Show the complete solution choices are on the second image. SOLUTIONa.y? = e-xC y?b.x2 =y= (2x – 3y)^3 (x + y)^2С.9 y + 2C =ln (sec(tan-1 G) +5) ·d.У — 11-1(tan- Csc3(у — 1)s = [C(x – C,)jе.f.C = x²y² + 4x – 6y13g.C =(x² + y²)7 – xy32 In |sin |+Ch.S =Cosi.C = s²t – tet + etj.y2 = -x?e-*(sin x + cos x ) + Cx?2y? + бу — 1c=V16 – x? - 8sin-) -k.42y + 3cos 2y +sin 2y41-cos y|I.x =y21m.= -x2sin 4y + (C – 4y)sin?(2y)x3y =+ (C, – 1)x + (2C, – 1) In |x – 1| + C2n. (y3 – 3y + 2)/9 + x²dx + x?dy = 0

Answered: Image /qna-images/answer/f879817c-3263-478b-9472-4d6ae1c28db0.jpg

Answered: (y³ – 3y + 2)/9 + x²dx + x²dy = 0

Math

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(y³ – 3y + 2)/9 + x²dx + x²dy = 0

Intermediate Algebra

10th Edition

ISBN:9781285195728

Author:Jerome E. Kaufmann, Karen L. Schwitters

Publisher:Jerome E. Kaufmann, Karen L. Schwitters

Chapter8: Conic Sections

Section8.2: More Parabolas And Some Circles

Problem 63.2PS: By expanding (xh)2+(yk)2=r2, we obtain x22hx+h22ky+k2r2=0. When we compare this result to the form...

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Question

Show the complete solution choices are on the second image.

SOLUTION
a.
y? = e-x
C y?
b.
x2 =
y
= (2x – 3y)^3 (x + y)^2
С.
9 y + 2
C =ln (sec(tan-1 G) +5) ·
d.
У — 1
1
-1
(tan
- Csc
3(у — 1)
s = [C(x – C,)j
е.
f.
C = x²y² + 4x – 6y
1
3
g.
C =
(x² + y²)7 – xy
3
2 In |sin |+C
h.
S =
Cos
i.
C = s²t – tet + et
j.
y2 = -x?e-*(sin x + cos x ) + Cx?
2y? + бу — 1
c=V16 – x? - 8sin-) -
k.
4
2y + 3
cos 2y +
sin 2y
4
1-cos y
|
I.
x =
y2
1
m.
= -
x2
sin 4y + (C – 4y)sin?(2y)
x3
y =+ (C, – 1)x + (2C, – 1) In |x – 1| + C2
n.

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