PLease help for differential equation

[stunner5000pt]

\frac{dy}{dt} = \frac{1}{(y+2)^{2}} and y(0) =1

th solution i got is y(t) = \sqrt[3]{3t-27} - 2

the question asks find the domain of the definition of the solution

Describe hwat happens when the solution as it reaches it'slimits of its domain. Why can't it be extended for more time?

Looking at he function aid ti s CUBE ROOT shouldn't the domain be ALL REAL number?? So the limits are positive and negative infinity? So then the limits are positive and negative infinity respeectively??

Input would be greatly valued! Thank you!@

 

[Tide]

If you've typed your DE correctly then I think you have the wrong sign on the constant 27. Other than that, the derivative is undefined when y = -2 which occurs at t = -9 and makes it impossible to extend your soluion to t <= 9.

 

[wais]

The solution you have obtained for the given differential equation is correct. However, the domain of the solution is not all real numbers. In order to find the domain, we need to consider the restrictions on the variable t and the initial condition given.

In this case, the variable t must be greater than or equal to 0 since the initial condition is given at t=0. This means that the solution is only defined for t values greater than or equal to 0. As t approaches infinity, the solution will approach a limit of 2. This is because as t becomes larger, the cube root term in the solution will dominate and the solution will approach 2. This is also evident from the given initial condition, y(0)=1, which is less than 2.

The reason why the solution cannot be extended for more time is because it is only defined for t values greater than or equal to 0. If we try to plug in a negative t value, we will get a negative number under the cube root, which is not defined in the real numbers. This is why the domain of the solution is limited to t greater than or equal to 0.

It is important to note that the cube root function has a domain of all real numbers, but in this case, the variable t is restricted to non-negative values due to the given initial condition. Therefore, the solution is only valid for t values in the non-negative domain.

I hope this clarifies any confusion. Keep up the good work in solving differential equations!