Learn First Order Problems with These Differential Equations Examples

The theorem of existence and uniqueness is: Let $y^{\prime }+p(x)y=g(x)$ be a first order linear differential equation such that p(x) and g(x) are both continuous for $a<x<b$. Then there is a unique solution that satisfies it.
When a differential equation has no solution that satisfies $y(x_0)=y_0$, what does this mean?? Can the theorem be verified??

$(x^3+x^2+x+1)(dy/dx{)}^2-<!--\; -\; -->(3x^2+2x+1)y(dy/dx)+2x{y}^2=0$

I was solving differential equation
$x\cos <!--\; \; -->x\frac{dy}{dx}+y(x\sin <!--\; \; -->x+\cos <!--\; \; -->x)=1$
which on dividing by $x\cos <!--\; \; -->x$ becomes FOLD(first order linear differential) equation.
But I am stuck at following integral. Can anyone help solve this integral? An alternate approach to the problem is also welcome.
$\int <!--\; \int \; -->\frac{e^{\cos <!--\; \; -->x}}{\cos <!--\; \; -->x}dx$

I am new to the differential equation and I need some ideas how to solve this problem.
$x^2{y}^{\prime }=x^2{y}^2-<!--\; -\; -->2$
$y^{\prime }=y^2-<!--\; -\; -->{\displaystyle \frac{2}{x^2}}$

I am currently doing a project about Growth and have found this really interesting Math Model by Dr. Geoffrey West et al in 2001 while researching.
I was wondering how does one integrate the differential equation
$\frac{dm}{dt}=a{m}^{0.75}[1-<!--\; -\; -->{\left(\frac{m}{M}\right)}^{0.25}]$
where $m$ is a function in terms of $t$ and $a$, $M$ are constants.
The end result (the integrated expression):
${\left(\frac{m}{M}\right)}^{0.25}=1-<!--\; -\; -->[1-<!--\; -\; -->{\left(\frac{m_0}{M}\right)}^{0.25}]e^{-<!--\; -\; -->at/(4M^{0.75})}$
where $m_0$ is another constant (in the context of the problem, the initial mass of organism when $t=0$).
I have tried solving by the integrating factors method but I couldn't express it in the form given... Anyone has any ideas?

I have the specific first order non-linear differential equation as shown below:
$\frac{d\mathrm{\Omega <!--\; \Omega \; -->}}{d\mathrm{\theta <!--\; \theta \; -->}}-<!--\; -\; -->M(\mathrm{\theta <!--\; \theta \; -->})\frac{1}{I\mathrm{\Omega <!--\; \Omega \; -->}}=\frac{D}{I}$
Where D and I are constants. And $M(\mathrm{\theta <!--\; \theta \; -->})=A\cdot <!--\; \cdot \; -->\sin <!--\; \; -->(2\mathrm{\theta <!--\; \theta \; -->})+B$, where A and B are constants. Could anyone advice me if this is solvable, and if so, what are the steps I should take?

If I have: $\stackrel{\dot{}<!--\; \dot{}\; -->}{\mathrm{\sigma <!--\; \sigma \; -->}}(t)=-<!--\; -\; -->\mathrm{\gamma <!--\; \gamma \; -->}\mathrm{\sigma <!--\; \sigma \; -->}(t)$
where $\mathrm{\gamma <!--\; \gamma \; -->}$ is a constant, the solution is given by:
$\mathrm{\sigma <!--\; \sigma \; -->}(t)=\mathrm{\sigma <!--\; \sigma \; -->}(0)e^{-<!--\; -\; -->\mathrm{\gamma <!--\; \gamma \; -->}t}$
Now what if I have the differential equation: $\stackrel{\dot{}<!--\; \dot{}\; -->}{\mathrm{\sigma <!--\; \sigma \; -->}}(t)=-<!--\; -\; -->\mathrm{\gamma <!--\; \gamma \; -->}x(t)\mathrm{\sigma <!--\; \sigma \; -->}(t)$
Then is the solution given by:
$\mathrm{\sigma <!--\; \sigma \; -->}(t)=\mathrm{\sigma <!--\; \sigma \; -->}(0)e^{-<!--\; -\; -->\mathrm{\gamma <!--\; \gamma \; -->}x(t)t}$
Or...?

Need help with this differential equation:
$\sqrt{3+y^2}dx-<!--\; -\; -->xdy=x^{2}dy$
I began with variable separable method. So, already had done this:
$\sqrt{3+y^2}dx=(x^2+x)dy$
$\frac{dx}{x^2+x}=\frac{dy}{\sqrt{3+y^2}}$
$\int <!--\; \int \; -->\frac{dx}{x^2+x}=\int <!--\; \int \; -->\frac{dy}{\sqrt{3+y^2}}$
Now I should solve that two integrals, but have some trouble with them and need your help. Also, I need help, explanation how to draw phase portrait for this equation, maybe some useful material about that?

I am given a single second order differential equation:
$\stackrel{¨<!--\; ¨\; -->}{x}-<!--\; -\; -->x^3-<!--\; -\; -->2x^2\stackrel{\dot{}<!--\; \dot{}\; -->}{x}+1=0$
.. and am asked to classify the critical points as stable, unstable or saddle points.
Finding the critical points is an easy task for first order differential equation(s), both single equation and system of equations. However, I have never done so for second order, and higher, equations.
I have an idea on how to solve it but not sure the approach is correct. Am I correct in saying I need to split the single differential equation into two differential equations and rename the relevant terms? Doing so on the above equation gives:
$\stackrel{\dot{}<!--\; \dot{}\; -->}{x_1}=x_2$
$\stackrel{\dot{}<!--\; \dot{}\; -->}{x_2}-<!--\; -\; -->x_1^3-<!--\; -\; -->2x_1^2{x}_2+1=0$
Thereafter, I follow the same process as starting off with a set of two first order differential equations. That is, set ${\stackrel{\dot{}<!--\; \dot{}\; -->}{x}}_1$ and ${\stackrel{\dot{}<!--\; \dot{}\; -->}{x}}_2$ to 0, and solve for the intersection of $x_1$ and $x_2$ to find the fixed points. The nature of the fixed points would then be determined by calculating the Trace and Determinant of the Jacobian at the specific fixed points.
Is my thinking on the right track?

I'm trying to solve the following differential equation
$\begin{array}{}& {\displaystyle y^{″}+(1+\frac{1}{y})y^{\prime 2}=0}\end{array}$
The book in which I found this question says that It can be solved by reducing it to a first order differential equation but I can't seem to find a way to do so.

How to find the particular the solution for this differential equation $x{y}^{\prime }(x)-<!--\; -\; -->y(x)(1+x^2)=\frac{x^2}{2}$ first i solved it as homogenous differential equation as follows $x{y}^{\prime }(x)-<!--\; -\; -->y(x)(1+x^2)=0$ thrn i got $y(x)=c{e}^{\frac{x^2}{2}}x$

Given the first-order differential equation:
$\frac{dy}{dx}=-<!--\; -\; -->6xy$
The textbook says you cannot differentiate both sides as y is on the right side and you have to use separation of variables. However, I did integrate both sides and arrived at this:
$\int <!--\; \int \; -->\frac{dy}{dx}dx=\int <!--\; \int \; -->-<!--\; -\; -->6xydx$
$y=-<!--\; -\; -->6y\int <!--\; \int \; -->xdx$
$y=-<!--\; -\; -->6y(\frac{x^2}{2}+C)$
$y=-<!--\; -\; -->3y{x}^2+C$
$y+3y{x}^2=C$
$y(1+3x^2)=C$
$y(x)=\frac{C}{1+x^2}$
And the second last step is valid since $1+3x^2$ can never be zero. However, this is not the correct answer, which is:
$y(x)=C{e}^{-<!--\; -\; -->3x^2}$. Why is this?

I was given the following second-order differential equation,
$y^{\mathrm{\prime <!--\; \text{'}\; -->}\mathrm{\prime <!--\; \text{'}\; -->}}+2y^{\mathrm{\prime <!--\; \text{'}\; -->}}+y=g(t),$
and that the solution is $y(t)=(1+t)(1+e^{-<!--\; -\; -->t})$. Using the solution I determined that
$g(t)=t+3.$
Following from this I transformed this second-order differential equation into a system of first-order differential equations, which is
$\left(\begin{array}{c}{x}_1\\ x_2\end{array}\right)=\left(\begin{array}{cc}0& 1\\ -<!--\; -\; -->1& -<!--\; -\; -->2\end{array}\right)\left(\begin{array}{c}y\\ y^{\mathrm{\prime <!--\; \text{'}\; -->}}\end{array}\right)+\left(\begin{array}{c}0\\ t+3\end{array}\right)$
Now I want to perform a single step with $\mathrm{\Delta <!--\; \Delta \; -->}t=1$ starting from t=0 with the Forward Euler method and after that with the Backward Euler method. Firstly with the Forward Euler method I use:
$w_{n+1}=w_n+\mathrm{\Delta <!--\; \Delta \; -->}tf(t_n,w_n)$
and I compute $w_0$ as
$w_0=\left(\begin{array}{c}y(0)\\ y^{\mathrm{\prime <!--\; \text{'}\; -->}}(0)\end{array}\right)=\left(\begin{array}{c}2\\ 1\end{array}\right)$
so therefore
$w_1=\left(\begin{array}{c}3\\ 0\end{array}\right)$
Now I want to perform the Backward Euler method.
$w_{n+1}=w_n+\mathrm{\Delta <!--\; \Delta \; -->}tf(t_{n+1},w_{n+1})$
so
$w_1=\left(\begin{array}{c}2\\ 1\end{array}\right)+\left(\begin{array}{cc}0& 1\\ -<!--\; -\; -->1& -<!--\; -\; -->2\end{array}\right)w_1+\left(\begin{array}{c}0\\ 4\end{array}\right)$
From which i get
$w_1=\frac{1}{4}\left(\begin{array}{c}11\\ 3\end{array}\right)$
y two results seems to be quite differnt and that gets me to believe that I have made a mistake somewhere. Could someone let me know if they believe this to be correct, or why this could be wrong?

I am studying numeric solutions of differential equations, and part of my reading is found in Simmonds' book, Differential Equations with Applications and Historical Notes. Although the chapter on numerical methods is written by John S. Robertson.
In the treatment of Euler's method, he states that the second derivative of the solution y is bounded by some constant M. What am I missing that makes it necessary that the solution of a first order differential equation even be twice differentiable?
For some context, the differential equations under consideration are those of the form y'=f(x,y), defined on some interval [a,b], with some initial value $y(a)=\mathrm{\alpha <!--\; \alpha \; -->}$
These are then transformed into the integral equation
$y(x_1)-<!--\; -\; -->y(x_0)={\int <!--\; \int \; -->}_{x_0}^{x_1}f(x,y)dx\text{.}$
Then, using Taylor's theorem and substituting back into that formula, the error is found to be
$\frac{h^2}{2}{y}^{″}(\mathrm{\xi <!--\; \xi \; -->})\text{,}$
which I understand to be true. For the remainder of the method, this quantity is neglected. But in the next section, it simply states that the quantity y''(x) is bounded on the entire interval, and so, by extension, is $y^{″}(\mathrm{\xi <!--\; \xi \; -->})$. I don't understand why this is necessarily true.

The difference between the definition offirst order DE and separable variable DE seem very near for me, so does separable variables DE are a special case of first order DE, if not is there some differential equation which obey both definitions, or they are totally opposite things?

Need some hints how to solve the following differential equation
$x{y}^{\prime }=\sqrt{y^2+x^2}+y$

I want to simulate the behaviour of a 2-DOF robotic manipulator, which is described by the following model:
$M(q)\stackrel{¨<!--\; ¨\; -->}{q}=-<!--\; -\; -->C(q,\stackrel{\dot{}<!--\; \dot{}\; -->}{q})\stackrel{\dot{}<!--\; \dot{}\; -->}{q}-<!--\; -\; -->G(q)+\mathrm{\tau <!--\; \tau \; -->}(1)$
Considering the fact that the 2x2 mass matrix M is positive definite, I could use the inverse matrix and break down the problem into the 4 first order ordinary differential equations and simulate it:
$x_1=q_1\Rightarrow <!--\; \Rightarrow \; -->\stackrel{\dot{}<!--\; \dot{}\; -->}{x_1}=x_2$
$x_2=\stackrel{\dot{}<!--\; \dot{}\; -->}{q_1}\Rightarrow <!--\; \Rightarrow \; -->\stackrel{\dot{}<!--\; \dot{}\; -->}{x_2}=-<!--\; -\; -->M^{-<!--\; -\; -->1}(1,:)\cdot <!--\; \cdot \; -->C\cdot <!--\; \cdot \; -->{\left[\begin{array}{cc}{x}_2& x_4\end{array}\right]}^T-<!--\; -\; -->M^{-<!--\; -\; -->1}(1,:)\cdot <!--\; \cdot \; -->G+M^{-<!--\; -\; -->1}(1,:)\cdot <!--\; \cdot \; -->\mathrm{\tau <!--\; \tau \; -->}$$x_3=q_2\Rightarrow <!--\; \Rightarrow \; -->\stackrel{\dot{}<!--\; \dot{}\; -->}{x_3}=x_4$
$x_4=\stackrel{\dot{}<!--\; \dot{}\; -->}{q_2}\Rightarrow <!--\; \Rightarrow \; -->\stackrel{\dot{}<!--\; \dot{}\; -->}{x_4}=-<!--\; -\; -->M^{-<!--\; -\; -->1}(2,:)\cdot <!--\; \cdot \; -->C\cdot <!--\; \cdot \; -->{\left[\begin{array}{cc}{x}_2& x_4\end{array}\right]}^T-<!--\; -\; -->M^{-<!--\; -\; -->1}(2,:)\cdot <!--\; \cdot \; -->G+M^{-<!--\; -\; -->1}(2,:)\cdot <!--\; \cdot \; -->\mathrm{\tau <!--\; \tau \; -->}$
Suppose I would like to use a solver that takes as an argument the mass matrix (a MATLAB ODE solver in particular) and don't use its inverse because this will also simplify the computation of the jacobian (I intend to simulate a 7-DOF robotic manipulator after that so providing the mass matrix would be great). How can I write the initial equation (1) as a series of first order ordinary differential equations and be able to simulate its response by using some software solvers ?

I'm supposed to find the solution of
$(xy-<!--\; -\; -->1)\frac{dy}{dx}+y^2=0$
The solution in the book manipulates it to
$\frac{dx}{dy}+\frac{x}{y}=\frac{1}{y^2}$
and proceeds to use an integrating factor to solve it to
$xy=\ln <!--\; \; -->y+c$
Which makes sense, but I did it by isolating the differential of y, substituting $xy=v$, and simplifying it to
$\frac{1-<!--\; -\; -->v}{v}dv=\frac{dx}{x}$
which gives, on solving,
$\ln <!--\; \; -->y+c=\ln <!--\; \; -->x^2+\frac{y}{x}$
Which isn't the same.
I can't see any flaw in what I did except for dividing variables throughout the equation (which I see everywhere in methods of solving differential equations, without a clue as to why.)
Can anyone help me with where I got it wrong?
EDIT: I'll add some detail on my steps;
$$\begin{gathered} xy=c \\ \frac{dy}{dx}=\frac{-<!--\; -\; -->y^2}{xy-<!--\; -\; -->1} \\ x\frac{dv}{dx}=\frac{v^2}{1-<!--\; -\; -->v}+v=\frac{v}{1-<!--\; -\; -->v} \\ \frac{dx}{x}=\frac{(1-<!--\; -\; -->v)dv}{v} \end{gathered}$$

Given a second order differential equation
$y^{″}+f(y)y^{\prime }+g(y)=0,$
write an equivalent system of first order equations with transformations
$x_1=y,x_2=y^{\prime }+{\int <!--\; \int \; -->}_0^{y}f(s)ds.$
This is what I did:
$x_1^{\prime }=y^{\prime }=x_2-<!--\; -\; -->{\int <!--\; \int \; -->}_0^{y}f(s)ds=x_2-<!--\; -\; -->{\int <!--\; \int \; -->}_0^{x_1}f(s)ds$
$x_2^{\prime }=y^{″}+f(y)=-<!--\; -\; -->f(y)y^{\prime }-<!--\; -\; -->g(y)+f(y)=f(y)(1-<!--\; -\; -->y^{\prime })-<!--\; -\; -->g(y)=f(x_1)(1-<!--\; -\; -->x_2+{\int <!--\; \int \; -->}_0^{x_1}f(s)ds)-<!--\; -\; -->g(x_1)$
I feel like this answer is wrong though, because I am not sure if I'm doing the standard procedure.

An ODE (Ordinary Differential Equation) of order n becomes a relation:
$F(x,y,y^{(1)},...,y^{(n)})=0$
Then $F(x,y,y^{(1)})=0$ defines an ODE of order one. In "basic standard texts", for purposes of simplicity, is assumed that some ODE of first order can take the form:
$y^{(1)}=f(x,y)$
for certain suitable f. Here is my "silly" question: What if that assumpion is not possible?
For example how I can deal with equations of the form:
${(y^{(1)})}^5+sen(y^{(1)})+e^{y^{(1)}}+x=0$
I appreciate any reference. Thanks in advance for your comments!