Solve System of Differential Equations with Expert Help

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.

Given non-commuting matrices $A$ and $B$ of order $n$, is there a closed-form solution to the differential equation
$\frac{dX}{dt}=AX+tBX$
with $X(0)=I$?
I know that for the reals, $x=a\exp <!--\; \; -->\int <!--\; \int \; -->f(t)$ is the general solution to $\stackrel{\dot{}<!--\; \dot{}\; -->}{x}=f(t)x$, but I'm also 99% certain this relies on the commutivity of the reals.
I'm more specifically looking to numerically compute $X(T)$ given the more general differential equation
$\frac{dX}{dt}=f(t)X,X(0)=I$
but in circumstances where $f^{\prime }(t)$ may be large and a 1st order piecewise approximation would be far more accurate than 0th order for any given $\mathrm{\Delta <!--\; \Delta \; -->}t$. Ultimately my concern is computing $X(T)$ as quickly as possible.
Are there better techniques for accomplishing this?

Consider
$\frac{dy}{dx}+p(x)y={\int <!--\; \int \; -->}_0^{\mathrm{\infty <!--\; \infty \; -->}}y(x)dx.$
I want to solve above differential equation. Can I consider right hand side as constant to solve this?
I know RHS is a constant but it also involves solution y, which might create trouble unless solution is known to us.
Also is it possible to find solution to given ordinary differential equation which is independent of y.

For a second order differential equation (many physical systems) in one variable, I know "procedures" to compute the energy. Given
$q^{″}(t)=f(q(t),q^{\prime }(t)),q(0)=q_0,q^{\prime }(0)=v_0,$
if we're lucky we can read off the related Lagrangian L, introduce $p=\frac{\mathrm{\partial <!--\; \partial \; -->}L}{\mathrm{\partial <!--\; \partial \; -->}q}$, do a Legendre transform and we got the Hamiltonian function H(q,p) for which $\frac{\text{d}}{\text{d}t}H(q(t),p(t))=0$ for solutions of the differential equation.
We can be more exact and give all the conditions for Noethers theorem to hold and the result is that along the flow $X(t)=⟨<!--\; ⟨\; -->q(t),p(t)⟩<!--\; ⟩\; -->$ in phase space given by a solution with initial conditions X(0), the function H(q,p) always takes the same values. It defines surfaces in phase space indexed by initial conditions X(0).
I wonder how to view this for a priori first order systems
$q^{\prime }(t)=f(q(t)),q(0)=q_0,$
where I think this must be even simpler. E.g. I thing some functional
$q(t)\mapsto <!--\; \mapsto \; -->q(t)-<!--\; -\; -->\text{an integral over}f(q(t))\text{something},$
should exists which will be constant for solutions of the equation, i.e. only depend on $q_0$. For each f, the functional dependence of this "energy" on "$q_0$" will be different.
However, I can't seem to find a general relation. What's the theory behind this, is there an energy'ish time integral of motion? What is the functional dependence on the intial condition, for a suitable constant of motion for for first order systems. And what would be the interpretation, given that we speak of a situation with only one initial condition?
If it is that case that the system is too restricted so that there is no meaningful geometrical interpretation, then let's think about a system of first order differential equations. This is like the one we generated from phase space, except that it doesn't really come from a second order situation and so the intial conditions aren't really e.g. intial position and velocity. I'm pretty sure there are situation where one considers such a directly generated flow (the equation might be more complicated than exponential flow $\stackrel{\dot{}<!--\; \dot{}\; -->}{x}\propto <!--\; \propto \; -->x$), but I don't recall any talk about the time-constant of motion in these systems, or how to interpret it.

I have to find the form of the function $h(t)$ in this equation:
$h^{\prime }(t)-<!--\; -\; -->\frac{\mathrm{\theta <!--\; \theta \; -->}h(t)}{2}+1=0$
with $h(T)=C$ where $T$ and $C$ are constants and $T\ge <!--\; \ge \; -->0$
At first I thought it was a Bernoulli differential equation and I tried:
$V(t)=\frac{1}{h(t)}\Rightarrow <!--\; \Rightarrow \; -->V^{\prime }(t)=\frac{h^{\prime }(t)}{h(t{)}^2}$
My equation becomes worse:
$V^{\prime }(t)-<!--\; -\; -->\frac{\mathrm{\theta <!--\; \theta \; -->}}{2}+\frac{1}{V(t{)}^2}=0$
Then I tried with the integrating factor:
$(e^{t}h(t){)}^{\prime }=2e^{t}h(t)-<!--\; -\; -->e^t$
$e^{t}h(t)=\int <!--\; \int \; -->2e^{t}h(t)-<!--\; -\; -->e^t$
I'm stuck here because $h(t)$ is not a variable, is a function and I have to find how it looks like.
Edit: I forgot: $\mathrm{\theta <!--\; \theta \; -->}$ is a constant.
How can I solve this problem?

Proceed estimating this quantity using Linear Approximation
${\displaystyle \frac{1}{\sqrt{95}}}-<!--\; -\; -->{\displaystyle \frac{1}{\sqrt{99}}}$

I'm working with the Schrodinger equation
$\frac{-<!--\; -\; -->{\mathrm{\hslash <!--\; \hslash \; -->}}^2}{2m}\frac{{\mathrm{\partial <!--\; \partial \; -->}}^2\mathrm{\psi <!--\; \psi \; -->}}{\mathrm{\partial <!--\; \partial \; -->}{x}^2}=E\mathrm{\psi <!--\; \psi \; -->}$
and trying to use the Euler Method to approximate a wavefunction. The SEQ can be rewritten as
${\mathrm{\psi <!--\; \psi \; -->}}^{″}=-<!--\; -\; -->\frac{2mE}{{\mathrm{\hslash <!--\; \hslash \; -->}}^2}\mathrm{\psi <!--\; \psi \; -->}$
My question is 2-fold: How do I rewrite the SEQ as a set of first order differential equations? Am I using Euler Method correctly?
My attempt:
I think it is this, but I'm not sure:
Let $w={\mathrm{\psi <!--\; \psi \; -->}}^{\prime }$ and $w^{\prime }={\mathrm{\psi <!--\; \psi \; -->}}^{″}$. Now we have
$w^{\prime }=\frac{-<!--\; -\; -->2mE}{{\mathrm{\hslash <!--\; \hslash \; -->}}^2}\mathrm{\psi <!--\; \psi \; -->}$
and to use Euler method it is
$w_{i+1}=w_i+h{w}^{\prime }\to <!--\; \backslash rightarrow\; -->{\mathrm{\psi <!--\; \psi \; -->}}_{i+1}^{\prime }={\mathrm{\psi <!--\; \psi \; -->}}_i^{\prime }+h{\mathrm{\psi <!--\; \psi \; -->}}_i^{″}$
${\mathrm{\psi <!--\; \psi \; -->}}_{i+1}={\mathrm{\psi <!--\; \psi \; -->}}_i+hw\to <!--\; \backslash rightarrow\; -->{\mathrm{\psi <!--\; \psi \; -->}}_{i+1}={\mathrm{\psi <!--\; \psi \; -->}}_i+h{\mathrm{\psi <!--\; \psi \; -->}}_{i+1}^{\prime }$