Learn First Order Problems with These Differential Equations Examples

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?

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
$w_{i+1}=w_i+h{w}^{\prime }\to <!--\; \backslash rightarrow\; -->{\mathrm{\psi <!--\; \psi \; -->}}_{i+1}^{\prime }={\mathrm{\psi <!--\; \psi \; -->}}_i^{\prime }+h{\mathrm{\psi <!--\; \psi \; -->}}_i^{″}$
${\mathrm{\psi <!--\; \psi \; -->}}_{i+1}={\mathrm{\psi <!--\; \psi \; -->}}_i+hw\to <!--\; \backslash rightarrow\; -->{\mathrm{\psi <!--\; \psi \; -->}}_{i+1}={\mathrm{\psi <!--\; \psi \; -->}}_i+h{\mathrm{\psi <!--\; \psi \; -->}}_{i+1}^{\prime }$

The integral surface of the first order partial differential equation
$2y(z-<!--\; -\; -->3)\frac{\mathrm{\partial <!--\; \partial \; -->}z}{\mathrm{\partial <!--\; \partial \; -->}x}+(2x-<!--\; -\; -->z)\frac{\mathrm{\partial <!--\; \partial \; -->}z}{\mathrm{\partial <!--\; \partial \; -->}y}=y(2x-<!--\; -\; -->3)$
passing through the curve $x^2+y^2=2x,z=0$ is
1. $x^2+y^2-<!--\; -\; -->z^2-<!--\; -\; -->2x+4z=0$
2. $x^2+y^2-<!--\; -\; -->z^2-<!--\; -\; -->2x+8z=0$
3. $x^2+y^2+z^2-<!--\; -\; -->2x+16z=0$
4. $x^2+y^2+z^2-<!--\; -\; -->2x+8z=0$
My effort:
I find the mulipliers x, 3y, -z and get the solution $x^2+3y^2-<!--\; -\; -->z^2=c_1$. How to proceed further?

$x^{\prime }=\frac{t-<!--\; -\; -->x}{t},t>0$
I tried to solve it by:
$x^{\prime }=1-<!--\; -\; -->\frac{x}{t}$
$\int <!--\; \int \; -->dx=\int <!--\; \int \; -->(1-<!--\; -\; -->\frac{x}{t})dt$
$x=t-<!--\; -\; -->x\ln <!--\; \; -->t+C$
$x=\frac{t}{1+lnt}+C$
But this doesn't seem right. What am I doing wrong? Also I'm supposed to determine the maximal interval around t=1, but how do I get rid of $C$?

For a given differential equation
$x{y}^{\prime }+2y\log <!--\; \; -->y-<!--\; -\; -->4x^{2}y=0;$
$y(1)=1$
I want to use the substitution $v=\log <!--\; \; -->y$. Which implies that
$v^{\prime }=\frac{dv}{dy}\frac{dy}{dx}=\frac{1}{y}{y}^{\prime }$
Hence:
$y^{\prime }=y{v}^{\prime }$
That being said, solving the equation I try as follows:
$xy{v}^{\prime }+2yv-<!--\; -\; -->4x^{2}y=0$
$x{v}^{\prime }+2v=4x^2$
$v^{\prime }+\frac{2}{x}v=4x$
$v^{\prime }=4x-<!--\; -\; -->\frac{2}{x}v$
How can I use seperation of variables here? Or how can I solve this thing ?

Given the inhomogeneous first order differential equation below:
$\stackrel{\dot{}<!--\; \dot{}\; -->}{x}(t)+ax(t)=q(t)$
where $a\in <!--\; \in \; -->\mathbb{R}$ and q(t) is a continuous function.
It is given that the differential equation has the following 2 particular solutions:
$\begin{array}{rl}{x}_1(t)& =e^{4t}+3e^{9t}\\ x_2(t)& =-e^{4t}+3e^{9t}\end{array}$
State below the value of the constant a and an expression for the function q(t).
When I apply the solution formula, I get 1 equation with 2 different unknowns. Can someone please help me?

$\frac{dv(t)}{dt}+a{v}^2(t)+bP(t)=c$
This is a first order non-linear differential equation. Unfortunately, I have no experience solving non-linear differential equations. From the research I have done, this type of equation looks similar to the Ricatti equation. Is there a closed form solution to the above equation? How can I solve this equation? I'm interested in getting a function that shows how the pressure or velocity of the system decays with time.
FYI: v(t) is velocity, P(t) is pressure, (a,b,c) are constants. Thanks!

Suppose u is a twice continuously differentiable function with linear growth,
$\underset{x\to <!--\; \backslash rightarrow\; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}{u}^{\prime }(x)-<!--\; -\; -->\frac{1}{g(x)}u(x)=0$
and g is a Lipschitz continuous function with Lipschitz constant $L<1$.
Consider the first order linear homogenous differential equation
$y^{\prime }(x)-<!--\; -\; -->\frac{1}{g(x)}y(x)=0.$
The general solution is
$y(x)=c\exp <!--\; \; -->(\int <!--\; \int \; -->\frac{1}{g(x)}dx)$
for constant $c\in <!--\; \in \; -->\mathbb{R}$. In any solution with linear growth, $\underset{x\to <!--\; \backslash rightarrow\; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}y(x)=0$
Is it possible to conclude that $\underset{x\to <!--\; \backslash rightarrow\; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}u(x)=0$?

I'm working through a question where the differential equation is
$y^2(y^{\prime 2}-<!--\; -\; -->1)(3y^{\prime 2}+1)=c,y(0)=0$
and the answer proceeds with two cases
$c=0⟹<!--\; ⟹\; -->y(x)=0\vee <!--\; \vee \; -->y(x)=\pm <!--\; \pm \; -->x$
$c\ne <!--\; \ne \; -->0⟹<!--\; ⟹\; -->\underset{x\to <!--\; \backslash rightarrow\; -->0}{lim}{y}^2{y}^{\prime 4}=c/3$
I'm fine with (1), but I can't see how to get (2). I've tried multiplying out the brackets and eliminating terms involving y only, but that doesn't get the answer so far as I can see. I'm not very familiar with how to manipulate limits, especially in the context of differential equations, and would appreciate some help.

My question is to find the solutions to the following
$\frac{df(x)}{dx}=f^{-<!--\; -\; -->1}(x)$
where $f^{-<!--\; -\; -->1}(x)$ refers to the inverse of the function f. The domain really isn't important, though I am interested in either (-inf, inf) or (0, inf), so if any solutions are known for more restricted domains then they are welcome.
I cannot find any material relating to this type of question in any of my calculus and differential equations textbooks and references; it seems quite unorthodox. Any material which covers this type of diff equation would be wlecome

Let the first order differential equation be
$dy/dx+P(x)y=Q(x)$
To solve this we multiply it by function say v(x) then it becomes
$v(x)dy/dx+P(x)v(x)y=Q(x)v(x)$
$\therefore <!--\; \therefore \; -->d(v(x)y)/dx=Q(x)v(x)$
where $d(v(x))/dx=P(x)v(x)$
I don't write full solution here but now here to obtain integrating factor v(x) from above equation we get $ln|v(x)|=\int <!--\; \int \; -->P(x)dx$.
Now my question is why we neglect absolute value of v(x) and obtain integrating factor as $e^{\int <!--\; \int \; -->P(x)dx}=v(x)$

$x{e}^y\cdot <!--\; \cdot \; -->y^{\prime }-<!--\; -\; -->2e^y=x^2$
Solve the equation using the proper substitution

I would like some help on comprehending this question as well as a push in the right direction. The question gave a system of first-order differential equation.
$x(t{)}^{\prime }=4x(t)-<!--\; -\; -->3y(t)+6e^{2t}$
$y(t{)}^{\prime }=4x(t)-<!--\; -\; -->6y(t)$
The question asked me to find the 2nd inhomogeneous equation that satisfies x(t). Does this mean the answer should all be in terms of x? I tried focusing on the x and differentiating it with respect to t.
so $x(t{)}^{\prime }=4x(t)-<!--\; -\; -->3y(t{)}^{\prime }+6e^{2t}$ becomes:
$x(t{)}^{″}=4x(t{)}^{\prime }-<!--\; -\; -->3y(t{)}^{\prime }+12e^{2t}$
for simplicity sake, I will write x(t) as x and y(t) as y.
After that step, I replaced the y' in the x'' equation with the rearranged y' from the original question into the differentiated x' equation. This gives:
$x^{″}=4x^{\prime }-<!--\; -\; -->3(4x-<!--\; -\; -->6(\frac{1}{-<!--\; -\; -->3}(x^{\prime }-<!--\; -\; -->4x-<!--\; -\; -->6e^{2t})+12e^{2t}$
this cancels down to:
$x^{″}=4x^{\prime }-<!--\; -\; -->12x+2x^{\prime }-<!--\; -\; -->8x$
but if you move everything to one side, it becomes
$x^{″}-<!--\; -\; -->6x^{\prime }+20x=0$
this is a second-order homogenous equation, so I don't quite know where I went wrong

I'm trying to solve an initial value problem. I know how to do the problem once integrated but solving the differential equation is where I'm finding trouble.
$dy/dx=3+\sqrt{2y+17x-<!--\; -\; -->3}$
I'd thought that maybe squaring both sides will get rid of the square root but that doesn't work so I was hoping someone would point me in the right direction of how to go about seperating $y$ and $x.$

I would like to solve:
$y-<!--\; -\; -->y^{\prime }x-<!--\; -\; -->y^{\prime 2}=0$
In order to do so, we let $y^{\prime }=t$, and we assume x as a function of t. Now, we take derivative with respect to t from the differential equation, and obtain
$\frac{dy}{dt}-<!--\; -\; -->x-<!--\; -\; -->t\frac{dx}{dt}-<!--\; -\; -->2t=0$
By the chain rule, we have: $dy/dt=tdx/dt$. So, the above simplifies to
$x=-<!--\; -\; -->2t$
That is, we have: $x=-<!--\; -\; -->2dy/dx$. Thus, we obtain
$y=-<!--\; -\; -->\frac{x^2}{4}+C$
Now, if we want to verify the solution, it turns out that C must be zero, in other words, $y=-<!--\; -\; -->x^2/4$ satisfies the original differential equation.
I have two questions:
1) What happens to the integration constant? That is, what is the general solution of the differential equation?
2) If we try to solve this differential equation with Mathematica, we obtain
$y=C_{1}x+C_1^2$,
which has a different form from the analytical approach. How can we also produce this result analytically?

1. Can someone help me solve the following system of differential equations?
$\begin{array}{rcl}\frac{d{P}_0}{dt}& =& -<!--\; -\; -->C_1\mathrm{\lambda <!--\; \lambda \; -->}{P}_0{P}_1+\frac{C_1}{2}{P}_1^2\\ \frac{d{P}_1}{dt}& =& -<!--\; -\; -->C_2{P}_1+C_1\mathrm{\lambda <!--\; \lambda \; -->}{P}_0{P}_1-<!--\; -\; -->\frac{1}{2}(1+\mathrm{\lambda <!--\; \lambda \; -->})C_1{P}_1^2+C_1{P}_1{P}_2\\ \frac{d{P}_2}{dt}& =& C_2{P}_1+\frac{C_1}{2}\mathrm{\lambda <!--\; \lambda \; -->}{P}_1^2-<!--\; -\; -->C_1{P}_1{P}_2\end{array}$
2. (Extending above) Suppose we have $K$ first order non-linear differential equations (similar to above), is there any simple method to solve them analytically?

I am trying to evaluate the equation:
$y^{\prime }=y(y^2-<!--\; -\; -->\frac{1}{2})$
I multiplied the y over and tried to solve it in seperable form (M and N). The partial deritives did not work out to be equal to eachother so I am now stuck finding an integrating factor. Is this the right approach?