ps1 – solutions – 2014
TRANSCRIPT
PS1 – Solutions – 2013 1a) !
!!!𝑑𝑥 sub u = x –a !
!𝑑𝑢 = 𝑙𝑛 𝑢 + 𝐶! = 𝑙𝑛 𝑥 − 𝑎 + 𝐶!
du = dx 1b) 𝑥𝑒!!! 𝑑𝑥 Integration by parts 𝑢𝑣 − 𝑣𝑑𝑢 u = x and dv = e-‐2xdx du = dx v = !!
!!!
!
!!!𝑥𝑒!!! − !!
!𝑒!!!𝑑𝑥 !!
!𝑥𝑒!!!-‐!
!𝑒!!!+C1 !!
!𝑒!!!(2𝑥 + 1)+C1
1c) sin𝑤𝑥 𝑑𝑥 = -‐!!cos𝑤𝑥 + 𝐶!
1d) !
!!!!!𝑑𝑥 = !
!!!𝑑𝑥 + !
!!!𝑑𝑥 A = !
!! and B = !!
!!
12𝑎
𝐴𝑥 − 𝑎 𝑑𝑥 −
𝐵𝑥 + 𝑎 𝑑𝑥
= !!!𝑙𝑛 𝑥 − 𝑎 − 𝑙𝑛 𝑥 + 𝑎 + 𝐶! =
!!!
!" !!!!" !!!
+ 𝐶! 1e) !
!!!!!!!𝑑𝑥 = !
(!!!)(!!!)𝑑𝑥 = !
!!!𝑑𝑥 + !
!!!𝑑𝑥 A = -‐1; B = 1
!!!!!
𝑑𝑥 + !!!!
𝑑𝑥 = !" !!!!" !!!
+ 𝐶!
2a) !!"𝑒!!! = -‐2𝑒!!!
2b) !
!"!
!!! = !!
(!!!)!
2c) !
!"(𝑡 𝑥 )! = !
!"(𝑡)! !"
!" = 3𝑡! !"
!" = 3𝑡! !
! !cos 𝑥 − 𝑒!! =
3(sin 𝑥 + 𝑒!!)! !! !
cos 𝑥 − 𝑒!! 2d) !
!"!(!)!(!)
Quotient Rule = !!!!!!!!!
2e) !
!"!(!)
(! ! )! (same as above) = !
!!!!!!!!!!!
= !!!!!!!!!!
2f) !
!"𝑞 𝜆 𝑑𝜆!(!)
!(!) Step 1: Split up !
!"𝑞 𝜆 𝑑𝜆!
!(!) + !!"
𝑞 𝜆 𝑑𝜆!(!)!
Step 2: Recall the 2nd Fundamental Theorem of Calculus (FToC) that says:
!!"
𝑞 𝜆 𝑑𝜆 = 𝑞(𝑥)!!
This is similar to what we have above in step 1: Step 3: Get Step 1 in the form of Step 2 to use 2nd FToC Take the first term in Step 1, and let h = 𝑞 𝜆 𝑑𝜆!
!(!) so the first term is !!!"
Step 4: We know that !!!"= !!
!"!"!" we need to find !!
!"
!!"
𝑞 𝜆 𝑑𝜆 = −!!(!)
!!"
𝑞 𝜆 𝑑𝜆 = −𝑞(𝑔 𝑥 )!(!)! if we take a = 0
so !!
!"= −𝑞 𝑔 𝑥 !"(!)
!" for the first term
Step 5: We can do the same for the second term in Step 1, this gives us a final answer: !!"
𝑞 𝜆 𝑑𝜆!(!)!(!) = 𝑞 𝑓 𝑥 !"(!)
!"− 𝑞 𝑔 𝑥 !"(!)
!"
3. a) Derive the differential equation describing the system. Step 1: We care about [S], so determine the rate of reaction that you want to use in the stuff equation wrt [S]: 2S + B D so, rD = ![!]!"
= 𝑘[𝑆]! = !!!![!]!" ![!]
!" !"#= 2𝑘[𝑆]!
From now on, we can leave out the brackets on [S] and just call it S. Step 2: Use the stuff equation to model the reactor (the CV):
The stuff we are concerned with is the amount of S in the CV. There is no S produced in the CV, just consumed by reaction. We can also assume that the reactor is well-‐mixed so the [S] leaving the CV is the same as the [S] in the CV. The stuff equation for this CV is as follows: 𝑉 !!!
!"= 𝑄!𝑆! − 𝑄!𝑆! − 2𝑘𝑉𝑆!! where S1 is the concentration of S coming in and St is
the concentration of S at any time in the CV. Re-‐arranging and substituting we can isolate for !!!
!":
𝑑𝑆!𝑑𝑡 =
𝑄!𝑆!𝑉 −
(𝑄! + 𝑄!)𝑆!𝑉 − 2𝑘𝑆!!
b) Solve for SSS at SS, !!!
!"= 0 so:
0 = !!!!!− (!!!!!)!!
!− 2𝑘𝑆!! and rearrange for St2 to get a quadratic:
𝑆!! +(𝑄! + 𝑄!)𝑆!
2𝑘𝑉 −𝑄!𝑆!2𝑘𝑉 = 0
Input values and solve the quadratic using the quadratic equation: St,1,SS = 0.122 mol/L and St,2,SS = -‐0.324 mol/L One is an extraneous root because a concentration can’t be negative, so: St,SS =0.122 mol/L
4. Equation ODE /
PDE Order Linear /
Nonlinear
a) ( )2
22
d yx sin(x) ydx
= ⋅ ODE 2 L
b) x xx yy 'xe y ''' e ye
− −−+ = − ODE 3 NL
c) PDE 2 NL
d) ( )2
22
d y d dycos(y)sin(y) 2sin (y)dx dx dx
− = ODE 2 L
e)
( )3 2
2 2 xy2 2
U U U(x y ) e Usin(x y)x x y y x∂ ∂ ∂ ∂
+ + + = +∂ ∂ ∂ ∂ ∂
PDE 3 L
Explanations: 4a) !
!!!!!
𝑦𝑥! = sin 𝑥 𝑦 y is the function expand the first term: 𝑑𝑑𝑥
𝑑𝑦𝑑𝑥 𝑥
! +𝑑(𝑥!)𝑑𝑥 𝑦 = sin 𝑥 𝑦
carry through the outer derivative: !!!!!!
𝑥! + 4 !"!"𝑥 + 2𝑦 = sin 𝑥 𝑦 This is linear wrt y
4b) make ex-‐y = ex/ey so that the e-‐x in the first two terms cancels out leaving you with: 𝑥𝑒!𝑦!!! + 𝑦𝑒! = −𝑦! First and second terms are non linear wrt y 4c) The function is u, so the u2 term in the second term makes it non-‐linear wrt u 4d) !
!!!!!
− !!"(cos𝑦 sin𝑦) = 2 sin! 𝑦 !"
!" sub in for cos(y) sin(y) = 1/2sin(2y)
𝑦!! −12𝑑𝑑𝑥 (sin 2𝑦) = 2 sin
! 𝑦𝑦′ 𝑦!! − cos 2𝑦𝑦′ = 2 sin! 𝑦𝑦′ let 2sin2y = 1-‐cos2y 𝑦!! − cos 2𝑦𝑦′ = 1− cos 2𝑦 𝑦′ cancel out (-‐cos2yy’) and you get y’’ – y’ = 0 Which is linear wrt y
4e) ( )3 2
2 2 xy2 2
U U U(x y ) e Usin(x y)x x y y x∂ ∂ ∂ ∂
+ + + = +∂ ∂ ∂ ∂ ∂
all terms are LINEAR wrt U