Which of the following functions are solutions of the differential equation y^''+y=sinx ? (

Which of the following functions are solutions of the...

Key Concepts

Second Order Linear Differential Equations

These are differential equations that involve the second derivative of the unknown function and can be written in a linear form. Their general structure typically involves constant or variable coefficients, and the solution methods depend on whether the equation is homogeneous or non-homogeneous.

Homogeneous versus Non-homogeneous Equations

A homogeneous linear differential equation has the right-hand side equal to zero, while a non-homogeneous equation has a forcing term. Understanding the distinction is important because the solution to a non-homogeneous equation is the sum of the general solution to the homogeneous part and a particular solution that accounts for the non-zero forcing function.

Particular Solution

A particular solution is a specific solution to the non-homogeneous differential equation that directly satisfies the entire equation, including any forcing terms. Finding a particular solution is essential in constructing the complete solution to a non-homogeneous differential equation.

Method of Undetermined Coefficients

This is a technique used to find a particular solution for non-homogeneous linear differential equations with constant coefficients when the forcing term is of a type that allows for an educated guess. This method involves proposing a form for the particular solution and then determining the unknown coefficients by substituting the guess into the differential equation.

Verification of Candidate Solutions

Verifying whether a function is a solution of a differential equation involves computing the necessary derivatives and substituting them back into the equation. This step ensures that the function satisfies the differential equation, confirming its validity as a solution.

Transcript

00:01 In this video, we're going to be looking at verifying the solutions to a differential equation.

00:04 So the differential equation we are given is y double prime plus y equals sign of x, and we have a variety of solutions that we want to check as to whether or not they are solutions to this differential equation.

00:17 So let's start with this one here.

00:20 Y equals the sign of x.

00:22 So we need to plug this into our differential equation.

00:24 We know y, but we need to find y double prime.

00:26 So let's go ahead and do that.

00:27 Let's take the first derivative to find y prime first, and the derivative of sine of x is cosine of x.

00:34 Then if we take the derivative again to get y double prime, we will get negative sine of x.

00:41 So if we plug in this here and this here, what we will end up with is negative sine of x plus sine of x equals sine of x.

00:59 So our left -hand side simplifies to zero, so we get zero equals sign of x.

01:06 And this is not always a true statement, so we cannot say that this is necessarily a solution.

01:13 Let's move on to b.

01:18 So this is the function we're giving.

01:21 Let's find y double prime.

01:22 So y prime is going to be negative sine of x.

01:27 Y double prime is going to be negative cosine of x.

01:31 So if we plug that in, we get negative cosine of x plus cosine of x equals sine of x.

01:47 So again, we end up with zero equals sine of x.

01:52 And this is not a true statement for all values of x.

01:55 So we cannot say that this is a solution to our differential equation.

02:00 Moving on to this next one, let's check this.

02:06 So y prime.

02:08 You have to use the product rule.

02:09 Here...

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