Find all solutions to the following IVP:
$$\dot{x}=x^{2023}-x^{2022}$$ $$ x(0) = 1 $$
I am really not sure how to start this problem, my initial idea was to use the separation of the variables, but no matter how I tried I couldn´t solve it correctly, I would appreciate any kind of help or ideas.
On
The ODE can be solved as follows. Separate variables.
$$\frac{dx}{dt} = x^{2023} - x^{2022} \implies \frac{dx}{x^{2022} (x-1)} = dt$$
Expand the left side into partial fractions. It's easy to show (by induction, for instance) that
$$\frac1{x^n (x-1)} = \frac1{x-1} - \left(\frac1x + \frac1{x^2} + \frac1{x^3} + \cdots + \frac1{x^n}\right)$$
It follows that
$$\int \frac{dx}{x^{2022} (x-1)} = \int dt \implies \ln|x-1| - \ln|x| + \frac1{x} + \frac1{2x^2} + \cdots + \frac1{2021x^{2021}} = t + C$$
As pointed out in comments, however, the given initial value is problematic, since the $\ln|x-1|$ diverges to $-\infty$ as $x\to1$.
On
Starting from @user170231's answer, we can make the solution more compact $$\int \frac{dx}{x^{n} (x-1)}= \frac{x^{1-n} }{n-1}\,\, _2F_1(1,1-n;2-n;x)$$ where appears the Gaussian hypergeometric function.
This also shows the problem at $x=0$, the limit of the hypergeometric function being equal to $1$
When applying separation-of-variables, the first step is to check the right side for roots. These are easily found as $x^*_0=0$ and $x^*_1=1$. These roots correspond to constant solutions. As the right side is a polynomial, the solutions are unique, no branching or coalescing of solutions can occur.
Thus the initial condition $x(0)=1=x^*_1$ implies directly, without any further calculation, that the one and only solution of the IVP is the constant solution $x(t)=x^*_1=1$.