Verify that the indicated function y=ϕ(x) is an explicit solution of the given first-order diffe

step 1 Hello, welcome to this lesson. And this lesson will check whether the solution here is an explic

Verify that the indicated function y=ϕ(x) is an explicit...

Verify that the indicated function $y=\phi(x)$ is an explicit solution of the given first-order differential equation. Proceed as in Example $6,$ by considering $\phi$ simply as a function and give its domain. Then by considering $\phi$ as a solution of the differential equation, give at least one interval $I$ of definition. $$(y-x) y^{\prime}=y-x+8 ; \quad y=x+4 \sqrt{x+2}$$

Verify that the indicated function $y=\phi(x)$ is an explicit solution of the given first-order differential equation. Proceed as in Example $6,$ by considering $\phi$ simply as a function and give its domain. Then by considering $\phi$ as a solution of the differential equation, give at least one interval $I$ of definition.
$$(y-x) y^{\prime}=y-x+8 ; \quad y=x+4 \sqrt{x+2}$$

Key Concepts

Explicit Solution

An explicit solution to a differential equation is one where the dependent variable is written explicitly as a function of the independent variable. This makes it straightforward to differentiate and substitute back into the differential equation for verification. It contrasts with implicit solutions, where the relationship between variables isn’t directly solved for the dependent variable.

First-Order Differential Equation

A first-order differential equation involves only the first derivative of the unknown function. These equations express the rate of change of a variable with respect to another and are often solved using techniques like separation of variables or integrating factors, or verified by direct substitution of a proposed solution.

Verification by Substitution

Verification by substitution is a method used to confirm that a proposed function is indeed a solution of a given differential equation. The process involves differentiating the proposed function, substituting both the function and its derivative back into the differential equation, and checking if the resulting equation holds true for all values in the given domain.

Domain of a Function and Interval of Definition

The domain of a function refers to the set of all input values for which the function is properly defined. When dealing with differential equations, the interval of definition is a subset of the function’s domain in which the solution is not only well-defined but also satisfies the differential equation. Selecting an appropriate interval ensures that the solution behaves well and that the existence and uniqueness theorems for differential equations apply.

Transcript

00:01 Hello, welcome to this lesson.

00:01 And this lesson will check whether the solution here is an explicit solution for a differential equation.

00:12 This will be done by comparing the left -hand side to the right -hand side.

00:17 So for the left -hand side, y minus x times the first differential of y.

00:33 So y is x plus four square roots x plus two.

00:42 Okay, so the whole thing minus x then times the first differential of y.

00:52 Okay, so let's look at the differential of y.

00:59 So divide the x.

01:02 Now this one becomes one.

01:07 And this is.

01:11 4 x plus 2 to the power half okay so plus the 4 out then the half comes to multiply so 1 over 2 then you subtract 1 from the half damage the negative half then when you differentiate what we have here you have a 1 okay so this becomes x plus 2 to the power negative half then time 1 so the whole thing here becomes 1 plus 2 all over the square root of x plus 2 all right so this is the left -hand side and this becomes 4 plus of 4 square root of x plus 2 times 1 okay so i can make the whole thing as square root of x plus 2 times 1 okay so i can make the whole thing as square root of x plus 2 2 plus 2 all over.

02:36 All right.

02:37 And this later becomes 4 plus 8.

02:50 Because this will cross out that 4 multiplies the whole thing.

02:55 So this is the left hand side.

02:59 Let's look at the right hand side.

03:04 For the right side, we have y minus.

03:12 Minus x plus 8.

03:14 So this is y minus x plus 8.

03:19 Okay, so y is x.

03:22 The whole thing here is x plus x...