I wouldn't claim that there exists a 'right' way to study any subject, not just physics, but watching lecture videos on YouTube won't suffice in order to truly understand advanced physics. Textbooks can be boring but they provide formulas, derivations and practice problems. It's very important...
I have taken all of those courses/classes you've listed above, except the last one, 'Introduction to Numerical Methods'. I just wanted to know the quality of the online math classes/courses that are offered from this online college/university, but now, I think I know what I should do.
Hello, I want to know if anyone has taken an 'Introduction to Partial Differential Equations" class/course via UIUC (University of Illinois Urbana-Champaign) through NetMath. I am planning to take this course given the fact that I have taken ODE (Ordinary Differential Equations) and Nonlinear...
From the original equation of ## \ddot{x}+x+\frac{3}{2}\beta x\lvert x\rvert=0 ##,
I've got ## \ddot{x}\dot{x}+x\dot{x}+\frac{3}{2}\beta x\lvert x\rvert\dot{x}=0
\implies \frac{d}{dt}(\frac{1}{2}\dot{x}^2+\frac{1}{2}x^2+\frac{\beta}{2}x^2\lvert x\rvert)=0
\implies...
Proof:
Let ## \epsilon=0 ##.
Then the unperturbed equation is ## \ddot{x}+x=0 ## and the general solution is
## x(t)=A\sin\omega t+B\cos\omega t ## where ## \omega=1 ## is the angular frequency
with the constants ## A ## and ## B ##.
With the initial condition ## x(0)=0 ##, we have that ##...
So far, I've got that ## g(a)=(c-1)a+\frac{3}{4}a^3\implies g'(a)=c-1+\frac{9}{4}a^2 ##.
I know that if the first derivative of a function is positive (greater than ## 0 ##),
then that function is always/strictly increasing. However, how should I construct
this proof in order to show that...
Sorry, I think I've made some huge mistakes above from my initial/first post. The equation of the phase paths should be ## y=\pm\sqrt{2(C-V(x))} ## where ## \dot{x}=y, \dot{y}=f(x) ##. So I have ## y=\pm\sqrt{2(C-(\frac{x^2}{2}+\frac{\beta}{2}x^3\operatorname{sgn}(x)))} ## and ## y=0 ## for ##...
From the system of ## \dot{x}=y, \dot{y}=-x-\frac{3}{2}\beta x\lvert x \rvert ##, after setting ## \dot{x}=y=0 ## and ## \dot{y}=-x-\frac{3}{2}\beta x\lvert x \rvert=0 ##, we get that the fixed point occurs at ## (x, y)=(0, 0) ## and is stable centre.