Answer:
(y^2+3y-28)/(4y^2-15y-4)
Step-by-step explanation:
(4y-1)(y-3)(2y+3)(y+3)(y+7)(y-3)/(2y+3)(y-4)(4y+1)(4y-1)(y-3)(y+3)
= (y+7)(y-3)/(y-4)(4y+1)
= (y^2+3y-28)/(4y^2-15y-4)
1 is in the 'Ten Thousand's' place.
The first equation is linear:

Divide through by

to get

and notice that the left hand side can be consolidated as a derivative of a product. After doing so, you can integrate both sides and solve for

.
![\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac1xy\right]=\sin x](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dx%7D%5Cleft%5B%5Cdfrac1xy%5Cright%5D%3D%5Csin%20x)


- - -
The second equation is also linear:

Multiply both sides by

to get

and recall that

, so we can write



- - -
Yet another linear ODE:

Divide through by

, giving


![\dfrac{\mathrm d}{\mathrm dx}[\sec x\,y]=\sec^2x](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dx%7D%5B%5Csec%20x%5C%2Cy%5D%3D%5Csec%5E2x)



- - -
In case the steps where we multiply or divide through by a certain factor weren't clear enough, those steps follow from the procedure for finding an integrating factor. We start with the linear equation

then rewrite it as

The integrating factor is a function

such that

which requires that

This is a separable ODE, so solving for

we have



and so on.
Answer:
The first choice.
Step-by-step explanation:
hope this helps!