Learn First Order Problems with These Differential Equations Examples

I have difficulties to solve these two differential equations:
1) $y^{\prime }(x)=\frac{x-<!--\; -\; -->y(x)}{x+y(x)}$ with the initial condition $y(1)=1$ .I'm arrived to prove that
$y=x(\sqrt{2-<!--\; -\; -->e^{-<!--\; -\; -->2(\ln <!--\; \; -->x+c)}}-<!--\; -\; -->1)$
but I don't know if it's correct. If it's right, how do I find the constant $c$? Because WolframAlpha says that the solution is $y(x)=\sqrt{2}\sqrt{x^2+1}-<!--\; -\; -->x$.
2) $y^{\prime }(x)=\frac{2y(x)-<!--\; -\; -->x}{2x-<!--\; -\; -->y(x)}$. I'm arrived to prove that $\frac{z-<!--\; -\; -->1}{(z+1{)}^3}=e^{2c}{x}^2$ but I don't know if it's correct. If it's right, how do I explain $z$ to substitute it in $y=xz$? Then, how do I find the constant $c$ ?

How do we solve the following differential equation?
$(y{}^{\prime }{)}^2+p(x)(1+y^2{)}^3=0$

I want to solve the following first order linear system of differential equations:
$\begin{array}{rl}& \frac{d{y}_0}{dx}+x{y}_1=\mathrm{\lambda <!--\; \lambda \; -->}{y}_2\\ & \frac{d{y}_1}{dx}+x{y}_2=\mathrm{\lambda <!--\; \lambda \; -->}{y}_0\\ & \frac{d{y}_2}{dx}+\frac{\mathrm{\alpha <!--\; \alpha \; -->}}{x}{y}_2+x{y}_0=\mathrm{\lambda <!--\; \lambda \; -->}{y}_1\end{array}$
with intial conditions: $y_0(0)=1,y_1(0)=0,y_2(0)=0.$

We have a theorem that says: Let $h:I\to <!--\; \backslash rightarrow\; -->\mathbb{R}$ and $g:J\to <!--\; \backslash rightarrow\; -->\mathbb{R}$ be continuous functions, $t_0\in <!--\; \in \; -->I$ and $y_0\in <!--\; \in \; -->\text{int(J)}$, hen the differential equation y'(t)=g(y(t))h(t) has a unique solution in some surrounding of $t_0$ if $g(y_0)\ne <!--\; \ne \; -->0$
How I determine the magnitude of this surrounding? Especially, if I integrate the differential equation and solve it for y(t), am I only able to say: My solution is only unique in the surround of $y_0$ for which g(y(t)) is not zero, is this the condition that determines my surrounding? I mean, I could have determined a solution that gives me zero for some values of t, but is the only solution for my given problem? Am I correct, that in this case, my theorem is not able to say something about the uniqueness of this solution?

I solved the differential equation
$\frac{dy}{dx}=\frac{x}{x^2+1}$
to get the general solution
$y=\frac{ln|x+1|+c}{2}$
Im given the initial condition
$y{y}^{\prime }-2e^x=0,y(0)=3$
but I dont know what to do with it

I have a first order linear differential equation (a variation on a draining mixing tank problem) with many constants, and want to separate variables to solve it.
$\frac{dy}{dt}=k_1+k_2\frac{y}{k_3+k_{4}t}$
y is the amount of mass in the tank at time t, and for simplicity, I've reduced various terms to constants, $k_1$ through $k_4$.
Separation of variables is made difficult by $k_1$, and I've considered an integrating factor, but think I might be missing something simple.

The function $y(x)$ satisfies the linear equation $y^{″}+p(x)y^{\prime }+q(x)=0$.
The Wronskian $W(x)$ of two independant solutions, $y_1(x)$ and $y_2(x)$ is defined as $\left|\begin{array}{cc}{y}_1& y_2\\ y_1^{\prime }& y_2^{\prime }\end{array}\right|$. Let $y_1(x)$ be given, use the Wronskian to determine a first-order inhomogeneous differential equation for $y_2(x)$. Hence show that $y_2(x)=y_1(x){\int <!--\; \int \; -->}_{x_0}^x\frac{W(t)}{y_1(t{)}^2}dt$$(\ast <!--\; \ast \; -->)$
The only first-order inhomogeneous differential equation I can get for $y_2(x)$ is simply $y_1(x)y_2^{\prime }(x)-<!--\; -\; -->y_1^{\prime }(x)y_2(x)=W(x)$. Which is trivial and simply the definition of $W(x)$, so I don't think this is what it wants. I can easily get to the second result by differentiating $\frac{y_2(x)}{y_1(x)}$ and then integrating, but the wording suggests I'm supposed to get to this result from the differential equation that I can't determine.
So I would ask please that someone simply complete the above question, showing the differential equation we are to determine and how to get to the second result from this.
Edit: The question later goes on to give the differential equation $x{y}^{″}-<!--\; -\; -->(1-<!--\; -\; -->x^2)y^{\prime }-<!--\; -\; -->(1+x)y=0$ call this equation $(†)$, we have already confirmed that $y_1(x)=1-<!--\; -\; -->x$ is a solution
Hence, using $(\ast <!--\; \ast \; -->)$ with $x_0=0$ and expanding the integrand in powers of $t$ to order $t^3$, find the first three non-zero terms in the power series expansion for a solution, $y_2$, of $(†)$ that is independent of $y_1$ and satisfies $y_2(0)=0,y\prime {\prime }_2(0)=1$.
I'm stuck here too as I don't see how we could expand $W(t)$ as it is surely a function of the unknown $y_2$. Apologies for my ignorance, this is independant study, and aside from a brief look at some online notes that simply define $W(x)$, I've never worked on these problems before.

As part of my research I get the following differential equation. I need to solve for $\mathcal{V}(\mathrm{\gamma <!--\; \gamma \; -->})$. In fact the requirement is not to solve but to show that $\mathcal{V}(\mathrm{\gamma <!--\; \gamma \; -->})$ is monotonic in $a_j$$\mathrm{\forall <!--\; \forall \; -->}j$, (which I hope it is) where $a_j$ are positive valued constants which do not depend on $\mathrm{\gamma <!--\; \gamma \; -->}$. If it can be shown without solving the differential equation that is sufficient. Please provide some suggestions.
$\frac{\mathrm{\gamma <!--\; \gamma \; -->}}{\log <!--\; \; -->\left(e\right)}\frac{d}{d\mathrm{\gamma <!--\; \gamma \; -->}}\mathcal{V}\left(\mathrm{\gamma <!--\; \gamma \; -->}\right)=1-<!--\; -\; -->\mathrm{\eta <!--\; \eta \; -->}\left(\mathrm{\gamma <!--\; \gamma \; -->}\right)$
$\mathrm{\eta <!--\; \eta \; -->}\left(\mathrm{\gamma <!--\; \gamma \; -->}\right)=\frac{1}{1+\mathrm{\gamma <!--\; \gamma \; -->}\underset{j}{\sum <!--\; \sum \; -->}\frac{a_j}{1-<!--\; -\; -->a_j(-<!--\; -\; -->\mathrm{\gamma <!--\; \gamma \; -->}\mathrm{\eta <!--\; \eta \; -->}\left(\mathrm{\gamma <!--\; \gamma \; -->}\right))}}$
where $j=\{1,\dots <!--\; \dots \; -->,n\}$

Find differential equation $y^{\mathrm{\prime <!--\; \text{'}\; -->}}=f(t,y)$ satisfied by $y(t)=4e^{2t}+3$
Solution:
Compute derivative of y,
$y^{\mathrm{\prime <!--\; \text{'}\; -->}}=8e^{2t}$
Write right hand side above, in terms of the original function y, that is,
$y-<!--\; -\; -->3=4e^{2t}$
Get a differential equation satisfied by y, namely
$y^{\mathrm{\prime <!--\; \text{'}\; -->}}=2y-<!--\; -\; -->6$
So my issue with that last answer. How is this a solution? Does it mean that if you somehow take an integral of $2y-<!--\; -\; -->6$ you should end up with the original $y(t)=4e^{2t}+3$???
It seems that there two different derivatives of y(t)
one is:
$y^{\mathrm{\prime <!--\; \text{'}\; -->}}=8e^{2t}$
the other is:
$y^{\mathrm{\prime <!--\; \text{'}\; -->}}=2y-<!--\; -\; -->6$
and I don't get it, can someone explain?
Also a bit offtopic, but the way y'=f(t,y) is written kinda bugs me.
Shouldn't it be written like y'=f(t,y(t)) to show that the function f contains t as an independent variable and the function y(t) which contains variable t as an input to itself (dependent variable t) ??? That's kinda an essential information, so surprised it's omitted in the writings.

Solve the differential equation
$3ydx-<!--\; -\; -->x(3x^{n}y\sin <!--\; \; -->y+3)dy=0.$
I need to find the general solution for both $n=0$ and $n=3$. So, for $n=0$ is done because the differential equation becomes a separable equation. However, I can't find a solution for $n=3$ because it is not a Bernoulli equation nor an exact DE. If someone could help me, it would be great.

I'm writing down the differential equation that I was solving, but it should be relatable to any other first-order linear differential equation. The equation is:
$x\frac{dy}{dx}=x^2+3y$ where $x>0$.
At first I got the wrong integrating factor. I realized it could be wrong after verification of the solution found and after I had made sure that I did not make any algebraic errors. Then I changed the integrating factor and got the correct solution to the equation. However, I'm still not sure why the first integrating factor does not work, and why is the second method is the way to go to find the correct solution.
Here is what I did:
(1) Wrong Integrating Factor:
$v(x)=e^{\int <!--\; \int \; -->-<!--\; -\; -->3x^{-<!--\; -\; -->1}dx}=e^{-<!--\; -\; -->3\int <!--\; \int \; -->x^{-<!--\; -\; -->1}dx}=e^{-<!--\; -\; -->3\ln <!--\; \; -->x}=e^{-<!--\; -\; -->3}{e}^{\ln <!--\; \; -->x}=\frac{x}{e^3}$
With this integrating factor we find the solution to be $y=\frac{x^2}{3}+C{x}^{-<!--\; -\; -->1}$ which is wrong.
(2) Correct Integrating Factor:
$v(x)=e^{\int <!--\; \int \; -->-<!--\; -\; -->3x^{-<!--\; -\; -->1}dx}=e^{-<!--\; -\; -->3\int <!--\; \int \; -->x^{-<!--\; -\; -->1}dx}=e^{-<!--\; -\; -->3\ln <!--\; \; -->x}=e^{\ln <!--\; \; -->x^{-<!--\; -\; -->3}}=x^{-<!--\; -\; -->3}$
With this integrating factor we find the correct solution $y=-<!--\; -\; -->x^2+C{x}^3$

The question asks to find y given y(x) is a differentiable function satisfying:
$\frac{dy}{dx}$=-$2x{y}^4$ , y(0)=$\frac{1}{3}$ and $y(x)>0$ , and explain why it is unique.
I assume the steps are to integrate $\frac{dy}{dx}$ to get a function for y(x), then use $y(0)=1/3$ to find the constant and then set $y(x)>0$ to find y but i keep getting stuck.
A differential equation solution gives y(x)=$\frac{1}{\sqrt[3]{c_1+3x^2}}$ , which I can use $y(0)$ to find $c_1$ but then I'm not sure how to get y from that since the equation doesn't have any y's in it.
A similar question I found on this site showed to rearrange and integrate each side like $\int <!--\; \int \; -->\frac{dy}{y^4}=\int <!--\; \int \; -->-2xdx$ which gives $-<!--\; -\; -->\frac{1}{3y^3}+c_1$=$-<!--\; -\; -->x^2+c_2$ , but then I don't know how to get $y(x)$ from that frunction, it seems further from the solution than my first try.

I am aware that the solution to a homogeneous first order differential equation of the form $\frac{dy}{dx}=p(x)y$ can be obtained by simply by rearranging to:
$\frac{dy}{y}=p(x)dx$
Then it is simply a question of integrating both sides and the answer is straightforward. Now what would happen if RHS had a constant, how a can find a particular solution to this case: $\frac{dy}{dx}=p(x)y+C$
I know that the general solution would be the sum of the homogeneous equation and the particular solution

Transform the differential equation
$\{\begin{array}{ll}{u}^{‴}(t)=\sin <!--\; \; -->(u^{″}(t))-<!--\; -\; -->u^2(t),& t>0,\\ u^{(i)}=u_i\text{for }i\in <!--\; \in \; -->\{0,1,2\}\end{array}$
into an equivalent system of first order ordinary differential equations.
I know how to carry out this procedure for linear systems but I am stuck because of the $\sin <!--\; \; -->(u^{″}(t))$ term.
For far I tried setting $x:\equiv <!--\; \equiv \; -->u^{″}$ and $y:\equiv <!--\; \equiv \; -->u^{\prime }$ and then substituting into the equation to obtain multiple differential equations in x and y but don't know how to simplify them to obtain the desired result:
$\{\begin{array}{l}{y}^{\prime }(t)=x(t),\\ x^{\prime }(t)=\sin <!--\; \; -->(x(t))-<!--\; -\; -->u^2(t),\\ y^{″}(t)=\sin <!--\; \; -->(y^{\prime }(t))-<!--\; -\; -->u^2(t).\end{array}$
For context: This is the first part of a two part question where in the second part one is asked to find the iterative rule for Eulers method.

Problem:
Solve the following differential equation:
$\begin{array}{rcl}6x^{2}ydx-<!--\; -\; -->(x^3+1)dy& =& 0\end{array}$
Answer:
This is a separable differential equation.
$\begin{array}{rcl}\frac{6x^2}{x^3+1}dx-<!--\; -\; -->\frac{dy}{y}& =& 0\\ \int <!--\; \int \; -->\frac{6x^2}{x^3+1}dx-<!--\; -\; -->\int <!--\; \int \; -->\frac{dy}{y}& =& c_1\\ 2\ln <!--\; \; -->|x^3+1|-<!--\; -\; -->\ln <!--\; \; -->|y|& =& c_1\\ \ln <!--\; \; -->(x^3+1{)}^2-<!--\; -\; -->\ln <!--\; \; -->|y|& =& c_1\\ \ln <!--\; \; -->(\frac{(x^3+1{)}^2}{|y|})& =& c_1\\ (x^3+1{)}^2& =& c|y|\end{array}$
However, the book gets:
$\begin{array}{rcl}(x^3+1{)}^2& =& |cy|\end{array}$
Is my answer different from the book's answer? I believe it is.

I'm taking an elementary differential equations class and we got our first exam back and I don't understand why I was wrong on one of the questions. I got a 96, and the professor said there were quite a few d's and f's, so I didn't want to quibble about this.
Anyway, the problem was ${\displaystyle \frac{dy}{dx}}={\displaystyle \frac{1}{x+y+2}}$ and I chose the sub $u=x+y+2$, with ${\displaystyle \frac{dy}{dx}}={\displaystyle \frac{dy}{du}}{\displaystyle \frac{du}{dx}}={\displaystyle \frac{dy}{du}}(1+{\displaystyle \frac{1}{u}})$ and then plugged in the expression for ${\displaystyle \frac{dy}{dx}}$, separated the equation and solved for y in terms of u and then substituted back to get $y=\ln <!--\; \; -->(x+y+3)+c.$ The professor took issue with my expression for ${\displaystyle \frac{dy}{dx}}$, saying that I couldn't differentiate y w.r.t. u if y is part of u. Is this correct?

My question is
If u(x,y) and v(x,y) are two integrating factors of a diff eqn M(x,y)dx+N(x,y)dy, u/v is not a constant. then u(x,y)=cv(x,y)is a solution to the differential eqn for every constant c. I m totally stuck :(
Another doubt i have is how to derive the singular solution for the Clairaut's equation. i tried it we have y=px+f(p) diff wrt x and considering dp/dx=0 we get p=c, how to solve the other part?

I have the following exepression in my book:
$\frac{dx}{dt}+a_1(t)x=g(t),x(t_0)=x_0$
Then it says, multiply both sides of the differential equation by the integrating factor $I(t)$.
$I(t)\frac{dx(t)}{dt}+a_1(t)I(t)x(t)=I(t)g(t)$
So far so good. Hereafter it says, the left-hand side is an exact derivative.
$\frac{d[x(t)I(t)]}{dt}=I(t)g(t)$
And my question is, how does the book come to the last? Can anyone give a HINT.

If a particle of mass $m$ moves in the $x$-$y$ plane, then its equations of motion are
$m\frac{d^{2}x}{d{t}^2}=f(t,x,y)\text{and}m\frac{d^{2}y}{d{t}^2}=g(t,x,y).$
Here $f$ and $g$ represent the $x$ and $y$ components, respectively, of the force acting on the particle. Replace this system of two second-order equations by an equivalent system of four first order equations of the form:
$y_1^{\prime }=f_1(x,y_1,...,y_n)$
$y_2^{\prime }=f_2(x,y_1,...,y_n)$
$y_n^{\prime }=f_n(x,y_1,...,y_n)$

I understand how replace a differential equation by an equivalent system of first order equations when the differential looks something like
$x{y}^{″}-<!--\; -\; -->x^2{y}^{\prime }-<!--\; -\; -->x^{3}y=0$
An equivalent system is:
$y_0^{\prime }=y_1$
$y_1^{\prime }=x^2{y}_0+x{y}_1$
Therefore, I need someone to point me in the right direction for my question stated at the top.

How to solve the differential equation $(dy/dx{)}^2=(x-<!--\; -\; -->y{)}^2$ with initial condition $y(0)=0$?
I solved the equation by partitioning it into two differential equations.
1) $dy/dx=x-<!--\; -\; -->y$ The solution is $\to <!--\; \to \; -->$ $1-<!--\; -\; -->x+y=-<!--\; -\; -->\exp <!--\; \; -->(-<!--\; -\; -->x)$
and
2) $1+x-<!--\; -\; -->y=\exp <!--\; \; -->(x)$
How do we write combined solution of such equations.