Calculus 2 Equations: Expert Solutions and More

Solve the differential equation:
${\displaystyle \frac{d}{dx}\left(2y{y}^{\prime }\right)={\left(y^{\prime }\right)}^2}$

Linearization around the equilibrium point for a system of differential equations
${\displaystyle \frac{dN}{dt}=\mu N(1-\frac{N}{K})+N{\int }_0^{\mathrm{\infty }}p(a,t)da}$
${\displaystyle \frac{\partial p}{\partial t}+\frac{\partial p}{\partial a}=kp(a,t).}$
Linearizing it around ${\displaystyle (N=K,p=0)}$, I am getting the system
${\displaystyle \frac{dN}{dt}=h-\mu N+N{\int }_0^{\mathrm{\infty }}p(a,t)da}$
${\displaystyle \frac{\partial p}{\partial t}+\frac{\partial p}{\partial a}=kp(a,t),}$
where ${\displaystyle h=\mu K}$.
The reason I am getting this system is because I am considering ${\displaystyle N{\int }_0^{\mathrm{\infty }}p(a,t)da}$ as linear term. Is this linearisation right or the correct system is ${\displaystyle \frac{dN}{dt}=h-\mu N}$
${\displaystyle \frac{\partial p}{\partial t}+\frac{\partial p}{\partial a}=kp(a,t),}$
where ${\displaystyle h=\mu K}$.

Function whose minimum is independent of the input?
I have a question that asks me to show that the functional
${\displaystyle I\left(x\right)={\int }_{\left\{x_1\right\}}^{x_2}(x^2+3y^2)y^{\prime }+2xydx}$
has an extremal that is independent of the choice of y that joins two arbitrary points ${\displaystyle (x_1,y_1),(x_2,y_2)}$.
So I tried solving for the extremal, which give me the euler lagrange equation:
${\displaystyle 6y{y}^{\prime }+2x=2x+6y{y}^{\prime }}$
That's not an error it seems that ${\displaystyle \frac{d}{dx}\left(\frac{\partial F}{\partial y^{\prime }}\right)=\frac{\partial F}{\partial y}}$.
Ok then no wonder that the path doesn't matter. But now I am trying to compute the value of the functional for the two points, I tried a straight line between the 2, but this is giving me a messy expression, it's computable but ugly:
${\displaystyle J\left(y\right)={\int }_{\left\{x_1\right\}}^{x_2}s(x^2+3{(sx+k)}^2)+2x(sx+k)dx}$
(where s and k are the linear coefficients of the line connecting the two points).

Solve the equation:
${\displaystyle x(2y+1)dx=y(x^2-3x+2)dy}$

Solve the differential equation: ${\displaystyle \frac{d^{2}y}{{dx}^2}-8\frac{dy}{dx}=-16y}$? Discuss what kind of differential equation is this, and when it may arise?

Let u be a solution of the differential equation ${\displaystyle y^{\prime }+xy=0}$ and ${\displaystyle \varphi =u\psi }$ be a solution of the differential equation ${\displaystyle y^{″}+2x{y}^{\prime }+(x^2+2)y=0}$ such that ${\displaystyle \varphi \left(0\right)=1,{\varphi }^{\prime }\left(0\right)=0}$. Then ${\displaystyle \varphi \left(x\right)}$ is equal to
A) ${\displaystyle \left({\cos }^2\left(x\right)\right)e^{\frac{-x^2}{2}}}$
B) ${\displaystyle \left(\cos \left(x\right)\right)e^{\frac{-x^2}{2}}}$
C) ${\displaystyle (1+x^2)e^{\frac{-x^2}{2}}}$
D) ${\displaystyle \left(\cos \left(x\right)\right)e^{-x^2}}$

Let ${\displaystyle f\in C_o^{\mathrm{\infty }}\left({\mathbb{R}}^n\right)}$. Propose formulas of the form ${\displaystyle u\left(x\right)=displaysty\le {\int }_{\{}^{\{}\frac{\hat{f}\left(\delta \right)e^{ix\delta }}{p\left(\delta \right)}d\delta }$
to solve:
i) ${\displaystyle \sum _{j=1}^n\frac{{\partial }^{4}u\left(x\right)}{\partial x_j^4}+u\left(x\right)=f\left(x\right)}$
andii) ${\displaystyle \sum _{k,l}^{\{}\frac{{\partial }^{4}u\left(x\right)}{\partial x_k^2\partial x_l^2}-2\cdot displaysty\le \sum _{k=1}^n\frac{{\partial }^{2}u\left(x\right)}{\partial x_k^2}+u\left(x\right)=f\left(x\right)}$

Solve the ode using fixed-point iteration
${\displaystyle -{\left[(1+u^4)u_x\right]}_x=\sin \left(x\right)+\sin \left(5x\right)}$, where the domain is ${\displaystyle [0,2\pi ]u\left(0\right)=u\left(2\pi \right)=0}$ for boundaries.

Solve the following differential equation by the form of homogeneous equation. Letting ${\displaystyle y=vx}$
The equation: ${\displaystyle x^2\frac{dy}{dx}+xy+1=0}$

Solve ${\displaystyle xydx=(y^3+x^{2}y+x^2)dy}$ dy differential equation.

Can a point be in the solution curve of a differential equation when domain of that differential equation does not contain it?
I am given: ${\displaystyle \frac{dy}{dx}=\frac{2xy}{x^2-y^2}}$ with initial condition ${\displaystyle y\left(1\right)=1}$
(1,1) does not belong to the domain of this O.D.E: ${\displaystyle \frac{dy}{dx}=\frac{2xy}{x^2-y^2}}$
I have a very little knowledge of Differential equation. I can not understand how (1,1) is belonging to the solution curve when domain of O.D,E does not contain it.
Can anyone please help me ?

Calculate sum of an infinite series
${\displaystyle \sum _{n=1}^{\mathrm{\infty }}{(-1)}^{n-1}{n}^2{x}^n}$

Behaviour of the solutions of a system of differential equations at infinity.
Consider the following system of differential equations:
$$\{\begin{array}{l}{u}^{\prime }=-u+uv\\ v^{\prime }=-2v-u^2\end{array}$$
I'm able to prove that solutions must tend to 0 if ${\displaystyle t\to 0}$ by the use of Lyapunov function ${\displaystyle L(x,y)=x^2+y^2}$ but I'm unable to prove that the function:
${\displaystyle I\left(t\right)=\frac{x^2\left(t\right)+2y^2\left(t\right)}{x^2\left(t\right)+y^2\left(t\right)}}$
must admit limit for ${\displaystyle t\to +\mathrm{\infty }}$

Absolute signs in solving ODEs
${\displaystyle \frac{1}{y}\frac{\text{d}y}{\text{d}x}=\frac{1}{x}}$
The way I solve this:
for all x in ${\displaystyle (-\mathrm{\infty },0)}$
${\displaystyle \ln \left|y\right|=\ln \left|x\right|+A}$ where A is a real constant (since we know antiderivatives are separated by a constant)
${\displaystyle \left|y\right|=B\left|x\right|}$ where B is a positive constant equals exp(A)
${\displaystyle y=Bx}$ or ${\displaystyle -Bx}$
It seems to me ${\displaystyle y=Cx}$ where C is a real non-zero constant that equals either B or -B.
Since it seems like if y is equal to ${\displaystyle +\text{or}-Bx}$ for the entire interval ${\displaystyle (-\mathrm{\infty },0)}$, then it must be only ${\displaystyle +Bx\text{or}-Bx}$ i.e. Cx for some subinterval. If it is ${\displaystyle +Bx}$, then for values outside and near this interval, they must also be ${\displaystyle +Bx}$ since we know y is continuous and x is not zero. Vice versa.
But I don't know how to prove this. How to prove ${\displaystyle y=Cx}$ for ${\displaystyle (-\mathrm{\infty },0)}$?
If that is the case, then I can write the general form of the differential equation picewisely as ${\displaystyle y=C_1\text{for}(-\mathrm{\infty },0),C_{2}x\text{for}(0,\mathrm{\infty })}$, where ${\displaystyle C_1\text{and}{C}_2}$ are non-zero reals.

Changing to polar coordinates bring the differential equation to ${\displaystyle \mathrm{\Phi }(\varphi ,\rho ,{\rho }^{\prime }\left(\varphi \right))=0}$ form.
I have a differential equation ${\displaystyle y^{\prime }=d\frac{x+y}{x-y}}$ and the problem says to change to polar coordinate system by assuming ${\displaystyle x=\rho \cos \varphi ,y=\rho \sin \varphi \text{and}\rho =\rho \left(\varphi \right)}$ and bring the equation to ${\displaystyle \mathrm{\Phi }(\varphi ,\rho ,{\rho }^{\prime }\left(\varphi \right))=0}$ form. The answer is ${\displaystyle {\rho }^{\prime }\left(\varphi \right)=\rho \left(\varphi \right)}$. This problem is under "Implicit Functions" in the book, so I suppose I should use the Implicit-function theorem. But I don't see what I can get from that. I did substitute and tried to simplify, but in vain. The only way I see to solve this after substitution is by applying trigonometry, but every time either ${\displaystyle \rho \left(\varphi \right)\text{or}{\rho }^{\prime }\left(\varphi \right)}$ are multiplied by some other trigonometric function (that can't be simplified to a constant) and it seems that just trigonometry is not enough. Could you explain how such problems are solved? I couldn't find examples on the internet either.

Can't solve the differential equation
${\displaystyle \left(\frac{2x}{y^3}\right)dx+\frac{y^2-3x^3}{y^4}dy=0}$

Arriving at a particular solution of the ODE
${\displaystyle y{}^{″}-2y^{\prime }-2y=\sin x}$