Learn First Order Problems with These Differential Equations Examples

I have a first order PDE:
$x{u}_x+(x+y)u_y=1$
With the initial condition:
I have calculated result in Mathematica: $u(x,y)={\displaystyle \frac{y}{x}}$, but I am trying to solve the equation myself, but I had no luck so far. I tried with method of characteristics, but I could not get the correct results. I would appreciate any help or maybe even whole procedure.

In my differential equations book, I have found the following:
Let $P_0(\frac{dy}{dx}{)}^n+P_1(\frac{dy}{dx}{)}^{n-<!--\; -\; -->1}+P_2(\frac{dy}{dx}{)}^{n-<!--\; -\; -->2}+......+P_{n-<!--\; -\; -->1}(\frac{dy}{dx})+P_n=0$ be the differential equation of first degree 1 and order n (where $P_i$ $\mathrm{\forall <!--\; \forall \; -->}$ i $\in <!--\; \in \; -->0,1,2,...n$ are functions of x and y).
Assuming that it is solvable for p, it can be represented as:
$[p-<!--\; -\; -->f_1(x,y)][p-<!--\; -\; -->f_2(x,y)][p-<!--\; -\; -->f_3(x,y)]........[p-<!--\; -\; -->f_n(x,y)]=0$
equating each factor to Zero, we get n differential equations of first order and first degree.
$[p-<!--\; -\; -->f_1(x,y)]=0,[p-<!--\; -\; -->f_2(x,y)]=0,[p-<!--\; -\; -->f_3(x,y)]=0,........[p-<!--\; -\; -->f_n(x,y)]=0$
Let the solution to these n factors be:
$F_1(x,y,c_1)=0,F_2(x,y,c_2)=0,F_3(x,y,c_3)=0,........F_n(x,y,c_n)=0$
Where $c_1,c_2,c_3.....c_n$ are arbitrary constants of integration. Since all the c’s can have any one of an infinite number of values, the above solutions will remain general if we replace $c_1,c_2,c_3.....c_n$ by a single arbitrary constant c. Then the n solutions (4) can be re-written as
$F_1(x,y,c)=0,F_2(x,y,c)=0,F_3(x,y,c)=0,........F_n(x,y,c)=0$
They can be combined to form the general solution as follows:
$F_1(x,y,c)F_2(x,y,c)F_3(x,y,c)........F_n(x,y,c)=0(1)$
Now, my question is, whether equation (1) is the most general form of solution to the differential equation.I think the following is the most general form of solution to the differential equation :
$F_1(x,y,c_1)F_2(x,y,c_2)F_3(x,y,c_3)........F_n(x,y,c_n)=0(2)$
If (1) is the general solution, the constant of integration can be found out by only one IVP say, $y(0)=0$. So, one IVP will give the particular solution. If (2) is the general solution, one IVP might not be able to give the particular solution to the problem.

Find all functions f(x) defined on $(-<!--\; -\; -->\frac{\mathrm{\pi <!--\; \pi \; -->}}{2},\frac{\mathrm{\pi <!--\; \pi \; -->}}{2})$ which has a primitive F(x) such that
$f(x)+\cos <!--\; \; -->(x)F(x)=\frac{\sin <!--\; \; -->(2x)}{(1+\sin <!--\; \; -->(x){)}^2}.$
Hence F satisfies a first order linear differential equation in F with integrating factor $e^{\sin <!--\; \; -->(x)}$
So i have to integrate
$\frac{2e^{\sin <!--\; \; -->(x)}\sin <!--\; \; -->(x)\cos <!--\; \; -->(x)}{(1+\sin <!--\; \; -->(x){)}^2}.$
By putting $t=\sin <!--\; \; -->(x)$, it reduces to $\frac{t{e}^t}{(1+t^2)}$ but now I am not getting it.

I'm currently stuck with a problem where I'm supposed to find all solutions that are asymptotic to the line $y=3-<!--\; -\; -->t$ when $t\to <!--\; \backslash rightarrow\; -->\mathrm{\infty <!--\; \infty \; -->}$. This is the demand, from here I'm supposed to create a first order linear differential equation. Can someone help me get started with this problem? Unsure of how to start....
Asymptotic would mean that for example $x(t)=y(t)-<!--\; -\; -->1/t$ would satisfy the given demand, not sure how to go further with this although.
A first order linear differential equation means I should have some sort of connection between my function and the derivative of the function, I cant make that connection....
It's my first time using this forum so I've probably made every mistake you can make, hopefully my question is still relevant...

$t{y}^{\prime }+ty=1-<!--\; -\; -->y$
I am having trouble going about this question. I have tried different processes to answer the differential equation, including separable, exact, homogeneous and linear equations. I have ran into a wall through all these processes. Maybe I'm not following the correct procedure... Help!
I know the first step would be to divide both sides by t, which would give:
$y^{\prime }+y=(1-<!--\; -\; -->y)/t$

My Problem is this given System of differential Equations:
$\stackrel{\dot{}<!--\; \dot{}\; -->}{x}=8x+18y$
$\stackrel{\dot{}<!--\; \dot{}\; -->}{y}=-<!--\; -\; -->3x-<!--\; -\; -->7y$
I am looking for a gerenal solution.
My Approach was: i can see this is a System of linear and ordinary differential equations. Both are of first-order, because the highest derivative is the first. But now i am stuck, i have no idea how to solve it. A Transformation into a Matrix should lead to this expression:
$\stackrel{\to <!--\; \to \; -->}{y}=\left(\begin{array}{cc}8& 18\\ -<!--\; -\; -->3& -<!--\; -\; -->7\end{array}\right)\cdot <!--\; \cdot \; -->x$
or is this correct:
$\stackrel{\to <!--\; \to \; -->}{x}=\left(\begin{array}{cc}8& 18\\ -<!--\; -\; -->3& -<!--\; -\; -->7\end{array}\right)\cdot <!--\; \cdot \; -->y\text{ ?}$
But i don't know how to determine the solution, from this point on.

$g(x)$ is continuous on $[1,2]$ such that $g(1)=0$ and
$g^{\prime }(x)x^2=\sqrt{1-<!--\; -\; -->g^2(x)}$
Find $g(2)$
I found that $g(x)=\sin <!--\; \; -->(c-<!--\; -\; -->\frac{1}{x})$ and since $g(1)=0$ shouldn't my c be equal to 1, so that $\sin <!--\; \; -->(1-<!--\; -\; -->1)=0$.
But when I try for $g(2)$ with $c=1$, I am getting a different answer from the textbook.

is there a method to solve
${\displaystyle \frac{dy}{dx}}=f(x,y)$
where $f(x,y)$ is a homogeneous function. I found some examples like $f(x,y)=(x+y{)}^2$ where it can be solved after converting it to Ricatti's equation.
thanks

I'm trying to solve this differential equation:
$\frac{dy}{dx}=\frac{x-<!--\; -\; -->\exp <!--\; \; -->(y)}{y+\exp <!--\; \; -->(y)}$
I thought I could use separation of variable, but I'm unable to isolate x.
Could you please help me get started on this differential equation?

In studying a reflection-transmission problem involving exotic materials, I have come across the following linear first-order differential equation:
$\begin{array}{}\text{(1)}& A\frac{\mathrm{\partial <!--\; \partial \; -->}}{\mathrm{\partial <!--\; \partial \; -->}t}g(t)+Bg(t)=f(t),\end{array}$
where A and B are constants, g(t) is associated with the reflected wave, and f(t) is a (finite) driving function associated with the incident wave. Both A and B may be positive or negative. I am interested in the behavior of the solution in the limit that $A\to <!--\; \backslash rightarrow\; -->0$
In studying a reflection-transmission problem involving exotic materials, I have come across the following linear first-order differential equation:A∂∂tg(t)+Bg(t)=f(t),(1)where A and B are constants, g(t) is associated with the reflected wave, and f(t) is a (finite) driving function associated with the incident wave. Both A and B may be positive or negative. I am interested in the behavior of the solution in the limit that A\rightarrow0.
I know there is an exact solution to Eq. (1), which is
$g(t)=C{e}^{-<!--\; -\; -->Bt/A}+\frac{1}{A}{\int <!--\; \int \; -->}_{-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->}}^t{e}^{-<!--\; -\; -->B(t-<!--\; -\; -->t^{\prime })/A}f(t^{\prime })d{t}^{\prime },$
where C=0 because g(t)=0 if f(t)=0. However, I do not understand how this exact solution reduces to the case where A=0, which is $g(t)=B^{-<!--\; -\; -->1}f(t)$. Any insight would be greatly appreciated.
I've seen a lot of documents discussing asymptotic analyses of linear differential equations, but they all start with second-order equations. Is this because there is inherently problematic with first-order?

I am trying to solve the following differential equation using the Laplace Transform:
$L\frac{di}{dt}+Ri=E$
I have managed to reduce it to the following equation:
$I(s)=\frac{E}{sL(s+R/L)}$
But I'm having a problem with it from there. Can somebody help me solve this please? Any help would be much appreciated.

Recently I have met such an equation:
$$\begin{gathered} x\frac{dy}{dx}=y+\sqrt{x^2+y^2} \\ \frac{dy}{dx}-<!--\; -\; -->\frac{y}{x}=\frac{\sqrt{x^2+y^2}}{x} \end{gathered}$$
First of all what type it has? I can not refer it to any I am aware of(I am a newbie so it is likely that I'm missing something). Trying to solve it gave no result therefore I googled and youtube revealed that I had to use change of variables $y=v(x)x$. However it looks like a cheating. Why use this particular replacement and is there any way to solve it without it?

The system described by $x^{\prime }=2x^2-<!--\; -\; -->8$ is linearized about the equilibrium point -2. What is the resulting linearized equation?
Answer is $x^{\prime }=-<!--\; -\; -->8x-<!--\; -\; -->16$.
How? I have no idea how it went from the first equation to the 2nd. Thanks.

I have a question on the particular case of the Cauchy theorem for a first order differential equation with separable variables.
$y^{\prime }(x)=a(x)b(y(x))$
If I impose:
$a(x)$ continuous in a interval $I$
$b(y(x))$ continuous and with continuous derivative in a interval $J$
Can I say that there is only one solution in all the interval I that satisfies the condition of passage through a point $(x_0,y_0)\in <!--\; \in \; -->I\times <!--\; \times \; -->J$?
Or, instead of the continuity of the derivative of $b(y(x))$, should I impose that the derivative of $b(y(x))$ is limited on $J$?
Can anyone suggest me the correct conditions to impose in this particular case?
Thanks in adivce

I have to solve following system of equations:
$x^{″}=-<!--\; -\; -->x-<!--\; -\; -->z+e^{-<!--\; -\; -->t},z^{\prime }=-<!--\; -\; -->2x-<!--\; -\; -->2z+3e^{-<!--\; -\; -->t}$
I would like to have formula for x' in order to find fundamental matrix and so on. So from above quations I obtain $z^{\prime }-<!--\; -\; -->2x^{″}=e^{-<!--\; -\; -->t}$, integrate sides and have $2x^{\prime }=z+e^{-<!--\; -\; -->t}$
So I have system of two first-order differential equation
$2x^{\prime }=z+e^{-<!--\; -\; -->t},z^{\prime }=-<!--\; -\; -->2x-<!--\; -\; -->2z+3e^{-<!--\; -\; -->t}$
Is it correct?

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!