Domain & Interval of Solution for y=x+4\sqrt{x+2}

[find_the_fun]

The question asks
Verify that the indicated function \(\displaystyle y=\phi(x)\) is an explicit solution of the given first order differential equation. Consider \(\displaystyle \phi\) simply as a function, give its domain. Then by considering \(\displaystyle \phi \)as a solution of the differential equation, give at least one interval \(\displaystyle I\) of definition.

\(\displaystyle (y-x)y'=y-x+8\); \(\displaystyle y=x+4\sqrt{x+2}\)The first step in solving a problem is understanding the question.

Verify that the indicated function \(\displaystyle y=\phi(x)\) is an explicit solution of the given first order differential equation.

Understood.

Consider \(\displaystyle \phi\) simply as a function, give its domain.

Is it asking where the function \(\displaystyle x+4\sqrt{x+2}\) is defined?

Then by considering \(\displaystyle \phi \)as a solution of the differential equation, give at least one interval \(\displaystyle I\) of definition.

What is an interval of definition?So I took the derivative of y and was verifying the solution and got to the equation

\(\displaystyle 4\sqrt{x+2}+\frac{16\sqrt{x+2}}{2\sqrt{x+2}}=4\sqrt{x+2}\) and got a little worried. This equality is only true when x=-2.

The answer key gives: domain of fuctnion is \(\displaystyle [-2, \infty)\); largest interval of definition for solution is \(\displaystyle (-2, \infty)\)

 

[Prove It]

The question asks
Verify that the indicated function \(\displaystyle y=\phi(x)\) is an explicit solution of the given first order differential equation. Consider \(\displaystyle \phi\) simply as a function, give its domain. Then by considering \(\displaystyle \phi \)as a solution of the differential equation, give at least one interval \(\displaystyle I\) of definition.

\(\displaystyle (y-x)y'=y-x+8\); \(\displaystyle y=x+4\sqrt{x+2}\)The first step in solving a problem is understanding the question.Understood.Is it asking where the function \(\displaystyle x+4\sqrt{x+2}\(\displaystyle is defined?
What is an interval of definition?So I took the derivative of y and was verifying the solution and got to the equation

\(\displaystyle 4\sqrt{x+2}+\frac{16\sqrt{x+2}}{2\sqrt{x+2}}=4\sqrt{x+2}\) and got a little worried. This equality is only true when x=-2.

The answer key gives: domain of fuctnion is \(\displaystyle [-2, \infty)\); largest interval of definition for solution is \(\displaystyle (-2, \infty)\)\)\)

\(\displaystyle \(\displaystyle

Wait, what is your $\displaystyle \begin{align*} \phi \end{align*}$ function that you are trying to test?\)\)

 

[find_the_fun]

Wait, what is your $\displaystyle \begin{align*} \phi \end{align*}$ function that you are trying to test?

I don't really understand what $\phi$ is?

 

[I like Serena]

Hi find_the_fun,

The question asks
Verify that the indicated function \(\displaystyle y=\phi(x)\) is an explicit solution of the given first order differential equation. Consider \(\displaystyle \phi\) simply as a function, give its domain. Then by considering \(\displaystyle \phi \)as a solution of the differential equation, give at least one interval \(\displaystyle I\) of definition.

\(\displaystyle (y-x)y'=y-x+8\); \(\displaystyle y=x+4\sqrt{x+2}\)

Consider ϕ simply as a function, give its domain.

Is it asking where the function \(\displaystyle x+4\sqrt{x+2}\) is defined?

Yes.

What is an interval of definition?

The interval of definition is the interval for $x$ where the differential equation is properly defined.
In particular it means that $y'$ must be properly defined.

So I took the derivative of y and was verifying the solution and got to the equation

\(\displaystyle 4\sqrt{x+2}+\frac{16\sqrt{x+2}}{2\sqrt{x+2}}=4\sqrt{x+2}\) and got a little worried. This equality is only true when x=-2.

You should have an extra $+8$ on the right hand side.

 

[blue_raver22]

I would explain to the student that the question is asking for the domain of the function y=\phi(x), which is x+4\sqrt{x+2}. This means that the function is defined for all values of x greater than or equal to -2, since the square root function is only defined for non-negative numbers. Therefore, the domain of the function is [-2, \infty).

As for the interval of definition, it is the largest interval on which the solution y=x+4\sqrt{x+2} is valid for the given differential equation. In this case, since the domain is [-2, \infty), the largest interval of definition would be (-2, \infty). This interval excludes the point x=-2, where the derivative of the function becomes undefined.

Therefore, the function y=x+4\sqrt{x+2} is an explicit solution for the given differential equation on the interval (-2, \infty). It is important to note that this is just one possible interval of definition, as there could be other intervals where the solution is also valid.