Learn First Order Problems with These Differential Equations Examples

I've been working on this question but just can't crack it. It is as follows.
Using $\frac{dy}{dx}=\frac{a\sqrt{x^2+y^2}+by}{bx}$ where y(1)=0
and by letting $v=\frac{y}{x}$ transform the above differential equation to an equation in v and x. Find the implicit solution of the resulting separable equation. Then use the relationship y=xv to obtain $y+\sqrt{x^2+y^2}=x^{\frac{a}{b}+1}$
I've tried separate the upper terms into two fractions which makes the second term simply $\frac{by}{bx}=v$ but I am still left with the nasty square root with the y on the inside and I can't shift it easily to solve as a first order DE. Also I was thinking about solving it as a linear first order ODE but we have a squared term. Sorry about formatting as well, first time posting a question.

Solve the differential equation by the appropriate substitution
$x^2\frac{dy}{dx}+2xy=x^4{y}^2+1$

I have a linear first order ordinary differential equation
$\frac{dy}{dx}+\tan <!--\; \; -->(x)y=2{\cos }^2<!--\; \; -->x\sin <!--\; \; -->x-<!--\; -\; -->\sec <!--\; \; -->x$
with an initial condition as $y(\frac{\mathrm{\pi <!--\; \pi \; -->}}{4})=3\sqrt{2}$
My integrating factor $\mathrm{\mu <!--\; \mu \; -->}(x)=\sec <!--\; \; -->x$
After multiplication with the integration factor what I get is:
$(secxy{)}^{\prime }=sin2x-<!--\; -\; -->se{c}^{2}x$
or
$(secxy{)}^{\prime }=2sinxcosx-<!--\; -\; -->se{c}^{2}x$
If I use the first equation I get:
$y(x)=\frac{\frac{1}{2}cos2x-<!--\; -\; -->tanx+c}{secx}$
and using the second equation I get:
$y(x)=\frac{si{n}^{2}x-<!--\; -\; -->tanx+c}{secx}$
$\int <!--\; \int \; -->2sinxcosxdx=2\frac{si{n}^{2}x}{2}=si{n}^{2}x$
with the first equation I get c=7 and second equation I get $c=\frac{13}{2}$.
It is a very simple differential equation but when I solve it I get two different answers. Is this ok?

Construct an example of a first-order differential equation on $\mathbb{R}$ for which there are no solutions to any initial value problem.
Could anyone please get me started on this. I am struck as to which direction to go

Below is a problem I did. However, it did not match the back of the book. I would like to know where I went wrong.
Problem:
Solve the following differential equation.
$y^{\prime }=\frac{y-<!--\; -\; -->x}{x}$
Answer:
$\begin{array}{rl}\frac{dy}{dx}& ={\displaystyle \frac{y}{x}}-<!--\; -\; -->{\displaystyle \frac{x}{x}}={\displaystyle \frac{y}{x}}-<!--\; -\; -->1\\ y& =xv\\ {\displaystyle \frac{dy}{dx}}& =x{\displaystyle \frac{dv}{dx}}+v\\ x{\displaystyle \frac{dv}{dx}}+v& =v-<!--\; -\; -->1\\ x{\displaystyle \frac{dv}{dx}}& =-<!--\; -\; -->1\\ dv& =-<!--\; -\; -->{\displaystyle \frac{dx}{x}}\\ v& =-<!--\; -\; -->\ln <!--\; \; -->x+c\\ {\displaystyle \frac{y}{x}}& =-<!--\; -\; -->\ln <!--\; \; -->|x|+c\\ y& =-<!--\; -\; -->x\ln <!--\; \; -->|x|+cx\end{array}$
The book's answer is:
$y=x\ln <!--\; \; -->|\frac{k}{x}|$
Where did I go wrong?

I am trying to solve the differential equation
$y^{\prime }\left(x\right)-<!--\; -\; -->\frac{y\left(x\right)}{2x}=x\sin <!--\; \; -->\left(\frac{x}{y\left(x\right)}\right)$
I think it is separable variable differential equations. I tried to substitute:
$z=\frac{x}{y}$
and
$y^{\prime }=\frac{z-<!--\; -\; -->z^{\prime }x}{z^2}$
$\frac{z-<!--\; -\; -->z^{\prime }x}{z^2}-<!--\; -\; -->\frac{1}{2z}=x\sin <!--\; \; -->(z)$
multiply by $z^2$
$z-<!--\; -\; -->z^{\prime }x-<!--\; -\; -->\frac{z}{2}=z^{2}x\sin <!--\; \; -->(z)$
And now I have no idea how to manipulate this.

Consider $x{y}^{″}+2y^{\prime }+xy=0$. Its solutions are $\frac{\cos <!--\; \; -->x}{x},\frac{\sin <!--\; \; -->x}{x}$.
Neither of those solutions (as far as I could find) can be the solutions of a first order linear homogeneous differential equation. However
$\frac{e^{\pm <!--\; \pm \; -->ix}}{x}$
Can be the solution of a first order linear homogeneous DE ($y^{\prime }+(x\mp <!--\; \mp \; -->i)y=0$)
Can I always find a first order homogeneous linear DE whose solution also solves a second order homogeneous linear DE?
For example can I find a first order homogeneous linear DE whose solution is a particular linear combination of $J_1$ and $Y_1$?
(unrelated: also I'd like to know if there is a way of solving $x{y}^{″}+2y^{\prime }+xy=0$ without noticing that it is a spherical bessel function or using laplace transform.)

I have a first order differential equation question with initial value.
$\frac{dv}{dt}+\frac{v^2}{5}=9.8\text{ with }v(0)=0$
I tried to solve it but just didn't know how to remove the absolute sign in my solution.
Please let me know if whole question is needed to remove the absolute sign inside the solution.
Edit 1: Part where I got stuck:
After integration and since C = 0:
$\frac{5ln(lv+7l)}{14}-<!--\; -\; -->\frac{5ln(lv-<!--\; -\; -->7l)}{14}=t$
Edit 2: I think my solution is wrong

I am solving a differential equation ${\displaystyle \frac{dy}{dt}=\frac{y+1}{t+1}}$.
I got the solution $y=c(t+1)-<!--\; -\; -->1$, $c$ a constant.
But the handout by my professor says
"The solution is $y=c(t+1)-<!--\; -\; -->1$, $c$ a constant. where $t\ne <!--\; \ne \; -->-<!--\; -\; -->1$. But it does not mean that $y$ is not defined at $t\ne <!--\; \ne \; -->-<!--\; -\; -->1$. It means that y can take any value at $t=-<!--\; -\; -->1$ as long as it satisfies $y=c(t+1)-<!--\; -\; -->1$."
It does not make much sense to me because if $y$ satisfies the equation $y=c(t+1)-<!--\; -\; -->1$ at $t=-<!--\; -\; -->1$, then the solution will be just $y=c(t+1)-<!--\; -\; -->1$, $c$ a constant.
And I believe my solution indeed makes sense because $y=c(t+1)-<!--\; -\; -->1$, $c$ a constant is defined on $\mathbb{R}$ and differentiable, $y^{\prime }=c$ for any $t\in <!--\; \in \; -->\mathbb{R}$. On the other hand, plugging this in to ${\displaystyle \frac{y+1}{t+1}}$ gives me $c$ for any $t\in <!--\; \in \; -->\mathbb{R}$.
Would you explain what the handout is saying?

I need to find a series solution to the following simple differential equation
$x^2{y}^{\prime }=y$
Assuming the solution to be of the form $y=\sum <!--\; \sum \; -->a_n{x}^n$ and equating the coefficients on both the sides, all the coefficients turn out to be zero which is definitely wrong.
Any help is appreciated.

I have a differential equation of the form
$dy/dx+p(x)y=q(x)$
under the condition that $q(x)=300$ if $y<3312$ and $q(x)=0$ if $y\ge <!--\; \ge \; -->3312$.
I could not understand how to solve this differential equation with such heavy side function ?
Any hints?

I am trying to remember again the stuff I did about nonlinear differential equations.
I have
$\stackrel{\dot{}<!--\; \dot{}\; -->}{x}=\left(\begin{array}{c}{x}_1^2\\ -<!--\; -\; -->1\end{array}\right)$

For example, if I have the system,
$$\begin{gathered} y^{\prime }+y=3x \\ {y}^{\prime }-<!--\; -\; -->y=x \end{gathered}$$
I could then use elimination to minus the top equation from the bottom one to get,
$$\begin{gathered} 2y=2x \\ y=x \end{gathered}$$
Which is obviously wrong as then, $1+x=3x$ which is wrong.
So why are you not able to use elimination in solving a system of first order differential equations?

We know that every differential equation is equivalent to a first-order system. I am trying to prove or disprove the converse. For example in ${\mathbb{R}}^2$, if we have a system $\stackrel{\dot{}<!--\; \dot{}\; -->}{x}=f(x,y)$, $\stackrel{\dot{}<!--\; \dot{}\; -->}{y}=g(x,y)$. Can we always convert it to one differential equation (for example, only in terms of $x$)? Under what condition, this is possible?

Find a one parameter family of solutions of the following first order ordinary differential equation
$(3x^2+9xy+5y^2)dx-<!--\; -\; -->(6x^2+4xy)dy=0$
Hello. So I am stuck after I find out that they are not exact. Please help.

The equation is:
$e^x(1+x)dx=(x{e}^x-<!--\; -\; -->y{e}^y)dy$
I've tried solving this as a non-exact differential equation but it's definitely incorrect. Not sure if this can be classified as an Bernoulli/Linear Differential equation either.
Any help is appreciated!

I have the following differential equation problem but I couldn't proceed any further -
$\frac{dy}{dx}=\frac{\frac{-<!--\; -\; -->ay}{x}}{1-<!--\; -\; -->b{\left(\frac{1}{x}\right)}^n{\left(\frac{y}{1-<!--\; -\; -->y}\right)}^m}$
where, $x\in <!--\; \in \; -->[0,1]\text{and}y\in <!--\; \in \; -->[0,1]$
But I can't solve it down.
I have tried $y=u{y}_1\text{where},y_1=x^{\frac{n}{m}}$, but it didn't help.
Wolfram gives the solution as -
$y(x)=c_1\exp <!--\; \; -->(\int <!--\; \int \; -->\frac{a}{x-<!--\; -\; -->b\frac{x^{m-<!--\; -\; -->n+1}}{(-<!--\; -\; -->x+1{)}^m}}dx)$
How to simplify the integral?
I just wanted a hint that whether it can be solved? If yes, please just tell me what am I doing wrong.

I need to convert the following differential equation to standard form.
$T_n=2T_{n-<!--\; -\; -->1}+1$
(not quite sure how to really format it properly)
I was thinking it is
$T_n-<!--\; -\; -->2T_{n-<!--\; -\; -->1}-<!--\; -\; -->1$
If I'm incorrect (or correct) can I please have somebody just explain briefly about the decreasing order. I'm just confused about the whole number, in this case the number 1, I'm not quite sure if it's bigger or smaller than the others.

Given differential equations
$\stackrel{¨<!--\; ¨\; -->}{x}=G{m}_1\frac{y-<!--\; -\; -->x}{|y-<!--\; -\; -->x{|}^3}\stackrel{¨<!--\; ¨\; -->}{y}=G{m}_2\frac{x-<!--\; -\; -->y}{|y-<!--\; -\; -->x{|}^3}$
with constant G,m1,m2 I want to solve them with the Euler method. I know I have to reduce the equations to a system of first order. So I did for $\stackrel{¨<!--\; ¨\; -->}{x}$
$\frac{d}{dt}\left(\begin{array}{c}{v}_0\\ v_1\end{array}\right)=\left(\begin{array}{c}{v}_1\\ G{m}_1\frac{y-<!--\; -\; -->v_0}{|v_0-<!--\; -\; -->y{|}^3}\end{array}\right)$
and the same for $\stackrel{¨<!--\; ¨\; -->}{y}$. But now my problem is that the right side depends on the other equation and the other way round. How can you get a system of first order so you can apply Euler to it?

I am familiar with first-order differential equations and how to solve them, but only for the case when there is a single unknown function.
For the following case there are two unknown functions and I'm struggling to determine where to even begin. The two equations:
$\begin{array}{}\text{(1)}& f^{\prime }(t)=-<!--\; -\; -->Af(t)+Bg(t)\end{array}$
$\begin{array}{}\text{(2)}& g^{\prime }(t)=-<!--\; -\; -->Bg(t)+Af(t)\end{array}$
Where f and g are functions of the variable t, and A & B are constants. The initial conditions are: $f(0)=0$ and $g(0)=1$.
I already have knowledge of the answer for f(t), which is:
$\begin{array}{}\text{(3)}& f(t)=\frac{B}{A+B}(1-<!--\; -\; -->e^{-<!--\; -\; -->(A+B)t})\end{array}$
I would really like to understand how to tackle questions of this type: where there are two unknown functions. I would attempt to show my working out but I don't know where to start.
Therefore, I would like to ask if anyone could please provide a hint or suggestion for me, something which I can work off. If necessary, I will update my post with original working. Thank you for your time.