Expert Assistance for Separable Differential Equations: Comprehensive Resources and Practice Problems

1 degree on celsius scale is equal to
A) $\frac{9}{5}$ degree on fahrenheit scale
B) $\frac{5}{9}$ degree on fahrenheit scale
C) 1 degree on fahrenheit scale
D) 5 degree on fahrenheit scale

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?

Separation of variables differential equation
Solve this equation by separation of variables: xu'=3u.
I see that the answer should be u(x)=$c{x}^3$, but since I have no knowledge of differential equations, can someone provide the steps?

How to solve such fraction differential equation?
Here's my first-order differential equation:
$(x^3-<!--\; -\; -->2x{y}^2)dx+3y{x}^{2}dy=xdy-<!--\; -\; -->ydx$
I've tried to make it fraction, but it isn't separable differential equation, also it isn't differential equation in total differentials, so after it I lose any clue for answer.

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

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.

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?

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)

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.

Is there a solution for y for $\frac{dy}{dx}=ax{e}^{by}$
I have come up with the equation in the form
$\frac{dy}{dx}=ax{e}^{by}$
, where a and b are arbitrary real numbers, for a project I am working on. I want to be able to find its integral and differentiation if possible. Does anyone know of a possible solution for $y$ and/or $\frac{d^{2}y}{d{x}^2}$?

All alternative solution for an equation
I'm looking for all alternative solutions of this
$x^{\prime }=x(x-<!--\; -\; -->1)(x+1)$
But I absolutely don't know what I have to do! Thanks!

Separable Differential Equation explained
So i have the equation $\frac{dy}{dt}=1+y$
looking at the answers my lecture has given it states
$\frac{d}{dt}(1+y)=1+y$
then $1+y(t)=A{e}^t$ where A is a constant
Can someone explain these steps please?

How do I find the singular solution of the differential equation $y^{\prime }=\frac{y^2+1}{xy+y}$?
I start out with the separable differential equation,
$y^{\prime }=\frac{dy}{dx}=\frac{y^2+1}{xy+y}=\frac{y^2+1}{y(x+1)}$
Thus, $\frac{1}{x+1}dx=\frac{y}{y^2+1}dy$
Then integrating both sides of the equation, I get
$\ln <!--\; \; -->(x+1)=\frac{1}{2}\ln <!--\; \; -->(y^2+1)+C$
Now, $e^{\ln <!--\; \; -->(x+1)}$ = $e^{\frac{1}{2}\ln <!--\; \; -->(y^2+1)+C}$. So...
$(x+1)=e^C(y^2+1{)}^{\frac{1}{2}}$
I kind of wanted to know if this is indeed the correct general formula. And also, how would I find the singular solution, if there happens to be one in this case.

Solving separable equations with dy/dx on both sides?
I'm unsure on how to even start with this equation:
$y-<!--\; -\; -->x\frac{dy}{dx}=1+x^2\frac{dy}{dx}$
I thought it would be possible to cancel out but the answer was different. Could someone please provide a hint on how to start this equation? I thought about moving the $-<!--\; -\; -->x\frac{dy}{dx}$ over to the RHS and then group it under the common variable $\frac{dy}{dx}$ but the question was under the 'separable differential' section in my textbook so I don't think it's the correct method.

Separable differential equation $x^2{y}^{″}=2y$
Show that $x^2-<!--\; -\; -->x^{-<!--\; -\; -->1}$ is a solution of
$x^2\frac{{\text{d}}^{2}y}{\text{d}{x}^2}=2y$
I've tried separating the x and y terms and then integrating, but I could tell it wouldn't work. Any ideas? It's a simple question but I have not attempted them in a while.

a separable differential equation
given this
$\frac{d}{dx}=x(1-<!--\; -\; -->x)$
where
$x(0)=0.1$
is this correct:?
$\frac{dx}{dt}=x(1-<!--\; -\; -->x)$
$t-<!--\; -\; -->t_0=\int <!--\; \int \; -->\frac{dx}{x(1-<!--\; -\; -->x)}$
Where did the t and $t_0$ come from?

First order non linear differential equation , non separable
First time I run into an equation of this "form", for non linear equations I know separation of variables and substitution. It doesn't seem to be homogenous of degree $0$ so the substitution $v=\frac{x}{t}$ is out of the question? Applying $\ln$ to both sides didn't do much either
$x^{\prime }(t)=e^{t+x(t)}-<!--\; -\; -->1,x(0)=1$
Wolfram gives the solution
$x(t)=-<!--\; -\; -->\ln <!--\; \; -->(c_1-<!--\; -\; -->t)-<!--\; -\; -->t$

How do you solve the following separable differential equation: y'y = y + 1?
I just started learning about differential equations and encountered following equation:
$y^{\prime }y=y+1$
But I'm not sure how the integration is performed. What rules are used? How does $\int <!--\; \int \; -->\frac{y^{\prime }y}{y+1}dx$ become $-<!--\; -\; -->\log <!--\; \; -->(y+1)+y$?

2 separable O.D.Es.
a) $y^{\prime }sinx+ycosx=2cosx$
b) $\frac{1+y^{\prime }}{1-<!--\; -\; -->y^{\prime }}=\frac{1-<!--\; -\; -->\frac{y}{x}}{1+\frac{y}{x}}$
These are separable first order differential equations but i don't know where to start. I've tried substituting for u but it doesn't seem to work out. Thanks