Explain this words in form of trigonometry graphs

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ANSWER: Midline: is the horizontal line that passes exactly in the middle between the graphs maximum and minimum points.

Amplitude: is the vertical distance between the midline and one of the extremum points.

Period: is the distance between two consecutive maximum points or two consecutive minimum points (these distances must be equal). HOPE THIS HELPS!

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(X^2+y^2+x)dx+xydy=0 Solve for general solution

aksik [14]

Check if the equation is exact, which happens for ODEs of the form

$M(x,y)\,\mathrm dx+N(x,y)\,\mathrm dy=0$

if $\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$ .

We have

$M(x,y)=x^2+y^2+x\implies\dfrac{\partial M}{\partial y}=2y$

$N(x,y)=xy\implies\dfrac{\partial N}{\partial x}=y$

so the ODE is not quite exact, but we can find an integrating factor $\mu(x,y)$ so that

$\mu(x,y)M(x,y)\,\mathrm dx+\mu(x,y)N(x,y)\,\mathrm dy=0$

is exact, which would require

$\dfrac{\partial(\mu M)}{\partial y}=\dfrac{\partial(\mu N)}{\partial x}\implies \dfrac{\partial\mu}{\partial y}M+\mu\dfrac{\partial M}{\partial y}=\dfrac{\partial\mu}{\partial x}N+\mu\dfrac{\partial N}{\partial x}$

$\implies\mu\left(\dfrac{\partial N}{\partial x}-\dfrac{\partial M}{\partial y}\right)=M\dfrac{\partial\mu}{\partial y}-N\dfrac{\partial\mu}{\partial x}$

Notice that

$\dfrac{\partial N}{\partial x}-\dfrac{\partial M}{\partial y}=y-2y=-y$

is independent of x, and dividing this by $N(x,y)=xy$ gives an expression independent of y. If we assume $\mu=\mu(x)$ is a function of x alone, then $\frac{\partial\mu}{\partial y}=0$ , and the partial differential equation above gives