Solve System of Differential Equations with Expert Help

What method would you use to solve:
$(1+x^2)\frac{\mathrm{d}y}{\mathrm{d}x}=1+y^2;y(2)=3$
I am asking this because I only know two methods of solving the DEs - separation of variables and integrating factor. Since the separation of variables does not work here, I tried integrating factor, however, I don't know what to do with the y2, because for the IF to work I need to get y on its own $\frac{\mathrm{d}y}{\mathrm{d}x}+P(x)y=Q(x)$)
What method do I use to solve this?

Is there any possibility to convert first order differential equation to second order differential equation?
I have a system of first order differential equations as below and i need the right hand side of the second order form of this first order equation. (if possible)
$\frac{du(t)}{dt}=[t{u}_2(t);4u_1(t{)}^{3/2}]$
Any help would be appreciated.

What is the linear approximation to $\sin <!--\; \; -->(-<!--\; -\; -->0.5)$?

For every differentiable function $f:\mathbb{R}\to <!--\; \to \; -->\mathbb{R}$, there is a function $g:\mathbb{R}\times <!--\; \times \; -->\mathbb{R}\to <!--\; \to \; -->\mathbb{R}$ such that $g(f(x),f^{\prime }(x))=0$ for every $x$ and for every differentiable function $h:\mathbb{R}\to <!--\; \to \; -->\mathbb{R}$ holds that
being true that for every $x\in <!--\; \in \; -->\mathbb{R}$, $g(h(x),h^{\prime }(x))=0$ and $h(0)=f(0)$ implies that $h(x)=f(x)$ for every $x\in <!--\; \in \; -->\mathbb{R}$.
i.e every differentiable function $f$ is a solution to some first order differential equation that has translation symmetry.

Problem:
Solve the following differential equations.
$x\frac{dy}{dx}+y=y^{-<!--\; -\; -->2}$
Answer:
To solve this equation, we reduce it to a linear differential equation with the substitution $v=y^3$.
$\begin{array}{rl}x{y}^2\frac{dy}{dx}+y^3& =1\\ \frac{dv}{dx}& =3y^2\frac{dy}{dx}\\ 3x{y}^2\frac{dy}{dx}+3y^3& =3\\ \frac{dv}{dx}+3v& =3\end{array}$
Now we have a first order linear differential equation. To solve it, we use the integrating factor $I=e^{\int <!--\; \int \; -->P(x)}$. In this case, we have P(x)=3.
$\begin{array}{rl}I(x)& =e^{3x}\\ e^{3x}\frac{dv}{dx}+3e^{3x}v& =3e^{3x}\\ D\left(e^{3x}v\right)& =3e^{3x}\\ e^{3x}v& =e^{3x}+C\\ v& =1+e^{-<!--\; -\; -->3x}\\ y^3& =1+e^{-<!--\; -\; -->3x}\end{array}$
Now to check my answer.
$\begin{array}{rl}3y^2\frac{dy}{dx}& =-<!--\; -\; -->3C{e}^{-<!--\; -\; -->3x}\\ y^2\frac{dy}{dx}& =-<!--\; -\; -->C{e}^{-<!--\; -\; -->3x}\\ \frac{dy}{dx}& =-<!--\; -\; -->C{e}^{-<!--\; -\; -->3x}{y}^{-<!--\; -\; -->2}\\ x\frac{dy}{dx}+y& =x(-<!--\; -\; -->C{e}^{-<!--\; -\; -->3x}{y}^{-<!--\; -\; -->2})+{(1+e^{-<!--\; -\; -->3x})}^{\frac{1}{3}}\end{array}$
I cannot seem to get my answer to check. Where did I go wrong?
Here is my second attempt to solve the problem:
To solve this equation, we reduce it to a linear differential equation with the substitution $v=y^3$.
$\begin{array}{rl}x{y}^2\frac{dy}{dx}+y^3& =1\\ \frac{dv}{dx}& =3y^2\frac{dy}{dx}\\ 3x{y}^2\frac{dy}{dx}+3y^3& =3\\ x\frac{dv}{dx}+3v& =3\\ \frac{dv}{dx}+3x^{-<!--\; -\; -->1}v& =3x^{-<!--\; -\; -->1}\end{array}$
Now we have a first order linear differential equation. To solve it, we use the integrating factor $I=e^{\int <!--\; \int \; -->P(x)}$. In this case, we have $P(x)=3x^{-<!--\; -\; -->1}$.
$\begin{array}{rl}I& =e^{3\int <!--\; \int \; -->x^{-<!--\; -\; -->1}dx}=e^{3\ln <!--\; \; -->|x|}\\ I& =3x\\ 3x\frac{dv}{dx}+9v& =9\end{array}$
Now, I want to write:
$D(3xv)=9$
but that is wrong. What did I do wrong?
Here is my third attempt to solve the problem. Last time, I made a mistake in finding the integrating factor.
To solve this equation, we reduce it to a linear differential equation with the substitution $v=y^3$
$\begin{array}{rl}x{y}^2\frac{dy}{dx}+y^3& =1\\ \frac{dv}{dx}& =3y^2\frac{dy}{dx}\\ 3x{y}^2\frac{dy}{dx}+3y^3& =3\\ x\frac{dv}{dx}+3v& =3\\ \frac{dv}{dx}+3x^{-<!--\; -\; -->1}v& =3x^{-<!--\; -\; -->1}\end{array}$
Now we have a first order linear differential equation. To solve it, we use the integrating factor $I=e^{\int <!--\; \int \; -->P(x)}$. In this case, we have $P(x)=3x^{-<!--\; -\; -->1}$.
$\begin{array}{rl}I& =e^{3\int <!--\; \int \; -->x^{-<!--\; -\; -->1}dx}=e^{3\ln <!--\; \; -->|x|}\\ I& =x^3\\ x^3\frac{dv}{dx}+3x^{2}v& =3x^2\\ D(x^{3}v)& =x^3+C\\ x^{3}v& =x^3+C\\ v& =C{x}^{-<!--\; -\; -->3}+1\\ y^3& =C{x}^{-<!--\; -\; -->3}+1\end{array}$
Do I have it right now?

Use the Laplace transform to solve the following initial value problem:
${\displaystyle y{}^{″}+y=2t}$
with ${\displaystyle y\left(\frac{\pi }{4}\right)=\frac{\pi }{2}}$ and ${\displaystyle y^{\prime }\left(\frac{\pi }{4}\right)=2-\sqrt{2}}$

Trying to find the Laplace transform of
${\displaystyle \frac{\cos t}{t}}$
It comes out as infinity, but that doesn't make any sense.
Does this mean that this function doesn't have a Laplace transform or is something wrong here?

I'm having difficulties calculating a simple Laplace inverse :
${\displaystyle \frac{S-4}{S^2-2S-11}}$

Use linear approximation to estimate $\tan <!--\; \; -->(\frac{\mathrm{\pi <!--\; \pi \; -->}}{4}+0.05)$. Identify the differentials $dy$ and $d$ in the situation.

I want to find the laplace inverse of
${\displaystyle s^{-\frac{3}{2}}}$
the steps given in the solution manual are as follows:
${\displaystyle \frac{2}{\sqrt{\pi }}\frac{\sqrt{\pi }}{2s^{\frac{3}{2}}}=2\sqrt{\frac{t}{\pi }}}$
I know the first part ${\displaystyle \frac{2}{\sqrt{\pi }}}$ is obtained using the gamma function ${\displaystyle \mathrm{\Gamma }\left(\frac{3}{2}\right)}$ , but not quite sure how the rest is obtained.

How to prove that the inverse Laplace Transform of zero is zero itself?
${\displaystyle L^{-1}\left\{0\right\}=0}$
I know that the inverse Laplace Transform of a constant is Dirac's Delta. But I think that that applies only to positive constants.

Find a solution to ${\displaystyle x^{\prime }\left(t\right)+{\int }_{\left\{0\right\}}^t(t-s)x\left(s\right)ds=t+\frac{1}{2}{t}^2+\frac{1}{24}{t}^4}$

I am stuck with this equation. If you can help me
${\displaystyle y{}^{″}\left(t\right)+12y^{\prime }\left(t\right)+32y\left(t\right)=32u\left(t\right)}$ with ${\displaystyle y\left(0\right)=y^{\prime }\left(0\right)=0}$
I found the laplace transform for y(t)
${\displaystyle Y\left(p\right)=x=\frac{32}{\left(p(p^2+12p+32)\right)}}$
so i need the laplace transform of y(t) and then the solution for y(t).

I am having trouble finding the inverse Laplace transform of:
${\displaystyle \frac{1}{s^2-9s+20}}$
I tried writing it in a different way:
${\displaystyle L^{-1}\left\{\frac{1}{{(s-\frac{9}{2})}^2-{\left(\frac{1}{2}\right)}^2}\right\}=2L^{-1}\left\{\frac{\frac{1}{2}}{{(s-\frac{9}{2})}^2-{\left(\frac{1}{2}\right)}^2}\right\}}$

Solve the differential equation by using method of variation of parametrs
${\displaystyle y{}^{‴}+9y^{\prime }=\frac{1}{f\left(x\right)}}$
take ${\displaystyle f\left(x\right)=\cos 11x}$

${\displaystyle H\left(s\right)=\frac{s-5}{(s-3)(s-1)}}$
The inverse Laplace transform of H(s) is equal to
${\displaystyle f\cdot g}$