xy" + (3sin^2x)y' = 5cosxy
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2
Answer:
ANSWER
3
dx
d
(sin(xy))+4
dx
d
(cos(xy))=
dx
d
(5)
⇒3cos(xy)
dx
d
(xy)−4sin(xy)
dx
d
(xy)=0
⇒(3cos(xy)−4sin(xy))
dx
d
(xy)=0
⇒(3cos(xy)−4sin(xy))(x
dx
dy
+y)=0
⇒3cos(xy)−4sin(xy)=0 and x
dx
dy
+y=0
⇒3cos(xy)=4sin(xy) and x
dx
dy
=−y
⇒tanxy=
4
3
and
dx
dy
=
x
−y
∴
dx
dy
=
x
−y
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Answered by
0
Step-by-step explanation:
3
dx
d
(sin(xy))+4
dx
d
(cos(xy))=
dx
d
(5)
⇒3cos(xy)
dx
d
(xy)−4sin(xy)
dx
d
(xy)=0
⇒(3cos(xy)−4sin(xy))
dx
d
(xy)=0
⇒(3cos(xy)−4sin(xy))(x
dx
dy
+y)=0
⇒3cos(xy)−4sin(xy)=0 and x
dx
dy
+y=0
⇒3cos(xy)=4sin(xy) and x
dx
dy
=−y
⇒tanxy=
4
3
and
dx
dy
=
x
−y
∴
dx
dy
=
x
−y
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