Solve System of Differential Equations with Expert Help

Solve the following first-order nonlinear ordinary differential equation:
${\displaystyle y^{\prime }\left(x\right)+\log \left(y\left(x\right)\right)=-x-1}$

How to solve this first-order differential equation,I don't have any idea which type this is..
${\displaystyle y^{\prime }+{(x-y)}^2=0}$
I have ${\displaystyle y^{\prime }+x^2-2xy+y^2=0}$, and then I divide by ${\displaystyle x^2}$
${\displaystyle t^{\prime }+t(\frac{1}{x}-2x)+t^{2}x=-x}$
and replace ${\displaystyle \frac{y}{x}=t}$, so I have then
It seems like it's a Riccati's diff. equation,but I don't know how to do it beacuse I don't have any particular solution..
On Wolfram Alpha solution is
$y=\frac{1}{c_1{e}^{2ix}+\frac{i}{2}}+x+i$

I have first-order differential equation
${\displaystyle y=xy\prime +\frac{1}{2}{\left(y\prime \right)}^2}$
Maybe, with this someone will find way to solve it
${\displaystyle \frac{1}{2}{y}^{\prime }(2x+y^{\prime })=y}$
I thought I can use ${\displaystyle x^2+y=t}$ for subtitution and when I derivate, I have ${\displaystyle t^{\prime }=2x+y^{\prime }}$ which is acctualy the
${\displaystyle (t^{\prime }-2x)t^{\prime }=2t=2x^2}$
same as previous. Who knows?

Let ${\displaystyle (\bar{x},\bar{y},\bar{z})}$ be an equilibrium of autonomous first-order differential equation:
${\displaystyle \{\begin{array}{c}\dot{x}=f_1(x,y,z)\\ \dot{y}=f_2(x,y,z)\\ \dot{z}=f_3(x,y,z)\end{array}}$
Is it possible to say that ${\displaystyle (\bar{x},\bar{y},\bar{z})}$ is unique equilibrium point of the following system:
${\displaystyle \{\begin{array}{c}\dot{x}=f_1(x,y,z)-(x-\bar{x})\\ \dot{y}=f_2(x,y,z)-(y-\bar{y})\\ \dot{z}=f_3(x,y,z)-(z-\bar{z})\end{array}}$

a) Solve the differential equation:
${\displaystyle (x+1)\frac{dy}{dx}-3y={(x+1)}^4}$
given that ${\displaystyle y=16}$ and ${\displaystyle x=1}$, expressing the answer in the form of ${\displaystyle y=f\left(x\right)}$.
b) Hence find the area enclosed by the graphs ${\displaystyle y=f\left(x\right),y={(1-x)}^4}$ and the ${\displaystyle x-a\xi s}$.
I have found the answer to part a) using the first order linear differential equation method and the answer to part a) is: ${\displaystyle y={(1+x)}^4}$. However how would you calculate the area between the two graphs (${\displaystyle y={(1-x)}^4}$ and ${\displaystyle y={(1+x)}^4}$) and the x-axis.

How can I find at least one first order differential equation when the solution for it is ${\displaystyle y=x^4}$?

find ${\displaystyle a_n}$ and ${\displaystyle f\left(x\right)}$ so that ${\displaystyle \sum _{n=0}^{\mathrm{\infty }}\frac{a_n}{n!}{x}^n}$ satisfy ${\displaystyle f^{\prime }\left(x\right)-f\left(x\right)=x^2}$ and ${\displaystyle f\left(0\right)=1}$
here I tried to find ${\displaystyle f^{\prime }\left(x\right)=\sum _{n=0}^{\mathrm{\infty }}\frac{a_n}{(n-1)!}{x}^{n-1}}$ from ${\displaystyle \frac{df\left(x\right)}{dx}-f\left(x\right)=x^2}$ using first order differential equation I got ${\displaystyle y=x^2+2x+1}$
${\displaystyle \to f\left(x\right)?}$
but I don't know what is the relation of the summation and first order differential equation. and how to find ${\displaystyle f\left(x\right)}$ and ${\displaystyle a_n}$
can someone give me hint? thanks!

I am trying to solve a differential equation of the type ${\displaystyle x{}^{″}=-x+x^3}$. Now when I first integrate it with respect to time, ${\displaystyle t}$, then I am getting ${\displaystyle x^{\prime }=x(x^2-1)+C}$, which is a non-linear first order differential equation. Now if the constant happens to be zero then I can solve it by partial fraction method but as life is not easier that constant term tagged to the ${\displaystyle x^{\prime }}$ is actually not zero. So how to proceed forward in this case.
Thank you!

Make use of vectors to re-write the following second order differential equation into a 1st order differential equation. (note: I do not need to solve it).
${\displaystyle 2.1x^{-2}\frac{d^{2}y}{{dx}^2}-3e^{2x}{y}^4\frac{dy}{dx}=-6y}$
where ${\displaystyle y=4.2}$ and ${\displaystyle \frac{dy}{dx}=-3.1}$ when ${\displaystyle x=0.5}$.
Im not sure were to begin any help is greatly appreciated.

Let λ ∈ R be a parameter. We consider the ODE with conditions to the limits. y + λy = 0, y(0)= 0, y(1)=0 Explain why the existence and uniqueness theorem does not apply.

${\displaystyle y^{\prime }+e^{y^{\prime }}-x=0}$
that I have simplified like so
${\displaystyle e^{y^{\prime }}=x-y^{\prime }}$
${\displaystyle {\ln e}^{y^{\prime }}=\ln (x-y^{\prime })}$
${\displaystyle y^{\prime }=\ln (x-y^{\prime })}$
but I do not know how to solve this further to obtain the general solution. I have done first order linear differential equation strategies so far. How should I get about doing this question with the strategies I have?

I understand that it applies to second order and any other differential equation as long as they are Homogenous and Linear, but I cant

${\displaystyle \dot{x}=rx-h{x}^2+q}$, where ${\displaystyle \dot{x}=\frac{dx}{dt}}$. I'm assuming ${\displaystyle r,h,q}$ have different units.
I first tried ${\displaystyle \tau =\frac{t}{t_0}}$ and ${\displaystyle A=\frac{x}{x_0}}$. Then I got ${\displaystyle \frac{dA{x}_0}{d\tau t_0}=rA{x}_0-h{A}^2{x}_0^2+q}$. But I am not really sure where to go from there.
Need any help.

I have this:
${\displaystyle \ddot{x}+\frac{x}{{(x^2+y^2)}^{\frac{3}{2}}}=0}$
${\displaystyle \ddot{y}+\frac{y}{{(x^2+y^2)}^{\frac{3}{2}}}=0}$
How do I go on to create 4 coupled first order differential equations for ${\displaystyle x,y,\dot{x},\dot{y}}$?

Suppose we have
${\displaystyle \frac{dy}{dx}+f\left(x\right)y=r\left(x\right)}$
and it has two solutions ${\displaystyle y_1\left(x\right)}$ and ${\displaystyle y_2\left(x\right)}$ then how to prove that solution of differential equation
${\displaystyle \frac{dy}{dx}+f\left(x\right)y=2r\left(x\right)}$
Will be ${\displaystyle y_1\left(x\right)+y_2\left(x\right)}$? I think given differential equations is linear first order equation so its solution will be
${\displaystyle y.e^{\int f\left(x\right)dx}=\int r.e^{\int f\left(x\right)dx}dx}$
now do I establish two solution as ${\displaystyle y_1}$ and ${\displaystyle y_2}$ out of this equation?

I've tried many times to reach the solution of a first order differential equation (of the last equation) but unfortunately I couldn't. Could you please help me to know how did he get this solution.
The equation is:
${\displaystyle \frac{dX}{dt}=k_f-(k_f+k_b)X}$
The author assumed that
${\displaystyle X=0}$ at ${\displaystyle t=0}$
Then, he said in his first publication the solution is given as:
${\displaystyle X=\frac{k_f}{(k_f+k_b)}\{1-\exp [-(k_f+k_b)t]\}}$
I agree with him about the above equation as I am able to reach this form by using the normal solution of the first order differential equation.
In another publication, the same author dealt with the same equations, but when he wrote the solution of the first order differential equation, he wrote it in another form which I could't reach it and I don't know how did he reach it. This form is given as:
${\displaystyle X=\frac{k_f}{(k_f+k_b)}\{1-\exp [-(k_f+k_b)t]\}+X_0\exp [-(k_f+k_b)t]}$ and he said ${\displaystyle X_0}$ was taken to be zero in the first publication.
Can any of you help how did the author get this solution please?

I really dont know what to do whit this equation, i have tried to integrate it, but i can do it.
${\displaystyle y^{\prime 3}-y{y}^{\prime 2}-x^2{y}^{\prime }+x^{2}y=0}$

Find f(x) that satisfies the first-order differential equation:
${\displaystyle f^{\prime }\left({\sin }^{2}x\right)={\cos }^{2}x+{\tan }^{2}x\left(\text{for}0<x<1\right)}$
Thanks!

Throughout my engineering education, Ive

Given that ${\displaystyle y=\sin \left(x\right)}$ is an expicit function of the first-order differential equation ${\displaystyle \frac{dy}{dx}=\sqrt{1-y^2}}$. Find an interval I of definition, the solution interval.
So I got to the point that ${\displaystyle \cos \left(x\right)\frac{=}{\cos \left(x\right)}}$, what do I do next?