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

Integrating a non-separable differential equation
I need to integrate
$$dx=[k-<!--\; -\; -->\mathrm{\lambda <!--\; \lambda \; -->}x]dt/\mathrm{\tau <!--\; \tau \; -->}$$
to solve for x(t), where k, $\mathrm{\lambda <!--\; \lambda \; -->}$, and $\mathrm{\tau <!--\; \tau \; -->}$ are constants. However, the equation seems to be non-separable, so I'm not sure how to go about integrating it.

Is the differential equation $y^{\prime }=x+y$ separable?
I'm at a loss. My guess is no, but I'm new to doing these problems. It can not be solved with cross multiplication but there are other ways of solving these problems I'm sure. Thank you for any help.

Solving differential equation $(2x-<!--\; -\; -->y)dx+(4x+y-<!--\; -\; -->6)dy=0$
To solve this, I let y be the dependent variable so that the equation becomes $2x-<!--\; -\; -->y+(4x+y-<!--\; -\; -->6)y^{\prime }=0$ and isolating $y^{\prime }$ I got
$$y^{\prime }=-<!--\; -\; -->{\displaystyle \frac{2x-<!--\; -\; -->y}{4x+y-<!--\; -\; -->6}}$$
Here, I find difficulty on how to rewrite the equation into a first order separable ODE in the form
$$N(y)\cdot <!--\; \cdot \; -->y^{\prime }=M(x)$$
as to what value of y containing v should I substitute in order to rewrite the equation into separable form. I have used a software on how to solve this, it suggests on substituting
$$y={\displaystyle \frac{2x-<!--\; -\; -->4xv+6v}{v+1}}$$
However, I find it difficult to trace back how did the software find the value of y that will be used in the substitution.

Solve the differential equation $y=2\sqrt{x}{y}^2{y}^{\prime }+4x{y}^{\prime }$
The main problem for me is to understand what type of DE it is, since it is neither of these:
Separable
Homogeneous
Linear
Exact
Bernoulli
Riccati
Implicit
Lagrange
Perhaps I am missing something. But how should I approach this problem?

Integrating the separable, first-order ordinary differential equation $m\frac{dv}{dt}=mg-<!--\; -\; -->av$
I can't solve the very first problem from Slater & Frank's book, and have no one to help me (I'm self-studying it in these vacations):
1. A particle moves in a vertical line under the action of gravity and a viscous force (−av) where v is its velocity. Show that the velocity at any time is given by
$$v=(v_0+\frac{mg}{a})e^{-<!--\; -\; -->\frac{a}{m}t}-<!--\; -\; -->\frac{mg}{a}.$$
Show that this solution reduces [...].
What I did is:
The resultant force is $mg-<!--\; -\; -->av$, and so by Newton's Second Law,
$$m\frac{dv}{dt}=mg-<!--\; -\; -->av.$$
Since m, a, and g are constants in this case, this is a first-order ODE, which also happens to be a separable one.
We rewrite it as a relation between differentials:
$$m\frac{dv}{mg-<!--\; -\; -->av}=dt.$$
But I have no idea how to integrate the left side relative to v. Despite the fact that the exercise already gave the solution, v(t).
Am I doing it right? Or maybe there is a simpler way using methods specials from mathematical mechanics?

What are the steps to solve this calculus problem?
$$3)x{y}^{\prime }\cos <!--\; \; -->y+\sin <!--\; \; -->y=0$$
I understand that y' is the first derivative, but I'm not really sure how to apply it. Would I move everything to the other side, convert y' to dy/dx and use trig applications to sin/cos? I only took business calculus and have no idea how to solve this.
It would be extremely nice if somebody showed all the steps to solve this.

Separable but not exact equation
In class, my professor stated that all separable equations are exact, and we even proved it for homework, but I think I found an equation that is separable but not exact:
$$(x\ln <!--\; \; -->y+xy)+(y\ln <!--\; \; -->x+xy)y\prime =0$$
My Work:
$$\begin{array}{rl}M& =x\ln <!--\; \; -->y+xy\\ M_y& =\frac{x}{y}+x\\ N& =y\ln <!--\; \; -->x+xy\\ N_x& =\frac{y}{x}+y\\ M_y& \ne <!--\; \ne \; -->N_x\end{array}$$
But
$$\begin{array}{rl}x(\ln <!--\; \; -->y+y)+y\times <!--\; \times \; -->y^{\prime }\times <!--\; \times \; -->(\ln <!--\; \; -->x+x)& =0\\ x(\ln <!--\; \; -->y+y)& =-<!--\; -\; -->y\times <!--\; \times \; -->y^{\prime }\times <!--\; \times \; -->(\ln <!--\; \; -->x+x)\\ \frac{x}{\ln <!--\; \; -->x+x}& =\frac{-<!--\; -\; -->y\times <!--\; \times \; -->y^{\prime }}{\ln <!--\; \; -->y+y}\end{array}$$
Separated
So whats up? Am I doing something wrong?

How to prove an ordinary differential equation is not separable?
Some elementary exercises require one to determine whether or not an ordinary differential equation is separable. For example, it is understood that the equation
$$y^{\prime }=\frac{1-<!--\; -\; -->2xy}{x^2}$$
is not separable. An easier example is $y^{\prime }=x+y$. Usually this is intuitive understandable; however, can one give a strict proof that the right hand of these equations cannot be written as a product of the form $p(y)q(x)$?

Differential equation maximal interval and solution
Consider the differential equation $y^{\prime }=1-<!--\; -\; -->y^2$
First, is $y(x)=1$ the only constant solution?
I now want to solve the equation for the initial value problem $y(0)=y_0$, with $y_0>1$
Also, what's the maximal interval the solution function can be defined on? How does it behave at the edges (potentially for $x\to <!--\; \to \; -->\pm \mathrm{\infty <!--\; \infty \; -->}$?
Thanks in advance. Differential equations are new to me and I have trouble on how to visualize and solve an equation like this.

Solve the differential equation: $y^{\prime }=2{\left(\frac{y+2}{x+y-<!--\; -\; -->1}\right)}^2$
I think a substitution is required here to make it linear or separable but my attempt of y=ux does not work.

General solution of $(x^2-<!--\; -\; -->y^2)dx+(3xy)dy=0$
Find the general solution to the homogeneous differential equation
$$(x^2-<!--\; -\; -->y^2)dx+(3xy)dy=0$$
The differential equation does not seem to be separable, and I'm having a tough time to put it in the general form of
$$\frac{dy}{dx}+p(x)y=f(x)$$
Would someone mind offering me some assistance?
Thank you

Representing differential equations explicitly and implicitly
For the past two weeks I have been dealing with differential equations and I have stumbled upon a seemingly simple problem. Suppose I have a separable differential equation $\frac{dy}{dx}=y(x)$. This simply tells that the function value has to equal to the slope at a particular $x$ value. The solution is therefore simply $e^x$. So far so good. Now suppose I have similiar equation, namely $\frac{dy}{dx}=x^2$. Now this tells me that the slope of the function at a particular $x$ has to equal $x^2$. Hence, the solution is simply the integral, namely $\frac{1}{3}{x}^3$. One does not even need the conventional methods to solve separable differential equations.
However, let me now define $y(x)=x^2$. The two differential equations (and hence the solutions) should be now equivalent. As we have seen, that's not the case. I can sense that I am commiting a mathematical fallacy, however, I do not know what seems to be the problem when I solve the differential equation implicitly rather than explicitly as stated above. Could anyone help me please?

What are the orthogonal trajectories of the given family of curves: $(x-<!--\; -\; -->a{)}^2+y^2=a^2$
Take the derivative:
$$2(x-<!--\; -\; -->a)+2y{y}^{\prime }=0$$
$$x-<!--\; -\; -->a+y{y}^{\prime }=0$$
a can be isolated from the original equation
$x^2-<!--\; -\; -->2ax+a^2+y^2=a^2\to <!--\; \to \; -->x^2-<!--\; -\; -->2ax+y^2=0$
Yielding the following differential equation:
$$x-<!--\; -\; -->\frac{x^2+y^2}{2x}+y{y}^{\prime }=0$$
This is suppose to be a separable differential equation yielding $x^2+(y-<!--\; -\; -->C{)}^2=C^2$. I can't seem to come this conclusion. Could someone help?

A differential equation with homogeneous coefficients: $(x+y)dx-<!--\; -\; -->(x-<!--\; -\; -->y)dy=0$
Functions $P(x)=x+y$ and $Q(x)=-<!--\; -\; -->x-<!--\; -\; -->y$ are both homoegeneous functions of order 1. So I can use substitusion:
$$y=ux,dy=udx+xdu$$
(because it will lead to a differential equation in which the variables are separable)
So after this substitution I'm obtaining result:
$$(u+ux)dx-<!--\; -\; -->(x-<!--\; -\; -->ux)(udx+xdu)=0$$
After simplification I have:
$$x^2(u-<!--\; -\; -->1)du+x(1+u^2)dx=0$$
Now I'm dividing both sides by $x^2\cdot <!--\; \cdot \; -->(1+u^2)$ and I'm obtaining:
$$\frac{u-<!--\; -\; -->1}{1+u^2}du+\frac{dx}{x}=0$$
Next, because:
$$\int <!--\; \int \; -->\frac{u-<!--\; -\; -->1}{1+u^2}du=\frac{1}{2}\ln <!--\; \; -->|u^2+1|-<!--\; -\; -->\arctan <!--\; \; -->(u)+C$$
and:
$$\int <!--\; \int \; -->\frac{dx}{x}=\ln <!--\; \; -->|x|+C$$
I have a solution:
$$\frac{1}{2}\ln <!--\; \; -->|u^2+1|-<!--\; -\; -->\arctan <!--\; \; -->(u)+\ln <!--\; \; -->|x|=C$$
$$\frac{1}{2}\ln <!--\; \; -->|\frac{y^2}{x^2}+1|-<!--\; -\; -->\arctan <!--\; \; -->(\frac{y}{x})+\ln <!--\; \; -->|x|=C$$
After simplification, I can obtain result:
$$\frac{1}{2}\ln <!--\; \; -->(x^2+y^2)-<!--\; -\; -->\arctan <!--\; \; -->(\frac{y}{x})=C$$
In book from this exercise from there is a little diffrent answer :
$$\arctan <!--\; \; -->(\frac{y}{x})-<!--\; -\; -->\frac{1}{2}\ln <!--\; \; -->(x^2+y^2)=C$$
Is my solution incorrect? I will be grateful for an explanation

Differential equation with separable, probably wrong answer in book
I have a differential equation:
$$\frac{dy}{dx}=y\log <!--\; \; -->(y)\cot <!--\; \; -->(x)$$
I'm trying solve that equation by separating variables and dividing by $y\log <!--\; \; -->(y)$
$$dy=y\log <!--\; \; -->(y)\cot <!--\; \; -->(x)dx$$
$$\frac{dy}{y\log <!--\; \; -->(y)}=\cot <!--\; \; -->(x)dx$$
$$\cot <!--\; \; -->(x)-<!--\; -\; -->\frac{dy}{y\log <!--\; \; -->(y)}=0$$
Where of course y>0 regarding to division
Beacuse:
$$\int <!--\; \int \; -->\frac{dy}{y\log <!--\; \; -->(y)}=\ln <!--\; \; -->|\ln <!--\; \; -->(y)|+C$$
and:
$$\int <!--\; \int \; -->\cot <!--\; \; -->(x)dx=\ln <!--\; \; -->|\sin <!--\; \; -->(x)|+C$$
So:
$$\ln <!--\; \; -->|\sin <!--\; \; -->(x)|-<!--\; -\; -->\ln <!--\; \; -->|\ln <!--\; \; -->(y)|=C$$
$$\ln <!--\; \; -->|<!--\; |\; -->\frac{\sin <!--\; \; -->(x)}{\ln <!--\; \; -->(y)}|<!--\; |\; -->=C$$
$$\frac{\sin <!--\; \; -->(x)}{\ln <!--\; \; -->(y)}=\pm <!--\; \pm \; -->e^C$$
$d=\pm <!--\; \pm \; -->e^C$
$$\sin <!--\; \; -->(x)=d\ln <!--\; \; -->(y)$$
$$\frac{\sin <!--\; \; -->(x)}{d}=\ln <!--\; \; -->(y)$$
$$e^{\frac{\sin <!--\; \; -->(x)}{d}}=y$$
This is my final answer. I have problem because in book from equation comes the answer to exercise is:
$$y=e^{c\sin <!--\; \; -->(x)}$$
Which one is correct?
I will be grateful for explaining Best regards

Find y(2) given y(x) given a separable differential equation
Find what y(2) equals if y is a function of x which satisfies:
$x{y}^5\cdot <!--\; \cdot \; -->y^{\prime }=1$
given $y=6$ when $x=1$
I got $y(2)=\sqrt{6\ln <!--\; \; -->(2)-<!--\; -\; -->46656}$
but this answer is wrong can anyone help me figure out the right answer and how I went wrong?

A differential maximization problem
OK, I know how to solve maximization problems on numbers, and I know how to solve differential equations which are equations on functions, but how do I solve a 'maximization problem' over functions?
Here is a specific problem:
Find a positive real function $F(x)$, continuous and monotonically increasing on the real interval $[0,1]$, which maximizes:
$$\frac{F(x)}{F(1)}$$
Subject to:
$$F^{\prime }(x)=(1-<!--\; -\; -->x)F^{″}(x)$$
what is the function $F$ which attains this maximum?

$u_{xx}+y^{2}u=\sin <!--\; \; -->2x$
I want to solve the non homogeneous differential equation
$$\frac{{\mathrm{\partial <!--\; \partial \; -->}}^{2}u}{\mathrm{\partial <!--\; \partial \; -->}{x}^2}+y^{2}u=\sin <!--\; \; -->2x.$$
I have tried to solve it by method of separable of variables. But unfortunately, not able to find out the solution. Please give me some hints to solve it.

Differential equation Euler substitution
I was trying to solve this differential equation but can't figure out the final integral I get by variable separable method
The equation is
$$x^3{y}^{\prime }=y^3+y^2\sqrt{y^2-<!--\; -\; -->x^2}$$
I got the integral
$$\frac{dv}{v^3+v^2\sqrt{(v^2-<!--\; -\; -->1)}-<!--\; -\; -->v}$$
but can't figure out how to solve it.
The Euler substitution $v=\frac{u^2+1}{2u}$ could work but I can't seem to proceed with it.
Any help with this problem is much appreciated.

Solve the following differential equation: $\frac{(ydx+xdy)}{(1-<!--\; -\; -->x^2{y}^2)}+xdx=0$
Is there any way I can get this into the form of a separable, Bernoulli, exact, or any other form of differential equation that is easy to solve?
Solve the following differential equation:
$$\frac{(ydx+xdy)}{(1-<!--\; -\; -->x^2{y}^2)}+xdx=0$$
can someone show me the procedure to simplify or convert this differential equation into a form that is easy to solve?