Could someone please solve this?
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Answered by
4
Answer:
(1+x)
dx
dy
−xy=1−x
dx
dy
+(
1+x
−x
)y=
1+x
1−x
∫pdx=∫
1+x
−x
=−∫
1+x
1+x−1
dx
=−∫1−
1+x
1
dx
=−x+log∣1+x∣dx
I.Fe
∫pdx
=e
log(1+x)−x
=
e
x
1+x
4.5y.
e
x
1+x
=∫
e
x
1−x
y
e
x
(1+x)
=∫e
−x
(1−x)
y(1+x)=e
x
c
−x
(+x)
y(1+x)=x
Answered by
2
Answer:
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