Solve System of Differential Equations with Expert Help

We have $y^{\prime }=x/y$, which is a first-order homogeneous differential equation.
It can be solved by rearranging to y dy=x dx and then integrating both parts which yields that $y=\pm <!--\; \pm \; -->\sqrt{x^2+c}$.
Now if we use the substitution $y=ux⟹<!--\; ⟹\; -->y^{\prime }=u^{\prime }x+u,$, and rewrite the differential equation as
$u^{\prime }x+u=\frac{1}{u}$
and then rearrange to
$\left(\frac{1}{1/u-<!--\; -\; -->u}\right)du=\left(\frac{1}{x}\right)dx$
by integrating both parts we get that
$\begin{array}{}\text{(1)}& -<!--\; -\; -->\frac{1}{2}\ln <!--\; \; -->|u^2-<!--\; -\; -->1|=\ln <!--\; \; -->\left|x\right|+c\end{array}$
For $y=\pm <!--\; \pm \; -->x$ (a special solution for c=0) $\to <!--\; \backslash rightarrow\; -->u=\pm <!--\; \pm \; -->1$, and by plugging $\pm <!--\; \pm \; -->1$ into (1) we get that
$\begin{array}{}\text{(2)}& \ln <!--\; \; -->\left|0\right|=\ln <!--\; \; -->\left|x\right|+c\end{array}$
What does equation (2) mean? $\ln <!--\; \; -->\left|0\right|$ is undefined. Is this of any significance?
Edit 1:
As pointed out when rearranging from $u^{\prime }x+u=\frac{1}{u}$ to $\left(\frac{1}{1/u-<!--\; -\; -->u}\right)du=\left(\frac{1}{x}\right)dx$, we implicitly assumed that $u\ne <!--\; \ne \; -->\pm <!--\; \pm \; -->1$. Equation (1) does not hold for $u=\pm <!--\; \pm \; -->1$
Edit 2:
Solving equation (1) for u with $u\ne <!--\; \ne \; -->\pm <!--\; \pm \; -->1$, we arrive at the same family of equations but with $c\ne <!--\; \ne \; -->0$. The fact that c can be zero comes from setting $u=\pm <!--\; \pm \; -->1$

Let
$\left[\begin{array}{ccc}-<!--\; -\; -->2& 1& 0\\ 0& -<!--\; -\; -->2& 1\\ 0& 0& -<!--\; -\; -->2\end{array}\right]$
and
$|x(t)|=(x_1^2(t)+x_2^2(t)+x_3^2(t){)}^{1/2}$
Then any solution of the first order system of the ordinary differential equation
$\{\begin{array}{r}{x}^{\prime }(t)=Ax(t)\\ x(0)=x_0\end{array}$
satisfies
1. $\underset{t\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}|x(t)|=0}{lim}$
2. $\underset{t\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}|x(t)|=\mathrm{\infty <!--\; \infty \; -->}}{lim}$
3. $\underset{t\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}|x(t)|=2}{lim}$
4. $\underset{t\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}|x(t)|=12}{lim}$

Solve the differential equatio
$\frac{dy}{dx}={\left(\frac{x+2y-<!--\; -\; -->3}{2x+y+3}\right)}^2$
The answer is given as:
$(x+3{)}^3-<!--\; -\; -->(y+3{)}^3=c(x-<!--\; -\; -->y+6{)}^4$
My attempt :I tried to expand the terms in the numerator and the denominator, arrange them into differentials and then integrate then, but I couldn't arrange all the terms in integrable format.

I was given a problem, it has been worded as follows: Use the substitution $z=\frac{y}{x}$ to transform the differential equation ${\displaystyle \frac{dy}{dx}=\frac{-<!--\; -\; -->3xy}{y^2-<!--\; -\; -->3x^2}}$, into a linear equation. Hence obtain the general solution of the original equation.
Workings:
${\displaystyle \frac{dy}{dx}=\frac{-<!--\; -\; -->3(\frac{y}{x})}{(\frac{y}{x}{)}^2-<!--\; -\; -->3}}$
Multiplied by $\frac{\frac{1}{x^2}}{\frac{1}{x^2}}$
$\frac{dy}{dx}=\frac{-<!--\; -\; -->3z}{z^2-<!--\; -\; -->3}$
$z^{\prime }=\frac{y^{\prime }}{x}-<!--\; -\; -->\frac{y}{x^2}$
$(z^{\prime }+\frac{y}{x^2})x=\frac{-<!--\; -\; -->3z}{z^2-<!--\; -\; -->3}$
After a whole page of workings, I arrive at this
$z^{\prime }+\frac{z}{x}=\frac{3}{z}$
Technically, this is a linear equation since it takes the form y'+p(x)y=f(x) . Since z=f(y,x). However, I'm unable to compute this, since I won't know how to integrate z with respect to x. I do know this can easily be done by separating variables, however, we are basically told to solve it this way.

How can I solve this ODE of first order:
$\begin{array}{rl}{y}^{{}^{\prime }}=y^2+x,& \text{where }{y}^{{}^{\prime }}=\frac{dy}{dx}\end{array}$
Is there any exact mehod to solve it ?
Thanks!

Im clueless on how to solve the following question...
$x{e}^y\frac{dy}{dx}=e^y+1$
What i've done is...
$\frac{dy}{dx}=\frac{1}{x}+\frac{1}{x{e}^e};\frac{dy}{dx}-<!--\; -\; -->\frac{1}{x{e}^e}=\frac{1}{x}$
Find the integrating factor..
$$\begin{gathered} v(x)=e^{P(x)};whereP(x)=\int <!--\; \int \; -->p(x)dx\Rightarrow <!--\; \Rightarrow \; -->P(x)=\int <!--\; \int \; -->\frac{1}{x}dx=ln|x| \\ v(x)=e^{P(x)}=e^{ln|x|}=x; \\ y=\frac{1}{v(x)}\int <!--\; \int \; -->v(x)q(x)dx=\frac{1}{x}\int <!--\; \int \; -->x\frac{1}{x}dx=1+c \end{gathered}$$
I know I made a mistake somewhere. Would someone advice me on this?

I was wondering if I could get some advice on how to tackle this question:
Consider the differential equation
$x^2\frac{dy}{dx}+2xy-<!--\; -\; -->y^3=0(3)$
Make the substitution $u=y^{-<!--\; -\; -->2}$ and show that the differential equation reduces to
$-<!--\; -\; -->\frac{1}{2}{x}^2\frac{du}{dx}+2xu-<!--\; -\; -->1=0(4)$
Solve equation (4) for u(x) and hence write down the solution for equation (3).
I'm trying to do the first part of showing that the differential equation reduces to equation 4. I have started out by:
$\begin{array}{rl}u& =y^{-<!--\; -\; -->2}\\ & =\frac{1}{y^2}\\ \therefore <!--\; \therefore \; -->y^2& =\frac{1}{u}\\ ⟹<!--\; ⟹\; -->y& =\pm <!--\; \pm \; -->\sqrt{\frac{1}{u}}\end{array}$
I'm not sure where to continue on from here though.

I am given this:
$(2x+1)\frac{dy}{dx}+y=0$
I tried this:
$\frac{1}{(2x+1)}dx=\frac{-<!--\; -\; -->1}{y}dy$
Then integrated the above sum and got this:
$\frac{ln(2x+1)}{2}=-<!--\; -\; -->ln(y)$
The answer is: $y^2(2x+1)=C$.
I tried solving it by placing the like terms together and integrating them. However, my answer is wrong from the answer given. Could you point out my mistake? Or am i evaluating the entire thing incorrectly?
All suggestions and help are appreciated!

I've been trying to solve the following differential equation:
$3x+y(x)-<!--\; -\; -->2+\frac{dy(x)}{dx}(x-<!--\; -\; -->1)=0$
And I found out it's an exact differential equation, since it can be rearranged as $(3x+y(x)-<!--\; -\; -->2)dx+(x-<!--\; -\; -->1)dy=0$
Assuming that a function $U(x,y)=\frac{\mathrm{\partial <!--\; \partial \; -->}U}{\mathrm{\partial <!--\; \partial \; -->}x}dx+\frac{\mathrm{\partial <!--\; \partial \; -->}U}{\mathrm{\partial <!--\; \partial \; -->}y}dy=k$, (where $k\equiv <!--\; \equiv \; -->$ constant) exists, I calculated it as:
$U(x,y)=\int <!--\; \int \; -->(3x+y-<!--\; -\; -->2)dx+\int <!--\; \int \; -->(x-<!--\; -\; -->1)dy=k$
And I got
$y(x)=\frac{-<!--\; -\; -->3x^2}{2(2x-<!--\; -\; -->1)}+\frac{2x}{2x-<!--\; -\; -->1}+\frac{k}{2x-<!--\; -\; -->1}$
But Mathematica says the solution is
$y(x)=\frac{-<!--\; -\; -->3x^2}{2(x-<!--\; -\; -->1)}+\frac{2x}{x-<!--\; -\; -->1}+\frac{k}{1-<!--\; -\; -->x}$
So either I assumed something which isn't correct or I made a mistake along the process. Where did I go wrong?

I am curious if there are any examples of functions that are solutions to second order differential equations, that also parametrize an algebraic curve.
I am aware that the Weierstrass $\mathrm{\wp <!--\; \wp \; -->}$ - Elliptic function satisfies a differential equation. We can then interpret this Differential Equation as an algebraic equation, with solutions found on elliptic curves. However this differential equations is of the first order.
So, are there periodic function(s) $F(x)$ that satisfy a second order differential equation, such that we can say these parametrize an algebraic curve?
Could a Bessel function be one such solution?

In a course on partial differential equations I came through this theorem about the general solution of a first order quasi-linear partial differential equation.
1. The general solution of a first-order, quasi-linear partial differential equation
$a(x,y,u)u_x+b(x,y,u)u_y=c(x,y,u)$
is given by $f(\varphi ,\psi )=0$, where $f$ is an arbitrary function of $\varphi (x,y,u)$ and $\psi (x,y,u).$.
2. $\varphi =C1$ and $\psi =C2$ are solution curves of the characteristic equations
$\frac{dx}{a}=\frac{dy}{b}=\frac{du}{c}.$
Is there any geometric interpretation of both these points so that I can have a better intuitive understanding of the graphical representation of $f$,$\mathrm{\varphi <!--\; \varphi \; -->}$ and $\mathrm{\psi <!--\; \psi \; -->}$ ?

I have been solving an inexact first order ODE, but arriving to the step of finding an integrating factor /mu, I was enable to find either /mu of x or /mu of y. Is there another way of solving this ODE? or how can I come up with the solution.
$(e^{x+y}-<!--\; -\; -->y{e}^x)\mathrm{d}x+e^y\mathrm{d}y=0$

I've been working on this differential equation for days now but I can not find a solution. I understand that finding a solution in closed form is difficult to say (assuming there is). What I ask is if there is a substitution (change of variables) that can "transform" this differential equation from the second order to the first order so that it is not too constrained. Thank you for your interest and time.
$y^{″}(x)+ax\cos <!--\; \; -->y(x)=0$
where $a\in <!--\; \in \; -->\mathbb{R}$

Verify the given linear approximation at $a=0$. Then determine the values of $x$ for which the linear approximation is accurate to within $0.1$
$1/(1+2x{)}^4\approx <!--\; \approx \; -->1–8x$
how do I determine its accuracy?

Determine how accurate should we measure the side of a cube so that the calculated surface area of the cube lies within $3$% of its true value, using Linear Approximation.
Let $A(x)=TSA$; $x=side$
$A(x)=6x^2$

Let $f$ be a differentiable function such that $f(3)=2.345$ and $f^{\prime }(x)=ln(x^2+1)$. What is the value of $f(5)$?

Use the linear approximation of Use the linear approximation of f(x,y)=e2x2+3y at (0,0) to estimate f(0.01,−0.02). at $(0,0)$ to estimate $f(0.01,-<!--\; -\; -->0.02)$.

Linear approximation to find $\frac{1}{4.002}$