Question

Does anyone know how to do this

Answer 1

`(\mathrm dy)/(\mathrm dx)=(y+2)^2\sin\left(\frac xe\right)`

a. If
`y=k`, then
`(\mathrm dy)/(\mathrm dx)=0`, so

`0=(k+2)^2\sin\left(\frac 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
`(\mathrm dy)/(\mathrm dx)=0`. So we have

`0=(y+2)^2\sin\left(\frac we\right)\implies\sin\left(\frac we\right)=0\implies\frac 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:

`(\mathrm dy)/((y+2)^2)=\sin\left(\frac xe\right)\,\mathrm dx`

Integrate both sides to get

`-\frac1{y+2}=-e\cos\left(\frac 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

`-\frac1{y+2}=-e\cos\left(\frac xe\right)+e-1\implies y=\frac1{e\left(\cos\left(\frac xe\right)-1\right)+1}-2`