# Doomsday differential equation

• Trousers
In summary: Now, for the solution:We have that -\frac{1}{cy(t)^{c}}=kt+C in (') where our initial condition y(0)=y_{0} now can be used to determine C:-\frac{1}{cy_{0}^{c}}=C" what is C? "Inserting in (*) and simplifying, we have:\frac{1}{y(t)^{c}}=\frac{1}{y_{0}^{c}}-ckt=\frac{1-y_{0}^{c}ckt}{y_{0}^{c}}" So this equation is saying that if I integrate this solution
Trousers
1. Let c be a positive number. a differential equation of the form
dy/dx=ky^(1+c)
Determine the solution that satisfies the initial condition y(0)=y(subzero)

The solution is y(subzero)=(1)/((ckT)^(1/c))

but I can't understand how to find it. ok when i integrate

dy/y^(1+c)=kdt, I get y^(-c)/(-c)=kt+C and I don't know what do from there.

I googled the term "doomsday differential equation" and guess what? The top 2 entries were this thread itself and an older one which concerns the same DE:

EDIT: Does anyone know why it's called the doomsday equation?

that doesn't help me.

Try leaving it in the form

$$y^c = f(x)$$

and then solve for C (the integration constant, not the one in the form I just mentioned).

1. It is called the Doomsday equation because for any c>0, there exists a finite time T when y blows up (i.e, graphing y(t) with t as the argument axis, there would be a vertical asymptote to the graph at t=T, which the graph of y(t) will follow into infinity as t approaches T)

2. Now, for the solution:
We have that $$-\frac{1}{cy(t)^{c}}=kt+C in (')$$ where our initial condition y(0)=y_{0} now can be used to determine C:
$$-\frac{1}{cy_{0}^{c}}=C$$
Inserting in (*) and simplifying, we have:
$$\frac{1}{y(t)^{c}}=\frac{1}{y_{0}^{c}}-ckt=\frac{1-y_{0}^{c}ckt}{y_{0}^{c}}$$
Or, reciprocating, and taking the c-root on both sides:
$$y(t)=\frac{y_{0}}{(1-y_{0}^{c}ckt)^{c}}$$

Now, the blow-up instant T must fulfill the equation:
$$1-y_{0}^{c}ckT=0\to{T}=\frac{1}{cky_{0}^{c}}$$
Using the blow-up time T as our independent parameter, rather than y_0, we get from the same equation: $$y_{0}=\frac{1}{(ckT)^{\frac{1}{c}}}$$

The first result is simplest to use when you DO know the value of y_0, and wants to know WHEN the solution will blow up.

The second result is most convenient when you know when the solution is going to blow up, and wants to know what was the initial value of y, i.e, y_0.

thank you.

Trousers said:
1. Let c be a positive number. a differential equation of the form
dy/dx=ky^(1+c)
Determine the solution that satisfies the initial condition y(0)=y(subzero)

The solution is y(subzero)=(1)/((ckT)^(1/c))

but I can't understand how to find it. ok when i integrate

dy/y^(1+c)=kdt, I get y^(-c)/(-c)=kt+C and I don't know what do from there.

I have no idea what you are saying. "The solution is y(subzero)=(1)/((ckT)^(1/c))" can't possibly be right- there is no y(t) in it so it can't be a solution to the differenital equation. Also, is T supposed to be t are is it a specific value of t?

## 1. What is the Doomsday Differential Equation?

The Doomsday Differential Equation is a mathematical model that predicts the rate of change or growth of a catastrophic event, such as a natural disaster or global pandemic.

## 2. How does the Doomsday Differential Equation work?

The equation takes into account various factors, such as initial conditions, growth rate, and potential interventions, to calculate the likelihood and severity of a doomsday scenario.

## 3. Can the Doomsday Differential Equation accurately predict a doomsday event?

While the equation provides a mathematical framework for understanding and forecasting doomsday scenarios, it is not a foolproof prediction tool. It can only make predictions based on the data and variables included in the model.

## 4. What are some real-world applications of the Doomsday Differential Equation?

The equation has been used to study the potential impact of climate change, disease outbreaks, and other catastrophic events. It can also be used to evaluate the effectiveness of interventions and preventative measures.

## 5. What are the limitations of the Doomsday Differential Equation?

The equation relies on a number of assumptions and simplifications, and may not accurately capture the complexity of real-world scenarios. It also requires accurate and up-to-date data to make accurate predictions.

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