When plugging the differential equation $\frac{dP}{dt}=P(1-P)$ into a symbolab differential equation calculator I get the result $- \frac{e^{t + c_1}}{1 – e^{t+c_1}}$. I don't know if this is correct because the correct solution in the book should be a sigmoid function $\frac{1}{1+e^{-t}}$. Is the calculator solution correct?
Sigmoid function as result of differential equation
ordinary differential equations
Related Solutions
To check it's a solution: find $y'=dy/dx$, and check that it takes the value of $xy$. If it does, it's a solution.
You solve an equation like that by a method known as seperation of variables, which you should be able to find some lecture notes on by googleing. In this example:
$$\frac{dy}{dx}=xy\Leftrightarrow \frac{1}{y}\frac{dy}{dx}=x\Leftrightarrow\int\frac{1}{y}\frac{dy}{dx}{dx}=\int x\ dx$$ $$\Leftrightarrow\int \frac{dy}{y}=\frac{x^{2}}{2}+c\Leftrightarrow ln(y)=\frac{x^{2}}{2}+c $$
Then exponentiating, we get that the solution is $y=e^{x^{2}/2 + c}=Ae^{x^{2}/2}$, where the constant $A$ is $A=e^{c}$.
The general solution is actually
$$\int \frac{1}{y}dy=2\int\frac{1}{x}dx\\ \ln y = 2\ln x +C\\ y=e^Ce^{2\ln x}=Cx^2$$
Assuming $x_0\neq0$, you can choose any initial point $(x_0,y_0)$ and find a unique solution of this form by letting $$C=\frac{y_0}{x_0^2}$$ For example, the solution of the differential equation that passes through $(4,2)$ will be $y=\frac{x^2}{8}$.
But if $x_0=0$, then we have $$xy^\prime=2y\\ 0=2y\\ y=0$$ So any initial value $(0,y_0)$ where $y_0\neq 0$ cannot be a solution of the differential equation. Also, notice that every solution of the form $y=Cx^2$ must pass through $(0,0)$, so you cannot find a unique solution at this point. In other words, if the differential equation describes the path of a particle, placing the particle initially at $(0,0)$ could lead to an infinite amount of trajectories. So the solution makes no sense at $x_0=0$.
There is no rule or convention about expressing higher order differential equations in the way you suggest. However, the Wronskian can be used to check if the general solution of a differential equation is valid for certain initial values. Your book should cover this in later chapters.
Best Answer
Yes, it is correct. If $P(t)=-\frac{e^{t+c}}{1-e^{t+c}}$, then$$P'(t)=-\frac{e^{t+c}}{\left(e^{t+c}-1\right)^2},$$whereas$$P(t)\bigl(1-P(t)\bigr)=-\frac{e^{t+c}}{1-e^{t+c}}\times\frac1{1-e^{t+c}}=-\frac{e^{t+c}}{\left(e^{t+c}-1\right)^2}.$$So, yes, those functions are solutions of your differential equation.