Learn First Order Problems with These Differential Equations Examples

I have a system of first order defferential equation that has the coefficient matrix in the picture. I calculated the eigenvalues Eigenvalues: +i,-i
$Q=\left[\begin{array}{cccc}0& -<!--\; -\; -->1& 0& 0\\ 1& 0& 0& 0\\ 0& 0& 0& -<!--\; -\; -->1\\ 2& 0& 1& 0\end{array}\right]$
How should I handle this system? Can i find a exponential matrix for this system in order to solve it and how?

Please tell how one can identify 1st order
1) Homogeneous differential equations.
2) Homogeneous linear differential equations.
3) Non Homogeneous differential equations.
4) Non Homogeneous linear differential equations.

I am taking an Engineering Math class. When it comes of differential equations with y2 included in the question I am lost on the procedure used to solve it. Bernoulli method does work because its not in form required and homogeneous equations doesn't work because of different degrees.
These are a few that had me stumped (Solving any one will be appreciated.)
1. $y^{\prime }=(y^2+9)/(xy+3y)$
2. $y\cdot <!--\; \cdot \; -->\ln <!--\; \; -->(x)y^{\prime }=(y^2+36)/x$
3. $y^{\prime }=(x^2{y}^4-<!--\; -\; -->x^2)/y^3$
Thanks!

How can I start to solve this differential equation?
$y^{\prime }=\frac{y}{2y\ln <!--\; \; -->(y)+y-<!--\; -\; -->x}$

Is a first order differential equation categorized by $f(y^{\prime },y,x)=0$ or $y^{\prime }=f(y,x)$?
In the case of the second, why $\sin <!--\; \; -->y^{\prime }+3y+x+5=0$ isnt a first order differential equation?

I have been having a problem with this simple equation. It is asking me this: Find all values of $k$ for which the function $y=\sin <!--\; \; -->(kt)$ satisfies the differential equation $y\prime \prime +7y=0$.
I have found the second derivative, plugged it back into the differential equation, and found out $\sqrt{7}$ checked, but $-<!--\; -\; -->\sqrt{7}$ did not. Please help me solve this question.

I am looking to solve the following equations numerically:
$ax=\frac{d}{dt}(f(x,y,t)\frac{dy}{dt}),by=\frac{d}{dt}(g(x,y,t)\frac{dx}{dt})$
For arbitrary functions f and g and constants a and b. I am struggling to find a way to transform this into a system of first order differential equations that I can pass into a solver. It looks like I will need to define these implicitly, but I'm not sure how to do that.
My best attempt so far is the following:
$\begin{array}{rl}{z}_1& =f(x,y,t)\frac{dy}{dt}\\ z_2& =g(x,y,t)\frac{dx}{dt}\\ z_3& =ax\\ z_4& =by\end{array}$
${\left(\begin{array}{c}{z}_1\\ z_2\\ z_3\\ z_4\end{array}\right)}^{\prime }=\left(\begin{array}{cccc}0& 0& 1& 0\\ 0& 0& 0& 1\\ 0& \frac{a}{g(t,x,y)}& 0& 0\\ \frac{b}{f(t,x,y)}& 0& 0& 0\end{array}\right)\left(\begin{array}{c}{z}_1\\ z_2\\ z_3\\ z_4\end{array}\right)$
However, this seems fairly inelegant and assumes that you are always able to divide by f and g. I'm trying to keep this as general as possible, so don't want to make that assumption. Is there a better way to turn this into a system of first order differential equations implicitly?

I am having trouble isolating the x and y to separate side in the differential equations below. Could someone give me a hint as to how to to this.
Equation 1:
$\frac{dy}{dx}-<!--\; -\; -->\frac{x}{y}=\frac{1}{x}$
Equation 2:
$xy\frac{dy}{dx}=y^2$

How can I write the following equation as a first-order vector differential equation?
$\begin{array}{r}m\frac{{\mathrm{d}}^{2}x}{{\mathrm{d}}^{2}t}+2\mathrm{\gamma <!--\; \gamma \; -->}m\frac{\mathrm{d}x}{\mathrm{d}t}+kx=0.\end{array}$

If I can write out the characteristic equation for such circles, then I can find a desired first-order differential equation. The problem is how to write the characteristic equation?

My question is the following:
How should I factorize a differential equation from the first order but with higher degrees?
Imagine I have:
$x^2(y^{\prime }{)}^2+xy{y}^{\prime }-<!--\; -\; -->6y^2=0$
Is it okay if I replace $y^{\prime }$ with p and just solve the equation for $p$ with $x$ and $y$ being constants?
Thanks in advance!

I would need some help in solving the following differential equation:
$u^{\prime }(p)+\frac{r+N\mathrm{\lambda <!--\; \lambda \; -->}p}{N\mathrm{\lambda <!--\; \lambda \; -->}p(1-<!--\; -\; -->p)}u(p)=\frac{r+N\mathrm{\lambda <!--\; \lambda \; -->}p}{N\mathrm{\lambda <!--\; \lambda \; -->}(1-<!--\; -\; -->p)}g.$
I already know that the solution is of the form
$gp+C(1-<!--\; -\; -->p){\left(\frac{1-<!--\; -\; -->p}{p}\right)}^{r/\mathrm{\lambda <!--\; \lambda \; -->}N},$,
where C is a constant.
Now this is a standard linear first-order differential equation. Letting
$f(p)=\frac{r+N\mathrm{\lambda <!--\; \lambda \; -->}p}{N\mathrm{\lambda <!--\; \lambda \; -->}p(1-<!--\; -\; -->p)},q(p)=\frac{r+N\mathrm{\lambda <!--\; \lambda \; -->}p}{N\mathrm{\lambda <!--\; \lambda \; -->}(1-<!--\; -\; -->p)}g,\mathrm{\mu <!--\; \mu \; -->}(p)=e^{\int <!--\; \int \; -->f(p)dp},$
the solution should have the form
$u(p)=\frac{C}{\mathrm{\mu <!--\; \mu \; -->}(p)}+\frac{1}{\mathrm{\mu <!--\; \mu \; -->}(p)}\int <!--\; \int \; -->\mathrm{\mu <!--\; \mu \; -->}(p)q(p)dp,$
where C is a constant.
The calculation of $\mathrm{\mu <!--\; \mu \; -->}(p)$ was not much of a problem. Indeed, we can rewrite f(p) as
$f(p)=\frac{r}{N\mathrm{\lambda <!--\; \lambda \; -->}}\frac{1}{p}+(1+\frac{r}{N\mathrm{\lambda <!--\; \lambda \; -->}})\frac{1}{1-<!--\; -\; -->p}$
Using $\int <!--\; \int \; -->\frac{1}{p}dp=\ln <!--\; \; -->(p)$ and $\int <!--\; \int \; -->\frac{1}{1-<!--\; -\; -->p}dp=-<!--\; -\; -->\ln <!--\; \; -->(1-<!--\; -\; -->p)$, I found that
$\mathrm{\mu <!--\; \mu \; -->}(p)=p^{\frac{r}{N\mathrm{\lambda <!--\; \lambda \; -->}}}(1-<!--\; -\; -->p{)}^{-<!--\; -\; -->1-<!--\; -\; -->\frac{r}{N\mathrm{\lambda <!--\; \lambda \; -->}}}.$
Where I have difficulties is calculating the term ∫μ(p)q(p)dp, where
$\mathrm{\mu <!--\; \mu \; -->}(p)q(p)=\frac{rg}{N\mathrm{\lambda <!--\; \lambda \; -->}}{p}^{\frac{r}{N\mathrm{\lambda <!--\; \lambda \; -->}}}(1-<!--\; -\; -->p{)}^{-<!--\; -\; -->2-<!--\; -\; -->\frac{r}{N\mathrm{\lambda <!--\; \lambda \; -->}}}+p^{\frac{r}{N\mathrm{\lambda <!--\; \lambda \; -->}}+1}(1-<!--\; -\; -->p{)}^{-<!--\; -\; -->2-<!--\; -\; -->\frac{r}{N\mathrm{\lambda <!--\; \lambda \; -->}}}.$
How do I integrate this last term? Or did I miss something obvious, because in the paper I'm looking at they don't provide any details about how they get to the solution of the differential equation
Edit: small typo corrected in the last displayed equation.
Edit: there is a big typo in the differential equation: the rhs should have $r+N\mathrm{\lambda <!--\; \lambda \; -->}$ rather than $r+N\mathrm{\lambda <!--\; \lambda \; -->}p$ in the numerator. This probably makes it much simpler to solve. My bad.

I was trying to compute the solution for the following differential equation:
$x(2x^{2}ylog(y)+1)y^{\prime }=2y$
As I couldn't get anywhere I checked the hints in the textbook which are the following:
Reverse the way of thinking, namely view $x$ as a function and $y$ as a variable, considering that
$y=\frac{dy}{dx}=\frac{1}{\frac{dx}{dy}}=>y^{\prime }=\frac{1}{x^{\prime }}$. Then it goes to say that the equation now becomes
$x^{\prime }-<!--\; -\; -->\frac{x}{2y}=log(y)x^3$
This final equation is obviously simple enough to solve, but how on Earth did they arrive there?

If I have a differential equation $y^{\prime }(t)=Ay(t)$ where A is a constant square matrix that is not diagonalizable(although it is surely possible to calculate the eigenvalues) and no initial condition is given. And now I am interested in the fundamental matrix. Is there a general method to determine this matrix? I do not want to use the exponential function and the Jordan normal form, as this is quite exhausting. Maybe there is also an ansatz possible as it is for the special case, where this differential equation is equivalent to an n-th order ode. I saw a method where they calculated the eigenvalues of the matrix and depending on the multiplicity n of this eigenvalue they used an exponential term(with the eigenvalue) and in each component an n-th order polynomial as a possible ansatz. Though they only did this, when they were interested in a initial value problem, so with an initial condition and not for a general solution.
I was asked to deliver an example: so $y^{\prime }(t)=\left(\begin{array}{cc}3& -<!--\; -\; -->4\\ 1& -<!--\; -\; -->1\end{array}\right)y(t)$ If somebody can construct a fundamental matrix for this system, than this should be sufficient

I completed near all problems om a differential equations text chapter on reducing non-separable first order differential equations to separable by using an appropriate substitution for example u=y/x with y'=u+u'x and similar substations in y and x for making other similar problems separable.
I am not asking anyone to do the entire problem for me but I do need a little guidance to begin tackling this problem. I completely understand the procedure for making them separable. What I need help in is setting up the differential equation for the following word problem. Once this is done I can easily finish the problem.
Ch 1.4-29. Show that a straight line through the origin intersects all solution curves of a given differential equation y=g(y/x) at the same angle.

I'm try to solve this differential equation:
$y^{\prime }=x-<!--\; -\; -->1+xy-<!--\; -\; -->y$
After rearranging it I can see that is a linear differential equation:
$y^{\prime }+(1-<!--\; -\; -->x)y=x-<!--\; -\; -->1$
So the integrating factor is $l(x)=e^{\int <!--\; \int \; -->(1-<!--\; -\; -->x)dx}=e^{(1-<!--\; -\; -->x)x}$
That leaves me with an integral that I can't solve... I tried to solve it in Wolfram but the result is nothing I ever done before in the classes so I'm wondering if I made some mistake...
This is the integral:
$y{e}^{(1-<!--\; -\; -->x)x}=\int <!--\; \int \; -->(x-<!--\; -\; -->1)e^{(1-<!--\; -\; -->x)x}dx$

Trying to solve this seemingly simple first order non-linear differential equation:
$y^{\prime }+y^2=\cos <!--\; \; -->2x$
Considered separation of variables and bernoulli methods but figured it's not applicable. Please I need a hint.

How to solve this differential equation:
$x\frac{dy}{dx}=y+x\frac{e^x}{e^y}?$
I tried to rearrange the equation to the form $f\left(\frac{y}{x}\right)$ but I couldn't thus I couldn't use $v=\frac{y}{x}$ to solve it.