Antiderivatives
General Solutions
Particular Solutions
Where did they go wrong?
100

Find the antiderivative for...

∫〖1/x〗dx

ln |x| + C

100

Find the general solution for...

dy/dx = (y+5)(x+2)

y = C1e(x^2)/2 + 2x - 5

100

Find the particular solution for...

y = C1e(x^2)/2 + 2x - 5

initial condition: f(0)=1

y = 5e(x^2)/2 + 2x - 5

100

Slide 2

Step 1: She did not separate the variables correctly.

200

Find the antiderivative for... 

∫〖secxtanx〗dx

secx + C

200

Find the general solution for...

dy/dx=y2(1+x2)

y= (-1)/(x+(x3/3)+C)

200

Find the particular solution for... 

y= 〖-1〗/〖(1/6)x+ C〗+ 2

initial condition: f(1)=0

y= 2-〖6/x2+2〗

200

Slide 4

Step 2: She forgot to add +C

300

Find the antiderivative for... 

∫〖-1/|x|√ (x2-1)〗dx

cot-1x + C

300

Find the general solution for...

dy/dx = 4x3y - 6x2y

Y = (+/-) C1e(x^4)-(2x^3)

300

Find the particular solution for... 

y= (-3)/(3x+x3+C)

initial condition: f(3)=2

y= (-3)/(3x+x3-(75/2))

300

Slide 6

Step 3: A cube root does not include +/-

400

Find the antiderivavtive for... 

∫〖cotx〗dx

ln|sinx| + C

400

Find the general solution for...

dy/dx= (3 - y)cosx

y=C1e-sinx + 3

400

Find the particular solution for... 

y=C1e-sinx + 3

initial condition: f(0)=1

y= 3 - 2e-sinx

400

Slide 8

He is correct

500

Find the antiderivative for... 

∫〖bx

bx/lnb + C

500

(2013 AP Exam Q) Find the general solution for...

dy/dx=ey(3x- 6x) 

y= -ln(-x3+3x2+C)

500

(2010 AP Exam Q) Find the particular solution for...

y= (1)/(√-x2 + C)

initial condition: f(1)=2 

y= (1)/(√-x2 + 5/4)

500

Slide 10 

She is correct

M
e
n
u