Cracking The Ap Calculus ab Exam 2018

(Marvins-Underground-K-12) #1
Therefore,  y^4     =       −   236.

Now we  isolate y.  We  get the equation    y   =   .


  1. y =


We  solve   this    differential    equation    by  separation  of  variables.  We  want    to  get all of  the y

variables   on  one side    of  the equals  sign    and all of  the x   variables   on  the other   side.   We  can do

this    easily  by  dividing    both    sides   by  y   and multiplying both    sides   by  dx. We  get     =   5x^2 dx.

Next,   we  integrate   both    sides.

∫ = ∫^5 x


(^2) dx
ln y = + C 0
Now we isolate y: . We can rewrite this as y = .
Note that we are using the letter C in the last equation. This is to distinguish it from the C 0 in
the first equation. Now we solve for C. We plug in x = 0 and y = 6: 6 = Ce^0 = C. Therefore, the
equation is y = .



  1. y =


We  solve   this    differential    equation    by  separation  of  variables.  We  want    to  get all of  the y
variables on one side of the equals sign and all of the x variables on the other side. We can do
this easily by cross-multiplying. We get y^2 dy = ex dx. Next, we integrate both sides.

∫^ y


(^2) dy =


∫^ e


x dx

    =   ex  +   C
Free download pdf