Find the antiderivative for...
∫〖1/x〗dx
ln |x| + C
Find the general solution for...
dy/dx = (y+5)(x+2)
y = C1e(x^2)/2 + 2x - 5
Find the particular solution for...
y = C1e(x^2)/2 + 2x - 5
initial condition: f(0)=1
y = 5e(x^2)/2 + 2x - 5
Slide 2
Step 1: She did not separate the variables correctly.
Find the antiderivative for...
∫〖secxtanx〗dx
secx + C
Find the general solution for...
dy/dx=y2(1+x2)
y= (-1)/(x+(x3/3)+C)
Find the particular solution for...
y= 〖-1〗/〖(1/6)x2 + C〗+ 2
initial condition: f(1)=0
y= 2-〖6/x2+2〗
Slide 4
Step 2: She forgot to add +C
Find the antiderivative for...
∫〖-1/|x|√ (x2-1)〗dx
cot-1x + C
Find the general solution for...
dy/dx = 4x3y - 6x2y
Y = (+/-) C1e(x^4)-(2x^3)
Find the particular solution for...
y= (-3)/(3x+x3+C)
initial condition: f(3)=2
y= (-3)/(3x+x3-(75/2))
Slide 6
Step 3: A cube root does not include +/-
Find the antiderivavtive for...
∫〖cotx〗dx
ln|sinx| + C
Find the general solution for...
dy/dx= (3 - y)cosx
y=C1e-sinx + 3
Find the particular solution for...
y=C1e-sinx + 3
initial condition: f(0)=1
y= 3 - 2e-sinx
Slide 8
He is correct
Find the antiderivative for...
∫〖bx〗
bx/lnb + C
(2013 AP Exam Q) Find the general solution for...
dy/dx=ey(3x2 - 6x)
y= -ln(-x3+3x2+C)
(2010 AP Exam Q) Find the particular solution for...
y= (1)/(√-x2 + C)
initial condition: f(1)=2
y= (1)/(√-x2 + 5/4)
Slide 10
She is correct