Solve System of Differential Equations with Expert Help

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.

Find the linear approximation of the function
$f(x,y)=\ln <!--\; \; -->(e+x+y)$
at point $(0,0)$. Use it to approximate the value of the function at $(0.1,0.2)$

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 $e^{\int <!--\; \int \; -->P(x)dx}=v(x)$

Use Linear Approximation to estimate $\mathrm{\Delta <!--\; \Delta \; -->}f=f(3.02)-<!--\; -\; -->f(3)$

Linear approximation by rational number to $4\sqrt{3}$

$x{e}^y\cdot <!--\; \cdot \; -->y^{\prime }-<!--\; -\; -->2e^y=x^2$
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?

Use a linear approximation of
$f(x)=\sqrt[3]{x}$
at
$x=8$
to approximate
$\sqrt[3]{7}$
Express your answer as an exact fraction.

You're measuring the velocity of an object by measuring that it takes $1$ second to travel $1.2$ meters. The measurement error is $001$ meters in distance and the error in time is $01$ second.
What is the absolute value of the error in the linear approximation for the velocity?

Consider the function $w=z^3$ near the point $z=1+i,w=-2+2i$ What is the linear approximation near this point for this mapping? Is the linear approximation a one-to-one function?