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How to solve homogenous differential equation ${(x+y)}^2 dx=xy dy$?
First i find value of $\frac{dy}{dx}=\frac{{(x+y)}^2}{x}y \left(i\right)$
let ${\displaystyle y=vx}$
differentiating both sides: $\frac{dy}{dx}=v+x d\frac{v}{dx} \left(ii\right)$
I tried to solve using equation (i) and (ii) but I am stuck.
${(x+y)}^2 dx=xy dy$

Degree 3; zeros: 1,1-i

Suppose that $G\left(x\right)={\log }_3(2x+1)-2.$
(a) What is the domain of $G$?
(b) What is ${\displaystyle G\left(40\right)}$? What point is on the graph of $G$?
(c) If ${\displaystyle G\left(x\right)=3}$, what is x? What point is on the graph of $G$?
(d) What is the zero of $G$?

Let ${\displaystyle \alpha ,\beta }$ be the roots of ${\displaystyle a{x}^2+bx+c=0}$, where ${\displaystyle 1<\alpha <\beta }$
Then $\underset{x\to x_0}{\lim }\frac{|a{x}^2+bx+c|}{a{x}^2+bx+c}=1$ then which of the following statements is incorrect
a) ${\displaystyle a>0}$ and ${\displaystyle x_0<1}$
b) ${\displaystyle a>0}$ and ${\displaystyle x_0<\beta }$
c) ${\displaystyle a<0}$ and ${\displaystyle \alpha <x_0<\beta }$
d) ${\displaystyle a<0}$ and ${\displaystyle x_0<1}$

There was a sample of

300

milligrams of a radioactive substance to start a study. Since then, the sample has decayed by

2.3%

each year.

Let

t

be the number of years since the start of the study. Let

y

be the mass of the sample in milligrams.

Write an exponential function showing the relationship between

y

and

t

.

What is the correct notation for non simultaneous (asynchronous) equations?
most common practice is when we type simultaneous equations in form like that:
$\{\begin{array}{l}x+y=2\\ x-y=2\end{array}$
But, what if have equation like
${\displaystyle x^4-5x^2+6=0}$
and i want to denote it into 2 equations:
${\displaystyle x^2=2}$
and ${\displaystyle x^2=3}$

How do I get an estimate for this nonlocal ODE?
Consider the following nonlocal ODE on ${\displaystyle [1,\mathrm{\infty })}$:
${\displaystyle r^{2}f{ }^{″}\left(r\right)+2r{f}^{\prime }\left(r\right)-l(l+1)f\left(r\right)=-\frac{(f^{\prime }\left(1\right)+f\left(1\right))}{r^2}}$
${\displaystyle f\left(1\right)=\alpha }$
${\displaystyle \underset{r\to \mathrm{\infty }}{lim}f\left(r\right)=0}$
where l is a positive integer and ${\displaystyle \alpha }$ is a real number.
Define the following norm $\Vert ·\Vert$
$\Vert f{\Vert }^2:={\int }_1^{\infty }{r}^{2}f\text{'}(r{)}^{2}dr+l(l+1\left){\int }_1^{\infty }f\right(r{)}^{2}dr$
I want to prove the estimate:
$\Vert f\Vert \le C\sqrt{l(l+1)}\left|\alpha \right|$
for some constant C independent of ${\displaystyle \alpha }$, l and f. But I am stuck.
Here is what I tried. Multiply both sides by f and integrate by parts to get:
$$\begin{array}{rl}\Vert f{\Vert }^2={\int }_1^{\mathrm{\infty }}{r}^2{f}^2+{\int }_1^{\mathrm{\infty }}l(l+1)f^2& =f^{\prime }(1)f(1)+(f^{\prime }(1)+f(1)){\int }_1^{\mathrm{\infty }}\frac{f}{r^2}\\ & \le f^{\prime }(1)\alpha +\frac{(f^{\prime }(1)+\alpha )}{3}\sqrt{{\int }_1^{\mathrm{\infty }}{f}^2}\\ & \le f^{\prime }(1)\alpha +\frac{(f^{\prime }(1)+\alpha )}{3}\Vert f\Vert \end{array}$$
where I used Cauchy-Schwartz in the before last line. I am not sure how to continue and how to get rid of the f'(1) term.
Any help is appreciated.