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Chapter 10: Problem 20

Solve the following differential equations with the given initial conditions. $$ y^{\prime}=y^{2}-e^{3 t} y^{2}, y(0)=1 $$

Short Answer

Expert verified
Separate variables, integrate both sides, and use the initial condition to find the solution.

Step by step solution

01

Write the Differential Equation

The given differential equation is \[ y' = y^2 - e^{3t} y^2 \] with the initial condition \[ y(0) = 1. \]
02

Simplify the Differential Equation

Factor out the common term \( y^2 \) on the right-hand side: \[ y' = y^2 (1 - e^{3t}) \]
03

Separate Variables

Divide both sides by \( y^2 (1 - e^{3t}) \) and multiply both sides by \( dt \): \[ \frac{dy}{y^2 (1 - e^{3t})} = dt \]
04

Integrate Both Sides

Integrate the left side with respect to \( y \) and the right side with respect to \( t \): \[ \int \frac{1}{y^2 (1 - e^{3t})} dy = \int dt \] **Note:** Here, it might be useful to use partial fractions or a substitution method for the integral on the left.
05

Apply Initial Conditions

After integrating, introduce the constant of integration and apply the initial condition \( y(0) = 1 \) to solve for this constant.
06

Simplify the Solution

Express the solution in terms of \( y(t) \) and verify it satisfies the initial condition.

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Key Concepts

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

initial conditions
An initial condition in differential equations specifies the value of the function at a particular point. This information helps to determine the constant of integration when solving the equation. In our problem, the initial condition is given as \[ y(0) = 1 \].

This means when the value of \( t \) is 0, the function \( y(t) \) must equal 1. Including this condition is crucial because differential equations can have multiple solutions, but the initial condition helps narrow it down to one specific solution.

If you neglect the initial condition, your solution might not satisfy the requirements of the problem.Remember:
  • Always incorporate initial conditions after integrating.
  • Initial conditions help to find the specific solution in the family of solutions.
separation of variables
Separation of variables is a method used to solve differential equations where variables can be separated on different sides of the equation. This method works only if the equation can be rearranged to isolate each variable separately.

In our exercise, our goal is to separate \( y \) and \( t \) to opposite sides of the equation. Starting from the simplified form \[ y' = y^2 (1 - e^{3t}) \],we divide both sides by \( y^2 (1 - e^{3t}) \) and multiply by \( dt \) to get:

\[ \frac{dy}{y^2 (1 - e^{3t})} = dt \]
Now we have successfully separated the variables, making it easier to integrate both sides.
integration methods
Integration is the next step after separating the variables. We need to integrate both sides of the equation to solve for \( y \). For complex integrals, you might use techniques like partial fractions or substitution.

Our integral:
\[ \int \frac{1}{y^2 (1 - e^{3t})} dy = \int dt \]
To solve the integral on the left, we might need to use substitution, which simplifies the integrand. After integrating both sides, we add the constant of integration to the right side:
\[ \text{Result of left integral} = t + C \]
Here, \( C \) is the constant of integration.
differential equations
Differential equations involve functions and their rates of change. They are used to describe various phenomena in physics, engineering, economics, and other disciplines.

In our case, the given differential equation is \[ y' = y^2 - e^{3t} y^2 \],
where \( y' \) represents the derivative of \( y \) with respect to \( t \).

Key points to remember about differential equations:
  • Differential equations can be ordinary (ODE) or partial (PDE).
  • Ordinary differential equations (ODEs) involve functions of one variable and their derivatives.
  • Partial differential equations (PDEs) involve functions of multiple variables and their partial derivatives.
  • Solving differential equations often involves finding a general solution and then applying initial conditions or boundary conditions.

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Most popular questions from this chapter

In an autocatalytic reaction, one substance is converted into a second substance in such a way that the second substance catalyzes its own formation. This is the process by which trypsinogen is converted into the enzyme trypsin. The reaction starts only in the presence of some trypsin, and each molecule of trypsinogen yields one molecule of trypsin. The rate of formation of trypsin is proportional to the product of the amounts of the two substances present. Set up the differential equation that is satisfied by \(y=f(t)\), the amount (number of molecules) of trypsin present at time \(t .\) Sketch the solution. For what value of \(y\) is the reaction proceeding the fastest? [Note: Letting \(M\) be the total amount of the two substances, the amount of trypsinogen present at time \(t\) is \(M-f(t) .]\)

\(y^{\prime}=k(y-A)\), where \(k<0\) and \(A>0 .\) Sketch solutions where \(y(0)A\).

In certain learning situations a maximum amount, \(M\), of information can be learned, and at any time, the rate of learning is proportional to the amount yet to be learned. Let \(y=f(t)\) be the amount of information learned up to time \(t\). Construct and solve a differential equation that is satisfied by \(f(t)\).

The National Automobile Dealers Association reported that the average retail selling price of a new vehicle was $$\$ 30,303$$ in 2012 . A person purchased a new car at the average price and financed the entire amount. Suppose that the person can only afford to pay $$\$ 500$$ per month. Assume that the payments are made at a continuous annual rate and that interest is compounded continuously at the rate of \(3.5 \%\). (Source: The National Automobile Dealers Association, www.nada.com.) (a) Set up a differential equation that is satisfied by the amount \(f(t)\) of money owed on the car loan at time \(t\). (b) How long will it take to pay off the car loan?

A Savings Account A person deposits $$\$ 10,000$$ in a bank account and decides to make additional deposits at the rate of \(A\) dollars per year. The bank compounds interest continuously at the annual rate of \(6 \%\), and the deposits are made continuously into the account. (a) Set up a differential equation that is satisfied by the amount \(f(t)\) in the account at time \(\bar{t}\). (b) Determine \(f(t)\) (as a function of \(A\) ). (c) Determine \(A\) if the initial deposit is to double in 5 years.
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