# Initial Value Problem / Differential Equation

• enkrypt0r
In summary: Hi, Tim!In summary, you have a problem that you're stuck on and you need help to solve it. The general solution is y = (-3) + ln(y+3), but when you try to solve it for y(2), you get a zero in the denominator of the second term and ln(0) for the third term, both of which are undefined. If y = -3 or 0, y' = 0, which says that y is a constant. If you try to solve the IVP, you may find that more than one solution curve satisfies the initial condition.
enkrypt0r
Thanks for clicking!

So, I've got a problem here that I'm stuck on. I need to find the general solution to

y' = (y3 + 6y2 + 9y)/9

I found this to be

ln|y| + (3/(y+3)) - ln|y+3| = x + c

but I would appreciate it if you would check my work. Anywho, once I have the general solution I need to solve it for y(2) = -3. Obviously, plugging -3 in for y creates a problem in the second and third terms above.

Is there a way to rewrite the general solution or solve it in such a way that it won't contain any undefined terms when evaluated like it does now?

Thanks guys.

EDIT: Shoot, sorry for the non-descriptive title. I accidentally clicked post before finishing it. I meant it to say "[Calc II] Initial Value Problem / Differential Equation." My bad.

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I got the same general solution, what do you mean by undefined terms?

I mean that when I plug -3 in for y, I get a zero in the denominator of the second term and ln(0) for the third term, both of which are undefined. I was wondering if there is any way to rewrite this or to solve this differently so -3 will not cause this discontinuity.

hi enkrypt0r!
enkrypt0r said:
y' = (y3 + 6y2 + 9y)/9

erm look at the equation!

if y = -3, what is y' ?

tiny-tim said:
hi enkrypt0r!

erm look at the equation!

if y = -3, what is y' ?

Hi, Tim!

I never thought to look at the original equation! This evaluates to 0 = 0. I'm afraid I don't know what this says about the differential equation. Does this mean that it's undefined, as my attempts to evaluate it have shown?

oooh think!

if y' = 0, then … ?

Ummm... I'm feeling stupid here, haha. Sorry. I'm brainstorming but coming up with nothing beyond the obvious. We could integrate both sides and see that y is a constant, but we already know that. I'm sorry, I'm at a loss.

If y = 0 or y = -3, then y' = 0, which says that y is a constant.

In your work, the tacit assumption is that y != 0 and y != -3.

Yeah, y=0 and y=-3 are "equilibria" solutions to the de. This sort of de is known as autonomous, because there are no t's on the right side of the equation. So, no matter what t is, the slope at y=0 and y=-3 is always zero.

So, why can't we plug into solve the initial value problem?

If you look at the direction field, I think you might find that more than one solution curve would satisfy the initial condition. Thus, there is no particular value of c that would net you a unique solution curve. Someone correct me if this is wrong, because I have never seen this case before! That's my analysis though.

Mining_Engr said:
If you look at the direction field, I think you might find that more than one solution curve would satisfy the initial condition. Thus, there is no particular value of c that would net you a unique solution curve. Someone correct me if this is wrong, because I have never seen this case before! That's my analysis though.

For the initial value problem, y' = (1/9)(y3 + 6y2 + 9y), y(2) = -3, we see that y' = (1/9)y(y + 3)2.

If y = -3, then y' = 0, so y(x) $\equiv$ C. Since y(2) = -3, C = -3, so the solution to the IVP is y(x) $\equiv$ -3. Are you saying that you think there are other solutions to this initial value problem?

hi enkrypt0r!

(just got up :zzz:)

the solution comes in three parts …

y = -3 (for all t) is a solution

y = 0 (for all t) is a solution

and the other solutions are all the solutions of

dy/(y3 + 6y2 + 9y) = dt/9

(you multiplied by 0/0 when you got that equation, if y = -3 or 0, so you illegitmately lost those solutions )

if you draw a graph of some specimen solutions, you'll probably find that they "asymptote up to" the vertical lines at -3 and 0,

and (i'm guessing here ) that one of y = 0 and v = -3 is stable and the other is unstable

## 1. What is an initial value problem?

An initial value problem is a type of differential equation that involves finding a function or set of functions that satisfy a given equation and a set of initial conditions. The initial conditions typically specify the value(s) of the function(s) at a specific point or points.

## 2. What is a differential equation?

A differential equation is an equation that involves an unknown function and its derivatives. It describes the relationship between the function and its rate of change, and is often used to model physical or natural phenomena.

## 3. How do you solve an initial value problem?

The most common method for solving an initial value problem is by using a technique called separation of variables. This involves separating the equation into two parts, one involving only the function and the other involving only its derivative, and then integrating both sides to find the solution.

## 4. What are some applications of initial value problems?

Initial value problems are used in many fields of science and engineering, including physics, chemistry, biology, and economics. They can be used to model the behavior of systems such as population growth, chemical reactions, and electrical circuits.

## 5. Are there different types of initial value problems?

Yes, there are different types of initial value problems, including first-order and higher-order problems, linear and nonlinear problems, and boundary value problems. The type of problem depends on the order of the highest derivative in the equation and the number and type of initial conditions given.

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