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Calculus AB
Section 5/8 – 6/3
Review Worksheet answers
Find the general solution
1.
dy
= 6x2 -8x +3
dx
Since the right side is all in
terms of x,
y =  6 x 2  8 x  3  dx
y  ln x  C
Separating variables by
cross-multiplication,
3y2∙dy = 16x∙dx
y  ln x  ln C
y  ln Cx
y = (ln|Cx|)2
Taking the anti-derivative
of both sides,
2
 3 y  dy   16 x  dx
y3 = 8x2 + C
7.
y =
6  13  x 3  8  12  x 2  3  x1  C
y = 2x3 - 4x2 + 3x + C
dy
2.
= cos(2x)
dx
Since the right side is all in
terms of x,
y =  cos( 2 x)  dx
u = 2x
du = 2dx
½ du = dx
y =  cos(u )  12 du
y =
1
2
 cos(u )  du
y = ½ sin(u) + C
y = ½ sin(2x) + C
dy
2
=
dx
x
Since the right side is all in
terms of x,
2
y =   dx
x
1
y = 2    dx
x
y = 2∙ln|x| + C
y = ln|x2| + C
y = ln|x2| + ln|C|
y = ln|C∙x2|
3.
4.
16 x
dy
=
dx
3y 2
5.
dy
y
=
dx
sec x
Cross-multiplying,
sec x  dy  y  dx
Dividing to separate the
variables,
1
1
 dy 
 dx
y
sec
1
 dy  cos x  dx
y
Taking the anti-derivative
of both sides,
1
 y  dy   cos  dx
ln y = sin x + C
y = e sin x C = C  e sin x
6.
dy 2 y
=
dx
x
Cross-multiplying,
x  dy  2 y  dx
Dividing to separate the
variables,
1
1
 dy   dx
x
2 y
Taking the anti-derivative
of both sides,
1
1
 2 y  dy   x  dx
dy
x
=
dx
y
Separating variables by
cross-multiplication,
y∙dy = -x∙dx
Taking the anti-derivative
of both sides,
 y  dy    x  dx
½ y2 = -½ x2 + C
y2 = -x2 + C
x2 + y2 = C
8.
dy
= 2x  1 y2
dx
Separating variables by
dividing,
1
 dy  2 x  dx
1 y2
Taking the anti-derivative
of both sides,
1
 1  y 2  dy   2 x  dx
arcsin(y) = x2 + C
y = sin(x2+C)
Identify a solution of the
differential equation.
9. Differential equation:
y’ = y”∙x + 1
Possible solutions:
A. y = x3
B. y = 2x
C. y = x2 + x
Calculus AB
Section 5/8 – 6/3
Review Worksheet answers
3
A. y = x
y’ = 3x2
y” = 6x
y’ = y”∙x + 1
3x2 ? 6x∙x + 1
3x2 ? 6x2 + 1
-3x2 ? 1
x2 ? -1/3
Never true
B. y = 2x
y’ = 2
y” = 0
y’ = y”∙x + 1
2 ? 0∙x + 1
2 ? 1
Never true
2
C. y = x + x
y’ = 2x + 1
y” = 2
y’ = y”∙x + 1
2x+1 ? 2∙x + 1
2x+1 ? 2x + 1
Always true
C is the only solution of the
three.
10. Differential equation:
y + y” = 0
Possible solutions:
A. y = sinx
B. y = e-x
C. y = x4
A. y = sinx
y’ = cosx
y” = -sinx
y + y” = 0
sinx + -sinx ? 0
0
? 0
Always true
B. y = e-x
y’ = - e-x
y” = e-x
y + y” = 0
e-x + e-x ? 0
2e-x ? 0
e-x ? 0
Never true
C. y = x4
y’ = 4x3
y” = 12x2
y + y” = 0
x4 + 12x2 ? 0
x2(x2 + 12) ? 0
Only sometimes true (when
x = 0)
A is the only solution of the
three.
11. Differential equation:
y” = 4∙y
Possible solutions:
A. y = 2x2
B. y = e-2x
C. y = sinh(2x)
A. y = 2x2
y’ = 4x
y” = 4
y” = 4∙y
4 ? 4∙2x2
Only sometimes true (when
x =  22 ).
B. y = e-2x
y’ = -2e-2x
y” = 4e-2x
y” = 4∙y
4e-2x ? 4 ∙ e-x
Always true.
C. y = sinh(2x)
y’ = 2cosh(2x)
y” = 4sinh(2x)
y’ = 4∙y
4sinh(2x) ? 4 ∙ sinh(2x)
Always true.
B and C are both solutions.
12. Differential equation:
y = y(4)
Possible solutions:
A. y = sin x
B. y = e-x
C. y = ln(x)
A. y = sin x
y’ = cosx
y” = -sinx
y’’’ = -cos x
y(4) = sin x
y = y(4)
sin x ? sin x
Always true
B. y = e-x
y’ = -e-x
y” = e-x
y’’’ = -e-x
y(4) = e-x
y = y(4)
-x
e ? e-x
Always true
C. y = ln(x)
y’ = 1x = x-1
y” = -x-2
y’’’ = 2x-3
y(4) = -6x-4
y = y(4)
ln(x) ? -6x-4
Never true.
A and B are both solutions.
Calculus AB
Section 5/8 – 6/3
Review Worksheet answers
Find the derivative
13. y = sinh(3x)
dy
= cosh(3x)∙3
dx
= 3cosh(3x)
cosh x  1
(cosh x  1)2
1
=
cosh x  1
22. I =
Find the anti-derivative.
14. y =
x2∙sinhx
18. I =  (2 x  sec h x)  dx
2
dy
= x2∙coshx + sinhx∙2x
dx
= x2∙coshx + 2x∙sinhx
15. y = ln(cosh x)
dy
1
d (cosh x)

=
dx
cosh x
dx
1
 sinh x
=
cosh x
sinh x
=
cosh x
= tanh x
= x2 – tanhx + C
19. I =  6 cosh(2 x)  dx
u = 2x
du = ½ dx
 6cosh(u)  dx
 3 cosh x  dx
1
2
u = sinh x
du = coshx∙dx
sinh x
17. y = cosh x1
dy
=
dx

1
5
=

sinh x
 dx
cosh x
u = cosh x
du = sinhx dx
1
2
I   u  du
 2 u C
5
C
cosh x
 sinh x  1  dx
1
 sinh x  1  cosh x  dx
u = sinhx + 1
du = coshx∙dx
I =
1
= -cosh   + C
x
I  2u 2  C
 sinh x 
21. I =
= -cosh(u) + C
= -cosh  x 1  + C
1
 du
 15 u 5  C
 cosh x  1  cosh x  sinh x  sinh x
2
 cosh x  1
cosh 2 x  cosh x  sinh 2 x
=
(cosh x  1)2
cosh x  cosh 2 x  sinh 2 x
=
(cosh x  1)2
4
=   sinh( u)  du
1
4
u
I =  sinh( u )( du )
   cosh x  2  sinh x  dx
20. I =   sinh x   cosh x  dx
dy
d (tanh x)
= etanh x ∙
dx
dx
d (tanh x)
= etanh x ∙
dx
tanh x
=e
∙ sech2x
u = x-1
du = -1∙x-2∙dx
-du = x-2∙dx
23. I =
 3sinh x  C
16. y = etanh x

sinh  1x 
 dx
x2
=  sinh( x 1 )  x 2  dx
=
1
 u  du
= ln | u | + C
= ln(sinh x + 1) + C
 2 cosh x  C