Question

Could someone please solve this?

Answer 1

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

Answer 2

Answer:

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