Solve System of Differential Equations with Expert Help

Consider the differential equation
$\stackrel{\dot{}<!--\; \dot{}\; -->}{x}(t)=\mathrm{\alpha <!--\; \alpha \; -->}x(t)+\mathrm{\beta <!--\; \beta \; -->}u(t),$
with $\mathrm{\alpha <!--\; \alpha \; -->},\mathrm{\beta <!--\; \beta \; -->}>0$ positive constant scalars and initial condition $x(0)=x_0$. Function $u(\cdot <!--\; \cdot \; -->)$ is known. The Laplace solution for this equation should be
$x(t)=e^{\mathrm{\alpha <!--\; \alpha \; -->}(t-<!--\; -\; -->t_0)}{x}_0+\mathrm{\beta <!--\; \beta \; -->}{\int <!--\; \int \; -->}_0^t{e}^{\mathrm{\alpha <!--\; \alpha \; -->}(t-<!--\; -\; -->\mathrm{\tau <!--\; \tau \; -->})}u(\mathrm{\tau <!--\; \tau \; -->})d\mathrm{\tau <!--\; \tau \; -->}.$
However, if now I take the time derivative, I get
$\stackrel{\dot{}<!--\; \dot{}\; -->}{x}(t)=\mathrm{\alpha <!--\; \alpha \; -->}{e}^{\mathrm{\alpha <!--\; \alpha \; -->}(t-<!--\; -\; -->t_0)}{x}_0+\mathrm{\beta <!--\; \beta \; -->}u(t).$
Comparing to the first equation, this leaves me with
$x(t)=e^{\mathrm{\alpha <!--\; \alpha \; -->}(t-<!--\; -\; -->t_0)}{x}_0,$
which looks inconsistent with the Laplace solution. Anybody has a clue on what's going on here?

I have the differential equation
$x^2{y}^{\prime }(x)+2xy(x)=y^2(x)$
With initial condition y(1)=1 and I want to solve this, by observation, we can see the LHS is $x^{2}y(x)$ but I am unfamiliar with tackling the RHS and was wondering where to go from here.
$x^{2}y(x)=\int <!--\; \int \; -->y^2(x)dx$

A car is travelling at 100 km/h on a level road when it runs out of fuel. Its speed v (in km/h) starts to decrease according to the formula
$\frac{dv}{dt}=-<!--\; -\; -->kv(1)$
where k is constant. One kilometre after running out of fuel its speed has fallen to 50 km/h. Use the chain rule substitution
$\frac{dv}{dt}=\frac{dv}{ds}\frac{ds}{dt}=\frac{dv}{ds}v$
to solve the differential equation.
Note: Although I haven't solved it yet, the answers say that this isn't a reasonable model as the velocity is always positive; I didn't make a typo in the question.
What I'm trying to do is solve velocity as a function of displacement (s, in km), velocity as a function of time (t, in hours), and displacement as a function of time (I need these functions for later parts of the question).
So far I've found velocity as a function of displacement (v(s)):
$\frac{dv}{dt}=-<!--\; -\; -->k\frac{ds}{dt}\text{(from (1))}$
$\int <!--\; \int \; -->\frac{dv}{dt}dt=-<!--\; -\; -->k\int <!--\; \int \; -->\frac{ds}{dt}dt$
$v(s)=-<!--\; -\; -->ks+C$
$v(0)=100⟹<!--\; ⟹\; -->C=100,v(1)=50⟹<!--\; ⟹\; -->k=50$
$v(s)=-<!--\; -\; -->50s+100$
Then I've tried to find velocity as a function of time (v(t)), but I've got stuck. I can't find any differential equation I can use to get this, or to get displacement as a function of time (s(t)).
The answer key says $v(t)=100e^{-<!--\; -\; -->50t}$ and $s(t)=2(1-<!--\; -\; -->e^{-<!--\; -\; -->50t})$
I've solved such questions many times before, but it's been a while so I'm a bit rusty. So, even a hint might be enough for me to realise what to do.

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 $c$ ?

What is the standard method for finding solutions of differential equations such as this one? (if there is any)
$x{y}^{\prime }=y^2-<!--\; -\; -->(2x+1)y+x^2+2x$
where $y=ax+b$ is a particular solution.
Do I substitute $y$ with $ax+b+u(x)$ and then search for a solution or am I not noticing something and there's quicker way?

Linear approximation of quotient
$\frac{(2.01{)}^2}{\sqrt{.95}}$

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: $y_0(0)=1,y_1(0)=0,y_2(0)=0.$

Use a linear approximation to estimate the number ${8.07}^{2/3}$

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

If $f(x)=g(x)h(x)$
does the linear approximation of $f(x)$ equals the linear approximation of $g(x)$ times the linear approximation of $h(x)$?
is it true for quadratic approximations as well?

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.

Since we know that in a good linear approximation, $L(x)=f(a)+f^{\prime }(a)(x-<!--\; -\; -->a).$. But what if $f^{\prime }(a)$ does not exist? How to prove that if a function has a good linear approximation, then it must be differentiable?

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.