Question

Does anyone know how to do this

Answer 1

$\dfrac{\mathrm dy}{\mathrm dx}=(y+2)^2\sin\left(\dfrac xe\right)$

a. If $y=k$, then $\frac{\mathrm dy}{\mathrm dx}=0$, so

$0=(k+2)^2\sin\left(\dfrac xe\right)\implies (k+2)^2=0\implies k=-2$

b. When $x=w$, we're told $y$ has a horizontal tangent, which has slope $\frac{\mathrm dy}{\mathrm dx}=0$. So we have

$0=(y+2)^2\sin\left(\dfrac we\right)\implies\sin\left(\dfrac we\right)=0\implies\dfrac we=2n\pi\implies w=2ne\pi$

where $n$ is any integer, whose smallest positive value occurs for $n=1$, giving $w=2e\pi$.

c. The equation is separable:

$\dfrac{\mathrm dy}{(y+2)^2}=\sin\left(\dfrac xe\right)\,\mathrm dx$

Integrate both sides to get

$-\dfrac1{y+2}=-e\cos\left(\dfrac xe\right)+C$

$y=-1$ when $x=0$, so we find

$-1=-e+C\implies C=e-1$

Then the particular solution to the DE is

$-\dfrac1{y+2}=-e\cos\left(\dfrac xe\right)+e-1\implies y=\dfrac1{e\left(\cos\left(\frac xe\right)-1\right)+1}-2$