I had taken a long break from math and physics for months due to stress and an illness so when I returned to it and tried answering problems given, I had difficulty figuring out how to do some of them all over again. Here were the items I had encountered:
1) Evaluate $\sum_{n=0}^{\infty} {p^n}{sin(2nx)}$
2) Solve for x(t): $\frac{dx}{dt} = x_o =-cos(t)x + t^3, x(t=0)$
3) Evaluate $\int_{0}^{\pi} \frac{d\theta}{(a+cos\theta)^2}, a>1$
4) Evaluate $\int_{-\infty}^{\infty} {x^6 e^{-x^2}} dx$
5) Evaluate $L_n(0)$ given the generating function for Laguerre polynomials is $g(x,t) = \sum_{n=0}^{\infty} L_n(x) t^n = \frac {e^{xt/(1-t)}}{1-t}$
6) Solve: $y'' + x^2 y'+(x-1)y = 0$ -> For this, I tried the Frobenius method but didn't get far.
Some weren't taught in my previous class but I was expected to already know it for my course this coming semester. I have tried searching already but I have not gotten far for these. I am not expecting all of these to be answered. I just really need help getting started again so any help for any of these would be really appreciated. You can just choose one among them. If you could perhaps give me a way I can try, step-by-step, or a really good reference for me to look at so I can self-study.
HINTS:
For $1)$, write
$$\sum_{n=0}^\infty p^n\sin(2nx)=\text{Im}\left(\sum_{n=0}^\infty (p\,e^{i2x})^n\right)\tag1$$
Then sum the geometric series on the right-hand side of $(1)$ and take the imaginary part of that result.
For $3)$, note that
$$\int_0^\pi \frac{1}{(a+\cos(\theta))^2}\,d\theta=-\frac{d}{da}\int_0^\pi \frac{1}{a+\cos(\theta)}\,d\theta\tag 2$$
The integral on the right-hand side of $(2)$ can be evaluated using the classical tangent half-angle substitution or contour integration. Finish by taking the derivative with respect to $a$ and multiplying by $-1$.
For $4)$ note that
$$\int_0^\infty x^6e^{-x^2}\,dx=-\left.\left(\frac{d^3}{da^3}\int_0^\infty e^{-ax^2}\,dx\right)\right|_{a=1} \tag 3$$
The integral on the right-hand side of $(3)$ is the form of the classical Gaussian integral. Take its third derivative, evaluate at $a=1$, and multiply by $-1$.