Solve System of Differential Equations with Expert Help

Correct approach to this homogeneous differential equation
I am trying to find the general solution to the following equation, but the integral at the end is very complicated and leads me to believe I may have made a mistake somewhere.
$x{y}^2\frac{dy}{dx}=y^3+x{y}^2-<!--\; -\; -->x^{2}y-<!--\; -\; -->x^3$
Which, by the substitution $z=\frac{y}{x}$, can be rearranged into the equation
$x\frac{dz}{dx}=1-<!--\; -\; -->\frac{1}{z}-<!--\; -\; -->\frac{1}{z^2}$
This is a separable equation, which I separated into
$\int <!--\; \int \; -->\frac{1}{1-<!--\; -\; -->\frac{1}{z}-<!--\; -\; -->\frac{1}{z^2}}dz=\int <!--\; \int \; -->\frac{1}{x}dx$
The right-hand side is easy to solve, but the left-hand integral is giving me trouble. Assuming I did the steps leading up to it correctly, the integral has me stumped. Even WolframaAlpha is unhelpful. My first thought would be to try a partial fraction, but after a few attempts it does not seem to work.
Am I approaching this differential equation correctly? Is there an error I haven't caught?

How can i solve this separable differential equation?
Given Problem is to solve this separable differential equation:
$y^{\mathrm{\prime <!--\; \prime \; -->}}=\frac{y}{4x-<!--\; -\; -->x^2}.$
My approach: was to build the integral of y':
$\int <!--\; \int \; -->y^{\mathrm{\prime <!--\; \prime \; -->}}=\int <!--\; \int \; -->\frac{y}{4x-<!--\; -\; -->x^2}dy=\frac{y^2}{2(4x-<!--\; -\; -->x^2)}.$
But now i am stuck in differential equations, what whould be the next step? And what would the solution looks like? Or is this already the solution? I doubt that.
P.S. edits were only made to improve language and latex

Find Inverse Laplace Transform of $\frac{1}{(s^2+1)(s^2-<!--\; -\; -->2s+7)}.$

How to find the Laplace transform of $|\sin <!--\; \; -->(t)|$?

Is $(2x+y)dx-<!--\; -\; -->xdy=0$ a separable differential equation?
I was given the following differential equation in an assignment the other day:
$(2x+y)dx-<!--\; -\; -->xdy=0$
The problem specified to solve the equation using the method of separation of variables. My problem was setting the integral, I tried multiple manipulations with but nothing seemed to work. So, I have to ask can this equation be solved using separation of variables?

Show ${\int <!--\; \int \; -->}_s^{\mathrm{\infty <!--\; \infty \; -->}}f(x)dx=\mathcal{L}\{\frac{F(t)}{t}\}$ given $f(x)={\int <!--\; \int \; -->}_0^{\mathrm{\infty <!--\; \infty \; -->}}{e}^{-<!--\; -\; -->xt}F(t)dt$

Separable Differential Equation
How would you solve the next ODE?
$\frac{dy}{dt}=\frac{at+by+m}{ct+dy+n},$
where $a,b,c,d,m,n$ are constants and $ad=bc$
Corrected.

This integral sounds quite complex and I could not find an approximate equivalent.
${\int <!--\; \int \; -->}_0^{+\mathrm{\infty <!--\; \infty \; -->}}x\log <!--\; \; -->(1+x^2)e^{-<!--\; -\; -->Bx}dx$

Solve $y^{\prime }(t)=\sin <!--\; \; -->(t)+{\int <!--\; \int \; -->}_0^{t}y(x)\cos <!--\; \; -->(t-<!--\; -\; -->x)dx$ by Laplace transform
My try:
I applied Laplace transform on both sides of the equation.
$sL\{y(t)\}=\frac{1}{s^2+1}+L\{cos(t)\ast <!--\; \ast \; -->y(t)\}⟹<!--\; ⟹\; -->sL\{y(t)\}=\frac{1}{s^2+1}+L\{cos(t)\}\times <!--\; \times \; -->L\{y(t)\}$
Now, I'm stuck on applying the inverse Laplace transform on (*) to find $y(t)$

Laplace transform of $t^2{e}^{at}$??
try to prove that
$\mathcal{L}\{t^2{e}^{at}\}=\frac{2}{(s-<!--\; -\; -->a{)}^3}.$
I've gotten to the last integration by parts where
$\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}{\int <!--\; \int \; -->}_0^n\frac{1}{(a-<!--\; -\; -->s{)}^{2}2e^{(a-<!--\; -\; -->s)t}}dt={\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}\frac{2}{(a-<!--\; -\; -->s{)}^3}{e}^{(a-<!--\; -\; -->s)t}|}_0^n.$
Now what do I do?

Find the inverse Laplace transform for the following:
a. $F(s)={\displaystyle \frac{2s-<!--\; -\; -->3}{s^2-<!--\; -\; -->3s+2}}$
I used partial fractions and I ended up getting
$y(t)=e^t+e^{2t}.$
b. $F(s)={\displaystyle \frac{e^{-<!--\; -\; -->2s}}{s-<!--\; -\; -->9}}$
I ended up getting $e^{9t-<!--\; -\; -->9}h(t-<!--\; -\; -->2)$ and 0 being $\ne <!--\; \ne \; -->2$

Question is find the Laplace transform of this equation:
$x\frac{\mathrm{\partial <!--\; \partial \; -->}(w)}{\mathrm{\partial <!--\; \partial \; -->}(x)}+\frac{\mathrm{\partial <!--\; \partial \; -->}(w)}{\mathrm{\partial <!--\; \partial \; -->}(t)}=xt$
Boundary conditions :
$w(x,0)=0x\ge <!--\; \ge \; -->0$
$w(0,t)=0t\ge <!--\; \ge \; -->0$
I've got as far as
$\frac{d(W)}{d(x)}+\frac{s}{x}W=\frac{1}{s^2}$

$\underset{s\to <!--\; \to \; -->0^{+}}{lim}{\int <!--\; \int \; -->}_0^{\mathrm{\infty <!--\; \infty \; -->}}a(t)e^{-<!--\; -\; -->st}dt$
${\int <!--\; \int \; -->}_0^{\mathrm{\infty <!--\; \infty \; -->}}a(t)e^{-<!--\; -\; -->st}dt=f(s)$
What is the meaning of the limit of this integral as $s\to <!--\; \to \; -->0^{+}.$

Solving the IVP
$t{y}^{″}-<!--\; -\; -->t{y}^{\prime }+y=1$
where $y(0)=0$ and $y^{\prime }(0)=2$

The Laplace transform of $\mathcal{L}(t{e}^t\cos <!--\; \; -->t)$
How do I find it?

Separable Differential Equation dy/dt = 6y
The question is as follows:
$\frac{dy}{dt}=6y$
$y(9)=5$
I tried rearranging the equation to $\frac{dy}{6y}=dt$ and integrating both sides to get $(1/6)\ln <!--\; \; -->|y|+C=t$. After that I tried plugging in the 9 for y and 5 for t and solving but I can't quite seem to get it.

Completing partial derivatives to make them converge
For a function $f(x,y)$ of two independent variables we have an incomplete specification of its partial derivatives as follows:
$\frac{\mathrm{\partial <!--\; \partial \; -->}f(x,y)}{\mathrm{\partial <!--\; \partial \; -->}x}=\frac{1}{g(x,y)\sqrt{1-<!--\; -\; -->(\frac{ky}{x^{(1/3)}}{)}^2}}$
$\frac{\mathrm{\partial <!--\; \partial \; -->}f(x,y)}{\mathrm{\partial <!--\; \partial \; -->}y}=\left(\frac{3x}{4}\right){\left(\frac{k}{x^{(1/3)}}\right)}^2(2y)\frac{1}{g(x,y)\sqrt{1-<!--\; -\; -->(\frac{ky}{x^{(1/3)}}{)}^2}}$
Problem: finding a suitable $g(x,y)$ that makes the partial derivatives converge to a single function $f(x,y)$ that fulfills the condition $f(x,0)=x$
I will be grateful if people with many flight hours can offer suggestions for $g(x,y)$. Needless to say, I am not asking that they verify those suggestions, but in case someone would like, these are the inputs to Wolfram integrator:
$$\begin{gathered} \frac{1}{(g(x,y\text{ as }r)}\sqrt{1-<!--\; -\; -->(kr/x^{(}1/3){)}^2)} \\ \frac{(3t/4)(k/t^{1/3}{)}^2(2x)}{(g(x\text{ as }t,y\text{ as }x)\sqrt{1-<!--\; -\; -->(kx/t^{(}1/3){)}^2)}} \end{gathered}$$

How to solve $x{y}^{\prime }=2\sqrt{x^2+y^2}+y$?
And what would be the standard form to illustrate this situation? (e.g. $y^{\prime }+P(x)y=Q(x)$ would be standard form of first order linear differential equation)

How do you evaluate the inverse transform below using convolution ? ${\mathcal{L}}^{\mathcal{-<!--\; -\; -->}\mathcal{1}}\left[\frac{s}{(s^2+a^2{)}^2}\right]$
I tried
$\begin{array}{rl}{\mathcal{L}}^{\mathcal{-<!--\; -\; -->}\mathcal{1}}\left[\frac{s}{(s^2+a^2{)}^2}\right](t)& =\underset{0}{\overset{t}{\int <!--\; \int \; -->}}\sin <!--\; \; -->t\cdot <!--\; \cdot \; -->\cos <!--\; \; -->(at-<!--\; -\; -->a\mathrm{\tau <!--\; \tau \; -->})d\mathrm{\tau <!--\; \tau \; -->}=\sin <!--\; \; -->t\cdot <!--\; \cdot \; -->\underset{0}{\overset{t}{\int <!--\; \int \; -->}}\cos <!--\; \; -->(at-<!--\; -\; -->a\mathrm{\tau <!--\; \tau \; -->})d\mathrm{\tau <!--\; \tau \; -->}\\ & =\sin <!--\; \; -->t\cdot <!--\; \cdot \; -->{[-<!--\; -\; -->\frac{1}{a}\sin <!--\; \; -->(at-<!--\; -\; -->a\mathrm{\tau <!--\; \tau \; -->})]}_0^t=\sin <!--\; \; -->t\cdot <!--\; \cdot \; -->[0-<!--\; -\; -->(-<!--\; -\; -->\frac{1}{a}\sin <!--\; \; -->(at))]\\ & =\frac{1}{a}{\sin }^2<!--\; \; -->(at)\end{array}$
What have I done wrong?