Learn First Order Problems with These Differential Equations Examples

What is a solution to the differential equation ${\displaystyle (2+x)y\prime =2y}$?

How do you find the particular solution to ${\displaystyle yy\prime -e^x=0}$ that satisfies y(0)=4?

What is a solution to the differential equation ${\displaystyle \frac{dy}{dt}=\frac{e^t}{4y}}$?

What is a solution to the differential equation ${\displaystyle \frac{dy}{dx}=3}$?

What are separable differential equations?

General solution of the differential equation ${\displaystyle y\prime =\frac{y}{x}+x{e}^x}$?

Solve the differential equation ${\displaystyle \frac{dy}{dx}+x\tan (y+x)=1}$ ?

If F(x) and G(x) both solve the same initial value problem, then is it true that F(x)=G(x)?

What is a solution to the differential equation ${\displaystyle \frac{dy}{dx}=\frac{y^2}{x^2}+\frac{y}{x}+1}$?

How to you find the general solution of ${\displaystyle \frac{dy}{dx}=3x^3}$?

How can I solve this differential equation? : ${\displaystyle (2x^3-y)dx+xdy=0}$

How to you find the general solution of ${\displaystyle \frac{dy}{dx}=\frac{e^x}{1+e^x}}$?

What is a general solution to the differential equation ${\displaystyle y\prime =5x^{\frac{2}{3}}{y}^4}$?

What is the general solution of the differential equation? : ${\displaystyle y\prime =x(1+y^2)}$

How can I solve this first order linear differential equation?
$y^{\prime }=1-<!--\; -\; -->\frac{2}{x+y}$
I have tried turning it into an inexact differential equation, but I get an integration factor $\mathrm{\mu <!--\; \mu \; -->}(x,y)$ and I don't know how to apply it.

Please consider the following differential equation:
$\begin{array}{rcl}2xy\frac{dy}{dx}+(1+x)y^2& =& e^x\end{array}$
I do not know how to solve this problem. I do not believe it is a separable differential equation. Could somebody point me in the right direction?

Consider the differential equation
$\stackrel{\dot{}<!--\; \dot{}\; -->}{x}(t)=\mathrm{\alpha <!--\; \alpha \; -->}x(t)+\mathrm{\beta <!--\; \beta \; -->}u(t),$
with $\mathrm{\alpha <!--\; \alpha \; -->},\mathrm{\beta <!--\; \beta \; -->}>0$ positive constant scalars and initial condition $x(0)=x_0$. Function $u(\cdot <!--\; \cdot \; -->)$ is known. The Laplace solution for this equation should be
$x(t)=e^{\mathrm{\alpha <!--\; \alpha \; -->}(t-<!--\; -\; -->t_0)}{x}_0+\mathrm{\beta <!--\; \beta \; -->}{\int <!--\; \int \; -->}_0^t{e}^{\mathrm{\alpha <!--\; \alpha \; -->}(t-<!--\; -\; -->\mathrm{\tau <!--\; \tau \; -->})}u(\mathrm{\tau <!--\; \tau \; -->})d\mathrm{\tau <!--\; \tau \; -->}.$
However, if now I take the time derivative, I get
$\stackrel{\dot{}<!--\; \dot{}\; -->}{x}(t)=\mathrm{\alpha <!--\; \alpha \; -->}{e}^{\mathrm{\alpha <!--\; \alpha \; -->}(t-<!--\; -\; -->t_0)}{x}_0+\mathrm{\beta <!--\; \beta \; -->}u(t).$
Comparing to the first equation, this leaves me with
$x(t)=e^{\mathrm{\alpha <!--\; \alpha \; -->}(t-<!--\; -\; -->t_0)}{x}_0,$
which looks inconsistent with the Laplace solution. Anybody has a clue on what's going on here?

I have the differential equation
$x^2{y}^{\prime }(x)+2xy(x)=y^2(x)$
With initial condition y(1)=1 and I want to solve this, by observation, we can see the LHS is $x^{2}y(x)$ but I am unfamiliar with tackling the RHS and was wondering where to go from here.
$x^{2}y(x)=\int <!--\; \int \; -->y^2(x)dx$

A car is travelling at 100 km/h on a level road when it runs out of fuel. Its speed v (in km/h) starts to decrease according to the formula
$\frac{dv}{dt}=-<!--\; -\; -->kv(1)$
where k is constant. One kilometre after running out of fuel its speed has fallen to 50 km/h. Use the chain rule substitution
$\frac{dv}{dt}=\frac{dv}{ds}\frac{ds}{dt}=\frac{dv}{ds}v$
to solve the differential equation.
Note: Although I haven't solved it yet, the answers say that this isn't a reasonable model as the velocity is always positive; I didn't make a typo in the question.
What I'm trying to do is solve velocity as a function of displacement (s, in km), velocity as a function of time (t, in hours), and displacement as a function of time (I need these functions for later parts of the question).
So far I've found velocity as a function of displacement (v(s)):
$\frac{dv}{dt}=-<!--\; -\; -->k\frac{ds}{dt}\text{(from (1))}$
$\int <!--\; \int \; -->\frac{dv}{dt}dt=-<!--\; -\; -->k\int <!--\; \int \; -->\frac{ds}{dt}dt$
$v(s)=-<!--\; -\; -->ks+C$
$v(0)=100⟹<!--\; ⟹\; -->C=100,v(1)=50⟹<!--\; ⟹\; -->k=50$
$v(s)=-<!--\; -\; -->50s+100$
Then I've tried to find velocity as a function of time (v(t)), but I've got stuck. I can't find any differential equation I can use to get this, or to get displacement as a function of time (s(t)).
The answer key says $v(t)=100e^{-<!--\; -\; -->50t}$ and $s(t)=2(1-<!--\; -\; -->e^{-<!--\; -\; -->50t})$
I've solved such questions many times before, but it's been a while so I'm a bit rusty. So, even a hint might be enough for me to realise what to do.

What is the standard method for finding solutions of differential equations such as this one? (if there is any)
$x{y}^{\prime }=y^2-<!--\; -\; -->(2x+1)y+x^2+2x$
where $y=ax+b$ is a particular solution.
Do I substitute $y$ with $ax+b+u(x)$ and then search for a solution or am I not noticing something and there's quicker way?