Differential Equation In Exercises 51-56, solve the differential equation. (d r)/(d t)=((1+e^t)^

Differential Equation In Exercises 51-56, solve the...

Key Concepts

Ordinary Differential Equations

An ordinary differential equation (ODE) involves functions of one independent variable and their derivatives. Understanding ODEs is crucial as they model rates of change in various physical and abstract systems.

General Solution

The general solution of an ODE represents the complete set of solutions that satisfy the differential equation, typically expressed in terms of an arbitrary constant. This constant accounts for the infinite family of solutions derived from integrating the equation.

Separable Differential Equations

A separable differential equation is one in which variables can be separated on different sides of the equation, allowing the equation to be rewritten so that one side depends solely on the independent variable and the other side solely on the dependent variable. This property simplifies solving the equation through integration.

Integration Technique

Integration is a fundamental tool for solving differential equations. When an ODE is expressed in a form that allows for integration, the antiderivative is computed on each side, leading to the general solution of the equation. This process may involve techniques such as substitution, partial fractions, or manipulation of exponentials.

Transcript

00:02 Into the differential equation, d r d t equals 1 plus b to the t quantity squared over e to the 3t.

00:16 So we want to use the method of separation of variables.

00:20 So we want to get everything with r on one side and t to the other side.

00:25 So i'm going to multiply both sides by d t.

00:32 So this gives us d r equals 1 plus e to the t squared over e to 3t, dt.

00:49 So next, since we have the t and the r separated, we integrate both sides.

01:00 And so the left -hand side just gives us r plus c equals two.

01:11 And so let's work out this integral over here.

01:19 So i'm going to pick you to equal e to the t.

01:27 So that way i can write this bottom in terms of you and i can write this in terms of you.

01:36 So e to the 3t is the same as e to the t cubed, which is u cubed.

01:48 So we had integral of u cubed on the bottom.

01:58 And so let's see what our d .u would be.

02:00 So d -u is e to the t, d -t.

02:14 So d -t is 1 over e to the t, d -u, which is equal to 1 over u, d -u.

02:32 All right, so we have our u -cube on the bottom.

02:35 On the top, we have 1 plus u squared.

02:40 And instead of d -t, i'm going to put 1 over u, d u so we can combine the bottom we've got u to the fourth and then on top i'm going to expand this out i have one plus two u plus u squared so let's finish this integral on the next page so i had one plus two u plus u squared over u to the fourth because i had a u cubed and a u on the bottom so i just combined d .u...

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