Learn First Order Problems with These Differential Equations Examples

Im clueless on how to solve the following question...
$x{e}^y\frac{dy}{dx}=e^y+1$
What i've done is...
$\frac{dy}{dx}=\frac{1}{x}+\frac{1}{x{e}^e};\frac{dy}{dx}-<!--\; -\; -->\frac{1}{x{e}^e}=\frac{1}{x}$
Find the integrating factor..
$$\begin{gathered} v(x)=e^{P(x)};whereP(x)=\int <!--\; \int \; -->p(x)dx\Rightarrow <!--\; \Rightarrow \; -->P(x)=\int <!--\; \int \; -->\frac{1}{x}dx=ln|x| \\ v(x)=e^{P(x)}=e^{ln|x|}=x; \\ y=\frac{1}{v(x)}\int <!--\; \int \; -->v(x)q(x)dx=\frac{1}{x}\int <!--\; \int \; -->x\frac{1}{x}dx=1+c \end{gathered}$$
I know I made a mistake somewhere. Would someone advice me on this?

I was wondering if I could get some advice on how to tackle this question:
Consider the differential equation
$x^2\frac{dy}{dx}+2xy-<!--\; -\; -->y^3=0(3)$
Make the substitution $u=y^{-<!--\; -\; -->2}$ and show that the differential equation reduces to
$-<!--\; -\; -->\frac{1}{2}{x}^2\frac{du}{dx}+2xu-<!--\; -\; -->1=0(4)$
Solve equation (4) for u(x) and hence write down the solution for equation (3).
I'm trying to do the first part of showing that the differential equation reduces to equation 4. I have started out by:
$\begin{array}{rl}u& =y^{-<!--\; -\; -->2}\\ & =\frac{1}{y^2}\\ \therefore <!--\; \therefore \; -->y^2& =\frac{1}{u}\\ ⟹<!--\; ⟹\; -->y& =\pm <!--\; \pm \; -->\sqrt{\frac{1}{u}}\end{array}$
I'm not sure where to continue on from here though.

I am given this:
$(2x+1)\frac{dy}{dx}+y=0$
I tried this:
$\frac{1}{(2x+1)}dx=\frac{-<!--\; -\; -->1}{y}dy$
Then integrated the above sum and got this:
$\frac{ln(2x+1)}{2}=-<!--\; -\; -->ln(y)$
The answer is: $y^2(2x+1)=C$.
I tried solving it by placing the like terms together and integrating them. However, my answer is wrong from the answer given. Could you point out my mistake? Or am i evaluating the entire thing incorrectly?
All suggestions and help are appreciated!

I've been trying to solve the following differential equation:
$3x+y(x)-<!--\; -\; -->2+\frac{dy(x)}{dx}(x-<!--\; -\; -->1)=0$
And I found out it's an exact differential equation, since it can be rearranged as $(3x+y(x)-<!--\; -\; -->2)dx+(x-<!--\; -\; -->1)dy=0$
Assuming that a function $U(x,y)=\frac{\mathrm{\partial <!--\; \partial \; -->}U}{\mathrm{\partial <!--\; \partial \; -->}x}dx+\frac{\mathrm{\partial <!--\; \partial \; -->}U}{\mathrm{\partial <!--\; \partial \; -->}y}dy=k$, (where $k\equiv <!--\; \equiv \; -->$ constant) exists, I calculated it as:
$U(x,y)=\int <!--\; \int \; -->(3x+y-<!--\; -\; -->2)dx+\int <!--\; \int \; -->(x-<!--\; -\; -->1)dy=k$
And I got
$y(x)=\frac{-<!--\; -\; -->3x^2}{2(2x-<!--\; -\; -->1)}+\frac{2x}{2x-<!--\; -\; -->1}+\frac{k}{2x-<!--\; -\; -->1}$
But Mathematica says the solution is
$y(x)=\frac{-<!--\; -\; -->3x^2}{2(x-<!--\; -\; -->1)}+\frac{2x}{x-<!--\; -\; -->1}+\frac{k}{1-<!--\; -\; -->x}$
So either I assumed something which isn't correct or I made a mistake along the process. Where did I go wrong?

I am curious if there are any examples of functions that are solutions to second order differential equations, that also parametrize an algebraic curve.
I am aware that the Weierstrass $\mathrm{\wp <!--\; \wp \; -->}$ - Elliptic function satisfies a differential equation. We can then interpret this Differential Equation as an algebraic equation, with solutions found on elliptic curves. However this differential equations is of the first order.
So, are there periodic function(s) $F(x)$ that satisfy a second order differential equation, such that we can say these parametrize an algebraic curve?
Could a Bessel function be one such solution?

In a course on partial differential equations I came through this theorem about the general solution of a first order quasi-linear partial differential equation.
1. The general solution of a first-order, quasi-linear partial differential equation
$a(x,y,u)u_x+b(x,y,u)u_y=c(x,y,u)$
is given by $f(\varphi ,\psi )=0$, where $f$ is an arbitrary function of $\varphi (x,y,u)$ and $\psi (x,y,u).$.
2. $\varphi =C1$ and $\psi =C2$ are solution curves of the characteristic equations
$\frac{dx}{a}=\frac{dy}{b}=\frac{du}{c}.$
Is there any geometric interpretation of both these points so that I can have a better intuitive understanding of the graphical representation of $f$,$\mathrm{\varphi <!--\; \varphi \; -->}$ and $\mathrm{\psi <!--\; \psi \; -->}$ ?

I have been solving an inexact first order ODE, but arriving to the step of finding an integrating factor /mu, I was enable to find either /mu of x or /mu of y. Is there another way of solving this ODE? or how can I come up with the solution.
$(e^{x+y}-<!--\; -\; -->y{e}^x)\mathrm{d}x+e^y\mathrm{d}y=0$

I've been working on this differential equation for days now but I can not find a solution. I understand that finding a solution in closed form is difficult to say (assuming there is). What I ask is if there is a substitution (change of variables) that can "transform" this differential equation from the second order to the first order so that it is not too constrained. Thank you for your interest and time.
$y^{″}(x)+ax\cos <!--\; \; -->y(x)=0$
where $a\in <!--\; \in \; -->\mathbb{R}$

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.

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?

How to solve this first-order differential equation,I don't have any idea which type this is..
${\displaystyle y^{\prime }+{(x-y)}^2=0}$
I have ${\displaystyle y^{\prime }+x^2-2xy+y^2=0}$, and then I divide by ${\displaystyle x^2}$
${\displaystyle t^{\prime }+t(\frac{1}{x}-2x)+t^{2}x=-x}$
and replace ${\displaystyle \frac{y}{x}=t}$, so I have then
It seems like it's a Riccati's diff. equation,but I don't know how to do it beacuse I don't have any particular solution..
On Wolfram Alpha solution is
$y=\frac{1}{c_1{e}^{2ix}+\frac{i}{2}}+x+i$

Solve the following first-order nonlinear ordinary differential equation:
${\displaystyle y^{\prime }\left(x\right)+\log \left(y\left(x\right)\right)=-x-1}$

Let ${\displaystyle (\bar{x},\bar{y},\bar{z})}$ be an equilibrium of autonomous first-order differential equation:
${\displaystyle \{\begin{array}{c}\dot{x}=f_1(x,y,z)\\ \dot{y}=f_2(x,y,z)\\ \dot{z}=f_3(x,y,z)\end{array}}$
Is it possible to say that ${\displaystyle (\bar{x},\bar{y},\bar{z})}$ is unique equilibrium point of the following system:
${\displaystyle \{\begin{array}{c}\dot{x}=f_1(x,y,z)-(x-\bar{x})\\ \dot{y}=f_2(x,y,z)-(y-\bar{y})\\ \dot{z}=f_3(x,y,z)-(z-\bar{z})\end{array}}$

a) Solve the differential equation:
${\displaystyle (x+1)\frac{dy}{dx}-3y={(x+1)}^4}$
given that ${\displaystyle y=16}$ and ${\displaystyle x=1}$, expressing the answer in the form of ${\displaystyle y=f\left(x\right)}$.
b) Hence find the area enclosed by the graphs ${\displaystyle y=f\left(x\right),y={(1-x)}^4}$ and the ${\displaystyle x-a\xi s}$.
I have found the answer to part a) using the first order linear differential equation method and the answer to part a) is: ${\displaystyle y={(1+x)}^4}$. However how would you calculate the area between the two graphs (${\displaystyle y={(1-x)}^4}$ and ${\displaystyle y={(1+x)}^4}$) and the x-axis.

I have first-order differential equation
${\displaystyle y=xy\prime +\frac{1}{2}{\left(y\prime \right)}^2}$
Maybe, with this someone will find way to solve it
${\displaystyle \frac{1}{2}{y}^{\prime }(2x+y^{\prime })=y}$
I thought I can use ${\displaystyle x^2+y=t}$ for subtitution and when I derivate, I have ${\displaystyle t^{\prime }=2x+y^{\prime }}$ which is acctualy the
${\displaystyle (t^{\prime }-2x)t^{\prime }=2t=2x^2}$
same as previous. Who knows?