Solve the following initial value problems: (a) (d u)/(d t)=5 u, u(0)=-3, (b) (d u)/(d t)=2 u, u

step 1 The topic of this question is linear differential equations. This question asks us to solve this

Solve the following initial value problems: (a) (d u)/(d...

Key Concepts

Initial Value Problem (IVP)

An initial value problem is a differential equation paired with a specific value at a given point. This initial condition ensures that the solution is unique and allows us to determine the particular solution of the differential equation among the family of general solutions.

Separable Differential Equations

A separable differential equation is one where the variables can be separated on different sides of the equation. This technique involves rewriting the differential equation in a form where all terms involving one variable are on one side and all terms involving the other variable on the opposite side, allowing the use of integration to solve the equation.

Exponential Function as a Solution

Many first order linear differential equations with constant coefficients have solutions in the form of exponential functions. This arises from integrating the equation and applying the exponentiation process, which naturally yields functions that exhibit exponential growth or decay depending on the sign of the constant coefficient.

Constant Coefficient First Order Linear ODE

A constant coefficient first order linear ordinary differential equation is one where the rate of change of the unknown function is proportional to the function itself. The general solution is found by integrating a constant coefficient, leading to an exponential form, and then applying the initial condition to determine the specific constant factor.

Transcript

00:02 The topic of this question is linear differential equations.

00:07 This question asks us to solve this linear first order initial value problem, which is a differential equation with an initial condition.

00:20 So this differential equation might look a bit different than what we are used to.

00:26 Usually we have a d, y, d, x term, and functions of x, and...

00:36 Usually a y.

00:40 However, in this case, we have, instead of d -y -d -x, d -u -d -t.

00:46 And that means, well, that just means our independent variable is t and our dependent variable is u, instead of having a dependent variable y and an independent variable x.

01:04 But we still use these letters the same way we do y and x, of course.

01:10 So, since this is a first order linear differential equation, we want to get an expression, our goal is to get an expression on the left side that is a derivative with respect to the independent variable of the product of a function of the independent variable and y.

01:38 And on the right side, we want just a function of the independent variable.

01:47 So to do this, we first get the equation into standard form, in which all the dependent variable and all the derivatives of the dependent variable are on the same side, and the coefficient of the derivative is 1.

02:18 So here we want to divide by t.

02:21 Divide everything by t.

02:31 The next step is find the integrating factor, which is a function of the independent variable by which we multiply both sides to get the product rule on the left side that we wanted before, that we want.

02:51 So the formula for this integrating factor, which i will call g of t, is e to the exponent of this function that is the coefficient of u the integral of this coefficient of you.

03:13 So since this is an exponent, i will find the integral off to the side.

03:26 Since the derivative of lone of t is 1 over t, this will be my antiderivivative...

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