)t + C_1}\] By properties of exponents, this simplifies to: \[y = e^{C_1}e^{(3/4)t}\] Rewrite \(e^{C_1}\) as \(C_2\) (another constant) to simplify. This gives: \[y = C_2e^{(3/4)t}\]
04
Solve for the constant using the initial condition
Substitute the point (0,10) into the equation from Step 3 to solve for \(C_2\). When t=0, y=10. Therefore: \[10 = C_2e^0\] Which gives \(C_2 = 10\). So the function is: \[y = 10e^{(3/4)t}\]
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Exponential Functions
Exponential functions are mathematical expressions where a constant base is raised to a variable exponent. This feature causes these functions to exhibit rapid growth or decay patterns.
- In the equation we solved, the exponential function appears as part of the solution: \(y = 10e^{(3/4)t}\).
- The base \(e\) is a mathematical constant approximately equal to 2.71828. It is often used in continuous growth processes.
- This type of function is crucial in modeling phenomena where change occurs continuously, such as population growth or radioactive decay.
Initial Value Problems
An initial value problem is a type of differential equation along with a specified value, known as the initial condition, for the function at a particular point.
- In the exercise, we are given the initial condition \((0, 10)\).
- This condition helps determine the specific solution out of the infinite solutions to the differential equation.
Separation of Variables
Separation of variables is a common and effective technique for solving ordinary differential equations, particularly when the equation can be structured into a form that isolates each variable for integration.
- Initially, the given differential equation was \(\frac{d y}{d t}=\frac{3}{4} y\).
- By rewriting this as \(\frac{1}{y} dy = \frac{3}{4} dt\), we isolate each variable on a different side.
- This separation allows us to integrate both sides individually, leading to the function's solution.