How to solve this linear pde for y ( x ) (other functions are known and &#x03BB;<

Find the linear approximation of the function $f(x)=\sqrt{4-x}$ at $a=0$
Use L(x) to approximate the numbers $\sqrt{3.9}$ and $\sqrt{3.99}$ Round to four decimal places

a) Simplify the expression ${\displaystyle 4x+6x}$ by combining the like terms.
b) Simplify the expression ${\displaystyle y-5y}$ by combining the like terms.
c) Simolify the expression ${\displaystyle 3x^2+5x^2-2}$ by combining the like terms.

What is $\sqrt{5}$ minus $\sqrt{5}$?

Solve absolute value inequality :
${\displaystyle |3-\frac{3}{4}x|>9}$

wo identitical pendula each of length $l$ and with bobs of mass $m$ are free to oscillate in the same plane. The bobs are joined by a spring with spring constant $k$, by looking for solutions where $x$ and $y$ vary harmonically at the same angular frequency $\omega$, form a simultaneous equation for the amplitudes of oscillation $x_0$ and $y_0$.
Considering the forces acting on each pendulum we can derive the following coupled-differential equations:
$\begin{array}{}\text{(1)}& m\ddot{x}& =k(y-x)-\frac{mgx}{\ell }\text{(2)}& m\ddot{y}& =-k(y-x)-\frac{mgy}{\ell }\end{array}$
Where $x$ and $y$ are the displacements of each of the pendulum as functions of time. If we assume they oscillate harmonically with angular frequency $\omega$ then we can write $\mathrm{\exists }\omega ,{\varphi }_1,{\varphi }_2\in \mathbb{R}$:
$\begin{array}{rl}x(t)& =x_0\cos (\omega t+{\varphi }_1)\\ y(t)& =y_0\cos (\omega t+{\varphi }_2)\end{array}$
Substituting these solutions back into $(1)$ and $(2)$ we get:
$\begin{array}{rl}-m{\omega }^2{x}_0\cos (\omega t+{\varphi }_1)& =k(x_0\cos (\omega t+{\varphi }_1)-y_0\cos (\omega t+{\varphi }_2))-\frac{mg}{\ell }\cos (\omega t+{\varphi }_1)\\ -m{\omega }^2{y}_0\cos (\omega t+{\varphi }_2)& =-k(x_0\cos (\omega t+{\varphi }_1)-y_0\cos (\omega t+{\varphi }_2))-\frac{mg}{\ell }\cos (\omega t+{\varphi }_2)\end{array}$
However, without assuming that ${\varphi }_1={\varphi }_2$, in which case everything factors out nicely to leave a simultaneous equation in $x_0$ and $y_0$, I cannot see a way of making it linear in $x_0$ and $y_0$. So am I expected to use this assumption or is there a mathematical way of simplifying it?
If it is the former, then what would the physical justification for this assumption be?

Consider the following simultaneous equations in $x$ and $y$:
$x+y+axy=a$
$x-2y-x{y}^2=0$
where $a$ is a real constant. Show that these equations admit real solutions in $x$ and $y$.

${\displaystyle x(k+2)+y(k-1)+z\left(k\right)=2}$
${\displaystyle y(k+2)+2x=0}$
${\displaystyle z(k^2+k-2)=k+2}$
Is there any value of k for which this system of linear equations would have infinite solutions? I mean, it seems as if it does when k = -2, but it also seems as if it has no solutions when k = -2. Im