The functions satisfy the differential equation and are...

The functions satisfy the differential equation and are linearly independent since $$W\left(e^{x / 2}, x e^{x / 2}\right)=e^{x} \neq 0$$ for $-\infty<x<\infty .$ The general solution is $$y=c_{1} e^{x / 2}+c_{2} x e^{x / 2}.$$

Key Concepts

Second Order Homogeneous Linear Differential Equations

These differential equations are of the form y'' + p(x)y' + q(x)y = 0, where the solution space is spanned by two linearly independent functions. Because the equation is homogeneous, any linear combination of its solutions will also be a solution, and the specific form of the general solution reflects this principle.

General Solution

The general solution of a differential equation represents the family of all possible solutions and includes arbitrary constants corresponding to the order of the differential equation. It is constructed as a linear combination of a set of independent solutions, ensuring that every possible behavior of the system under different initial conditions is captured.

Linear Independence and the Wronskian

Linear independence of solutions is crucial for forming a complete basis for the solution space. The Wronskian is a determinant formed from the functions and their derivatives; if it is nonzero on an interval, the functions are linearly independent over that interval. This guarantees that the general solution, expressed as a linear combination of these functions, is indeed the most general form.

Transcript

00:01 All right, so part a is saying that we should show that each of the functions, e to the x, e to the 4x, and 2 times e to the x minus 3e to the 4x are all solutions of this equation.

00:19 So first way to do that, let's check the first one, e to x.

00:33 So let's start by taking the derivative and the second derivative of these so that we can put those into here.

00:41 So, divide the x, or i'm just going to wait as y prime for shorthand.

00:50 Y prime, the derivative of e to the x is e to the x.

00:56 And second derivative, guess what? it's also e to the x.

01:01 Okay, so then if we just take all of these things, plug them in for their respective parts.

01:10 So e to the x minus 5, e to the x.

01:17 The x plus four, e to the x equals 0.

01:26 And 1 plus 4 is 5 minus 5 is 0.

01:30 So 0 equals 0.

01:34 Okay, so that one works.

01:35 Let's try the next one.

01:38 So let's look at e to the 4x.

01:40 So the derivative of that would be 4 times e to the 4x.

01:50 And the second derivative would be 16e to the 4x.

02:00 So then if we just plug those all back in, 16e to the 4x.

02:09 So once we substitute all that stuff in, we have 16 minus 20, which is negative 4 plus 4, which is 0.

02:18 So 0 equals 0.

02:21 So that one works as well.

02:23 Now let's try the next.

02:30 So the next one is 2.

02:34 So there are our first and second derivatives for when we are using this.

02:39 Of 2e to the x minus 3 e to the 4x and then let's put those in to the equation to make sure they work okay so now they've got all those stuff plugged in we can see that the like terms are 2 e to the x minus 10 e to the x plus 80 to x which is 0 and then negative 48 uh plus 60 minus 12 or plus, yeah, minus 12.

03:20 So negative 48 plus 60 minus 12, that's also 0.

03:22 So 0 equals 0.

03:24 They're all answers and solutions.

03:27 So now part b says that we want to prove or show that e to the x and each of the 4x are linearly independent, which basically means as long as we can't find constants c1 and c2, where this is true, then they are linearly independent.

03:57 So let's subtract c1e to the x, c2, e to the 4x, equals negative c1, e to the x.

04:15 If we divide everything by e to the x, then we're left with c2, e to the 3x, equals to negative c1...

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