Use the method of Exercise 8.1.7 to solve the following initial value problems: (a) (d u)/(d t)=

Use the method of Exercise 8.1.7 to solve the following...

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Key Concepts

First-Order Linear Differential Equations

These are differential equations of the form du/dt + p(t)u = q(t), where the unknown function u and its derivative appear linearly. This form allows the use of systematic methods to obtain solutions, particularly when p(t) and q(t) are continuous functions, ensuring that a unique solution exists for given initial conditions.

Integrating Factor Method

This method involves multiplying both sides of a first-order linear differential equation by an integrating factor, usually of the form exp(?p(t) dt), which transforms the equation into an exact derivative. This transformation simplifies the process of integration by allowing the equation to be written as the derivative of a product, making it easier to solve for the unknown function u.

Initial Value Problem

An initial value problem not only specifies a differential equation but also provides the value of the unknown function at an initial point. This additional information is crucial as it allows us to determine the constant of integration arising from the solution process, thereby ensuring that the solution is unique to the problem.

Transcript

00:01 For problem 53, we're given an initial value problem in which we need to solve and get the function s, right? so we're going to have to do that through integration.

00:09 So let's set this up for that.

00:12 So make ds equal to 12t times 3t squared minus 1 cubed, right? we shifted that dt over.

00:21 So now if we integrate both sides, we're actually going to get s and we're going to get it equal to the function that we want, right? so let's integrate this problem.

00:30 Let's make you equal to 3t squared minus 1.

00:35 D .u will equal 6t dt, right? but we do need it to be 12.

00:40 So we need to multiply both sides by two, right? we need to multiply this whole thing by two.

00:44 So 2, du equals 12t d t.

00:50 That looks like a z.

00:51 So let me just fix that equals.

00:55 Okay.

00:56 So now we can go ahead and make our substitution.

00:58 So we're going to pull that 2 out front, but we're going to do the integral of u cubed du.

01:06 And now when we integrate this, we will get two times u to the fourth times one -fourth plus c, which we can now distribute and make s equal to one -half u to the fourth plus c.

01:24 So now we can go ahead and back substitute in to get to t.

01:28 So s equals 1⁄2 times 3t squared minus 1 to the fourth power plus c...

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