Problem 1
Question
Find the general solution to the differential equation $$ \frac{d y}{d t}=2 t $$
Step-by-Step Solution
Verified Answer
The general solution is \( y = t^2 + C \).
1Step 1: Recognize the Equation Type
The given equation \( \frac{d y}{d t} = 2t \) is a first-order differential equation and a separable equation.
2Step 2: Separate Variables
Rewrite the equation by separating the variables: \( dy = 2t \, dt \). This separation allows us to consider both sides in terms of `y` and `t` independently for integration.
3Step 3: Integrate Both Sides
Integrate both sides of the equation: \( \int dy = \int 2t \, dt \). This results in \( y = t^2 + C \), where \( C \) is the constant of integration.
4Step 4: Write the General Solution
The general solution to the differential equation is \( y = t^2 + C \). This solution includes the constant \( C \), accounting for the family of solutions.
Key Concepts
Understanding First-order Differential EquationsSeparable Equations and Their RoleIntegration: Solving and Finding the General Solution
Understanding First-order Differential Equations
First-order differential equations are foundational in the study of calculus and differential equations. These types of equations involve derivatives of a function that are of first order or degree, meaning they include terms like \( \frac{d y}{d t} \). In simpler terms, it represents the rate of change of the function \( y \) with respect to the variable \( t \).
A key point to note is that first-order differential equations can often model real-world situations, such as population growth, cooling processes, or chemical reactions.
A key point to note is that first-order differential equations can often model real-world situations, such as population growth, cooling processes, or chemical reactions.
- They involve one dependent variable (\( y \)) and one independent variable (\( t \)).
- The order of the differential equation is determined by the highest order of derivative present.
- In applications, the solution to these equations helps us predict the behavior of the system over time.
Separable Equations and Their Role
Separable equations are a specific type of first-order differential equations that allow for an intuitive solution method. The principle behind separable equations is the potential to rewrite the equation such that all terms involving one variable are on one side of the equation, and all terms involving the other variable are on the opposite side.
This restructuring helps in simplifying the integration process, as each side contains only one single variable after separation.
Grasping this concept enables you to solve a broad range of differential equations efficiently!
This restructuring helps in simplifying the integration process, as each side contains only one single variable after separation.
- Often take the form \( \frac{d y}{d t} = g(t)h(y) \).
- The separation of variables results in \( \frac{1}{h(y)} dy = g(t) dt \).
- By isolating \( y \) in terms of \( t \), the equation becomes easier to manage.
Grasping this concept enables you to solve a broad range of differential equations efficiently!
Integration: Solving and Finding the General Solution
Integration is a vital concept when solving differential equations, especially separable ones. After separating the variables, the next step involves integrating both sides of the equation to find the general solution. The integration process involves finding the antiderivative of each side.
Understanding how to integrate correctly is essential in solving and interpreting these equations. It forms a key part of your toolkit when dealing with more complex problems in calculus.
- Left-hand integration: \( \int dy \) results in \( y \).
- Right-hand integration: \( \int 2t \, dt \) results in \( t^2 + C' \), where \( C' \) represents the constant of integration.
- The general solution takes the form \( y = t^2 + C \) (with \( C \) being arbitrary).
Understanding how to integrate correctly is essential in solving and interpreting these equations. It forms a key part of your toolkit when dealing with more complex problems in calculus.
Other exercises in this chapter
Problem 1
Find particular solutions \(\frac{d H}{d t}=3(H-75), \quad H=0\) when \(t=0\)
View solution Problem 1
Find solutions to the differential equations, subject to the given initial condition. $$ \frac{d w}{d r}=3 w, \quad w=30 \text { when } r=0 $$
View solution Problem 2
Give the rates of growth of two populations, \(x\) and \(y\), measured in thousands. (a) Describe in words what happens to the population of each species in the
View solution Problem 2
Find particular solutions \(\frac{d y}{d t}=0.5(y-200), \quad y=50\) when \(t=0\)
View solution