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

Seperable differential equation question
b) $(2x{y}^3)dx+(3x2y^2+y^4)dy=0$
c) $(2x{y}^3)dx+(3x2y^2+y^2)dy=0$
I know that c is a separable differential equation but b is not. Why? The only difference is the power of the final y.

Solving initial value problem of homogenous equation $y^{\prime }=\frac{7xy}{7x^2+2y^2}$ using substitution.
I am given the initial value problem
$\begin{array}{l}{y}^{\prime }={\displaystyle \frac{7xy}{7x^2+2y^2}}\\ y(1)=1\end{array}$
where I must answer in the form of $F(x,y)=\frac{7}{4}$
Here, I am also asked to use the substitution $y=xu$ to transform this differential equation into a separable differential equation in u. I am not sure how to go about doing this since I am not too familiar with differential equations. Any help would be greatly appreciated!

Solving the next differential equation
How can I solve the next differential equation?
$$\cos <!--\; \; -->(x+y)dx=x\sin <!--\; \; -->(x+y)dx+x\sin <!--\; \; -->(x+y)dy$$
I dont know what kind of equation it is. It's not homogeneous, separable differential equation or linear. Any suggestions?

Solve separable differential equation $\frac{dy}{dx}=2y^2$
Using the initial condition $y(2)=3$, find $y(1)$

Solve $x\frac{dy}{dx}-<!--\; -\; -->y=(x^2+y^2)$
Solve the following differential equation -
$x\frac{dy}{dx}-<!--\; -\; -->y=(x^2+y^2).$
It most probably involves a change of variables so that it becomes variable separable.

Question on ODE notation
I recently had a quiz on separable and exact differential equations. The question was this:
Answer using the exact method.No points will be awarded for any other method.
$y^{\prime }(x)+y(x)=y(x{)}^2$
My question is that I'm confused with this type of notation.
Whereas the usual exact differential equation would look like $Mdx+Ndy=0$
This looks nothing like it.
So what I did was as follows:
$y^{\prime }(x)+y(x)=y(x{)}^2$
$y^{\prime }+x=x^2$
$y^{\prime }+(x-<!--\; -\; -->x^2)=0$
$(x-<!--\; -\; -->x^2)dx+dy=0$
Then I proceeded to solving it with the exact method and found my answer:
$y=\frac{x^3}{3}-<!--\; -\; -->\frac{x^2}{2}+c$
Did I do this right? Thank you.

I have a problem solving this differential equation
$xdy-<!--\; -\; -->ydx=(x^2+y^2)dx$
How can I solve this differential equation? which type is this equation? I don't think it is separable and I don't think it's linear too, but it might be homogenous or exact maybe!
I tried to take partial derivatives of them by splitting them into P(x) and Q(x) to know if it's exact or not, but I didn't get any solutions.
which type this equation is and what's the answer with steps, please!

Solving separable differential equations via partial fractions: $\frac{dy}{dx}=y(y-<!--\; -\; -->1)$
I got
$$\frac{dy}{y(y-<!--\; -\; -->1)}=dx⟹<!--\; ⟹\; -->\ln <!--\; \; -->|y-<!--\; -\; -->1|-<!--\; -\; -->\ln <!--\; \; -->|y|=x+c$$
$$⟹<!--\; ⟹\; -->\ln <!--\; \; -->|\frac{y-<!--\; -\; -->1}{y}|=x+c$$
$$⟹<!--\; ⟹\; -->1-<!--\; -\; -->\frac{1}{y}=C{e}^x$$
$$⟹<!--\; ⟹\; -->y=\frac{1}{1-<!--\; -\; -->C{e}^x}$$

Having trouble with simple integral
sorry I'm having some trouble evaluating this integral
$\frac{dv}{dt}=-<!--\; -\; -->k(v-<!--\; -\; -->gt{)}^2-<!--\; -\; -->g$ where g and k are constants
I'm assuming you just separate and integrate but I cannot seem to get it to work out.

Solve $y^{\prime }=y^{2}x$
Given the differential equation $y^{\prime }=y^{2}x$, what are the solutions to y? Obviously this is not a linear differential equation, which I'm used to solving. I checked if it is separable, which would make it easier to solve, and I ended up with the solution $y=\frac{2}{x^2+c}$, where c is the constant. Are there more solutions to this I'm missing?

Find the equations of trajectories for the system of ODE
Find the equations of trajectories for the system
${\displaystyle \frac{dx}{dt}}=xy$, ${\displaystyle \frac{dy}{dt}}=-<!--\; -\; -->5x$
I know that the equation ${\displaystyle \frac{dy}{dx}}={\displaystyle \frac{\frac{dy}{dt}}{\frac{dx}{dt}}}$ is sometimes used to find equations for trajectories in the phase plane, so can I just say that the equations will all take the form:
${\displaystyle \frac{dy}{dx}}={\displaystyle \frac{-<!--\; -\; -->5x}{xy}}={\displaystyle \frac{-<!--\; -\; -->5}{y}}$ for varying values of y?

Transformation of a differential equation into variables u &x
Suppose we have $Mdx+Ndy=0$ We want to transform it into variable u&x such that $x=uy$
My attempt is below!
$$\frac{dy}{dx}=g(1/u)$$
$$\frac{dy}{dx}=\frac{u-<!--\; -\; -->x\frac{du}{dx}}{u^2}$$
After getting the new separable differential equation use it to solve
$$(x^2-<!--\; -\; -->3y^2)dx+2xydy=0$$

Calculus: Solving differential equation e^y(1+(dy)/(dx))=1
I'm assuming this is will break down into a separable differential equation once some algebra has been applied but I can't figure out how to go about it.
It's the first time I encounter a differential equation containing $e^y$
I started out distributing and then rewriting into $e^y+\frac{dy}{dx}{e}^y=1$, don't know if that's on the right track. I also tried rewriting with natural logarithm ln.
Also, to solve the equation explicitly: $y(0)=\ln <!--\; \; -->(3)$

Method Selection for Solving Differential Equations
I have a differential equation here:
$$\frac{dy}{dx}=\frac{4y-<!--\; -\; -->3x}{2x-<!--\; -\; -->y}$$
This equation can be solved relatively easily by dividing through by x, and then doing a y=vx substitution (making the equation separable).
However, my question is: can this be solved by taking a Laplace transform instead?
When I try to do so, I get this:
$$\mathcal{L}(2t{y}^{\prime })-<!--\; -\; -->\mathcal{L}(y{y}^{\prime })=\mathcal{L}(4y)-<!--\; -\; -->\mathcal{L}(3t)$$
The RHS is fairly simple when it comes to a laplace transform. On the LHS, I believe that the 2ty′ term can be translated to something like −F′(s). But that second term is causing me a bit of trouble.
Can a Laplace transform even be taken for such an equation? I feel that it should, I'm just not quite sure how, the integral ends up being quite messy. If so, could you show me how? (and if not, why not?)
Additionally, I'd like to know how to best select a method to solve a differential equation. In the case of this one, clearly, the easiest way would be doing a substitution. But how do I know that would be the most efficient way?

Solving a separable differential equation: $x{y}^{\prime }=(1-<!--\; -\; -->4x^2)\tan <!--\; \; -->(y)$

Seems straight-forward but I've been unable to get it right. Here are my steps:
$$y^{\prime }(x)=\sqrt{-<!--\; -\; -->2y(x)+28},y(-<!--\; -\; -->4)=-<!--\; -\; -->4$$
$$\int <!--\; \int \; -->\frac{1}{\sqrt{28-<!--\; -\; -->2y}}\text{d}y=\int <!--\; \int \; -->\text{d}x$$
$$-<!--\; -\; -->\sqrt{28-<!--\; -\; -->2y}=x+c$$
$$(28-<!--\; -\; -->2y)=(x+c{)}^2$$
$$y=-<!--\; -\; -->1/2x^2-<!--\; -\; -->cx-<!--\; -\; -->c^2/2+14$$
$$c=-<!--\; -\; -->2,10$$
$$\Rightarrow <!--\; \Rightarrow \; -->y=-<!--\; -\; -->1/2x^2-<!--\; -\; -->10x-<!--\; -\; -->36$$
I've checked it over many countless times but for the life of me I can't figure out why it won't work. I've tried plugging the result back into the original equation and it seems to me like it checks out if you take the negative of the square root..

How can i solve this separable differential equation with trigonometric function?
The given Problem is separable differential equation:
$\cos <!--\; \; -->ydx+(1+e^{-<!--\; -\; -->x})\sin <!--\; \; -->ydy=0$
$y(0)=\frac{\mathrm{\pi <!--\; \pi \; -->}}{4}$

When linear differential equation can also directly be solved by variable separable method?
Consider the linear differential equation $\frac{dy}{dx}+p(x)y=q(x).$ 1. What conditions may be imposed on functions $p(x)$ and/or $q(x)$ so that it could directly be solved by variable separable method rather than integrating factor $e^{\int <!--\; \int \; -->p(x)dx}$. 2. In the case if it can be solved by both variable separable and integrating factor method, which would be more suitable?

How to construct an exact but non separable differential equation?
So I can prove that all separable differential equations are exact, and I can intuitively figure out that not all exact differential equations are necessarily separable, but I'm having a hard time constructing a counterexample?
Is a differential equation still considered inseparable if you cannot initially separate it, but do so using an integrating factor?
Sorry for asking rudimentary questions, I'm just starting to learn diff eq's.
Thank you in advance.

Find the general solution to the differential eq: $(x^2+y^2)+xy{y}^{\prime }=0$