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Find the general solution to the differential equation $$ \frac{d y}{d t}=2 t $$

Short Answer

Expert verified
The general solution is \( y = t^2 + C \).

Step by step solution

01

Recognize the Equation Type

The given equation \( \frac{d y}{d t} = 2t \) is a first-order differential equation and a separable equation.
02

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.
03

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.
04

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.

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Key Concepts

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

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.
  • 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.
Ensure you understand whether any initial conditions have been specified. Initial conditions can convert the general solution into a specific solution. Otherwise, the solution will be expressed in terms of a constant \( C \), representing a family of possible solutions.
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.
  • 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.
In the example \( \frac{d y}{d t} = 2t \), notice how it's split into \( dy = 2t \, dt \), simplifying the integration process of both sides independently.
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.
  • 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).
The constant \( C \) signifies that there are multiple solutions differing only by a vertical shift.
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.

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Most popular questions from this chapter

A yam is put in a \(200^{\circ} \mathrm{C}\) oven and heats up according to the differential equation \(\frac{d H}{d t}=-k(H-200), \quad\) for \(k\) a positive constant. (a) If the yam is at \(20^{\circ} \mathrm{C}\) when it is put in the oven, solve the differential equation. (b) Find \(k\) using the fact that after 30 minutes the temperature of the yam is \(120^{\circ} \mathrm{C}\).

A quantity \(W\) satisfies the differential equation $$ \frac{d W}{d t}=5 W-20 $$ (a) Is \(W\) increasing or decreasing at \(W=10 ? W=2\) ? (b) For what values of \(W\) is the rate of change of \(W\) equal to zero?

In some chemical reactions, the rate at which the amount of a substance changes with time is proportional to the amount present. For example, this is the case as \(\delta\) glucono-lactone changes into gluconic acid. (a) Write a differential equation satisfied by \(y\), the quantity of \(\delta\) -glucono-lactone present at time \(t\). (b) If 100 grams of \(\delta\) -glucono-lactone is reduced to \(54.9\) grams in one hour, how many grams will remain after 10 hours?

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 absence of the other. (b) Describe in words how the species interact with one another. Give reasons why the populations might behave as described by the equations. Suggest species that might interact in that way. $$ \begin{array}{l} \frac{d x}{d t}=0.01 x-0.05 x y \\ \frac{d y}{d t}=0.2 y-0.08 x y \end{array} $$

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 absence of the other. (b) Describe in words how the species interact with one another. Give reasons why the populations might behave as described by the equations. Suggest species that might interact in that way. $$ \begin{array}{l} \frac{d x}{d t}=0.2 x \\ \frac{d y}{d t}=0.4 x y-0.1 y \end{array} $$

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