Solve System of Differential Equations with Expert Help

Elementary Separable Differential Equation with tricky integral
Dear Plaimath. The problem is to solve the separable differential equation $\sqrt{xy}\frac{dy}{dx}=\sqrt{4-<!--\; -\; -->x}$
My work thus far: $$\begin{gathered} \sqrt{xy}dy=\sqrt{4-<!--\; -\; -->x}dx \\ \sqrt{y}dy=\sqrt{\frac{4-<!--\; -\; -->x}{x}}dx \\ \int <!--\; \int \; -->\sqrt{y}dy=\int <!--\; \int \; -->\sqrt{\frac{4-<!--\; -\; -->x}{x}}dx \\ \frac{2}{3}{y}^{3/2}=\int <!--\; \int \; -->\sqrt{\frac{4-<!--\; -\; -->x}{x}}dx \\ \frac{2}{3}{y}^{3/2}=\int <!--\; \int \; -->\sqrt{\frac{4}{x}-<!--\; -\; -->1}dx \end{gathered}$$
Is there an easier approach I have missed, or should I persist in trying to solve the integral.

Solve the following differential equation using separation of variables
$f^{\prime }(x)+2x[f^2(x)-<!--\; -\; -->f(x)]=0$

Solving a separable differential equations
We have an equation:
$\frac{\mathrm{d}P}{\mathrm{d}t}=kP(1-<!--\; -\; -->\frac{P}{M}),$
and we need to find $P(t)$ given the initial $P(0)$ given the initial $P(0)$. Here k is a constant and P represents population, M represents maximum population.
I tried to use separable differential equations, but I am slightly confused. There is a hint:
$\frac{M}{P(M-<!--\; -\; -->P)}=\frac{1}{P}+\frac{1}{M-<!--\; -\; -->P},$
and I need to simplify as far as possible. I have tried to use separable equations but I cannot seem to get the hint equation. I think the main issue is trying to get all P's on to one side.
Thank you in advance!

Why does the method of separable equations work in differential equations?
In differential equations, the idea of multiplying by an infinitesimal, dx, is used in the method of separable equations. My confusion is in why this works as I've heard many times in the past that you shouldn't multiply and cancel out infinitesimals like that. I understand that in the setting of nonstandard analysis that there may not be something wrong with this, but in the usual setting is there a more rigorous interpretation of this?

A carousel has a radius of 17 ft and takes 28 seconds to make one complete revolution. What is the linear speed of the carousel at its outside edge?

For the transformation $L:C^1[a,b]\to <!--\; \to \; -->$ $C[a,b]$ definied by $L(f)=\frac{df}{dx}$ (means, derivative of function $f$ w.r.t $x$),
a) Show that $L$ is linear.
b) Find $Ker(L)$.
c) Find $nullity(L)$.
d) Is $L$ one-to-one? Explain.
e) is $L$ invertible? Explain.

Use Laplace transform to solve the given initial-value problem.
$\frac{dy}{dt}-<!--\; -\; -->y=1,y(0)=0$

Earth rotates on an axis through its poles. The distance from the axis to a location on Earth 40 degrees north latitude is about 3033.5 miles. Therefore, a location on Earth at 40 degrees north latitude is spinning on a circle of radius 3033.5 miles. Compute the linear speed on the surfuce of Earth at 40 degrees north latitude.

Find the Laplace Transform of $L\{(\sin <!--\; \; -->ht)/t\}$

Consider the equation $dy/dx=y^{1/3}$.

Having trouble solving this separable differential equation
I am having some trouble with the following separable differential equation
$\frac{dx}{dt}=x(x-<!--\; -\; -->1)(x-<!--\; -\; -->3)$
with initial condition $x(0)=2$. What is ${\displaystyle \underset{t\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}x(t)}$?
I am having some trouble with the logarithmic laws when solving for $x(t)$

Solving separable differential equations by formal computations
When solving a separable differential equation we move from the original equation
$N(y){\displaystyle \frac{dy}{dx}}=M(x)$
to
$N(y)dy=M(x)dx$
Then we integrate both sides. My question is how precise is the expression $N(y)dy=M(x)dx$ ? is it a formal writing to simplify computation and get quickly into integrating both sides or is it a precise mathematical expression that has a precise mathematical meaning but goes beyond an introductory course on differential equations? Thanks for your help !

Tricky Separable Differential Equation
Please guide me:
$y^{\prime }+ay+b=0$
(a not zero) is supposed to be separable and has solution
$y=c{e}^{-<!--\; -\; -->ax}-<!--\; -\; -->\frac{b}{a}$
Here is my start to this problem:
$\frac{dy}{dx}+ay=-<!--\; -\; -->b$ is as far as I can go with this. How should I go about separating x and

Using Laplace transforms calculate the solution of the ordinary differential equation $\frac{d^{2}y}{d{t}^2}-<!--\; -\; -->2\frac{dy}{dt}+y=e^{-<!--\; -\; -->t}$, subject to the initial conditions y(0)=0 and $\frac{dy}{dt}(0)=1$. Select the correct answer from the following list.
$$\begin{gathered} a)y(t)=\frac{3t{e}^t}{2}+\frac{e^{-<!--\; -\; -->t}}{4}-<!--\; -\; -->\frac{e^t}{4} \\ b)y(t)=t{e}^t+\frac{t^2{e}^t}{2} \\ c)y(t)=\frac{t{e}^{-<!--\; -\; -->t}}{2}-<!--\; -\; -->\frac{e^{-<!--\; -\; -->t}}{4}+\frac{e^t}{4} \\ d)y(t)=t{e}^{-<!--\; -\; -->t}+\frac{t^2{e}^{-<!--\; -\; -->t}}{2} \\ e)y(t)=\frac{t{e}^{-<!--\; -\; -->t}}{2}-<!--\; -\; -->\frac{e^{-<!--\; -\; -->t}}{2}+\frac{e^t}{2} \end{gathered}$$

Solve this separable differential equation
How would I go about solving the following separable differential equation?
$\frac{dx(t)}{dt}=8-<!--\; -\; -->3x$ with $x(0)=4$?
My solution thus far is the following:
$\int <!--\; \int \; -->\frac{dx(t)}{8-<!--\; -\; -->3x}=\int <!--\; \int \; -->dt+C$
$\Rightarrow <!--\; \Rightarrow \; -->\text{ln}|8-<!--\; -\; -->3x(t)|=-<!--\; -\; -->3(t+C)$
Now using the fact that $x(0)=4$ we can solve for $C=-<!--\; -\; -->\frac{1}{3}\text{ln}|4|$. From this, I would seem to get
$|8-<!--\; -\; -->3x(t)|=e^{-<!--\; -\; -->3t-<!--\; -\; -->\frac{1}{3}\text{ln}4}$
but something seems to go wrong. Am I making a mistake somewhere and how could I solve this entirely?

If $y=x{e}^{7x}$ is a solution of a linear homogeneous 2nd order DE, then another solution might be?

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?

Let $y=f(x)=(x_1^2+2x_2,x_1{x}_2-<!--\; -\; -->3x_1)$
Is the linear approximation just $f(y)=f(x)+A(y-<!--\; -\; -->x)$ whenever y is approximately near $x$?