Question

the particular solution of the differential equation is D + 1

Answer 1

Answer:

y=(2x

2

+1)

−1

dx

dy

=−4xy

2

∴

y

2

dy

=−4xdx

∴∫y

−2

dy=−4∫xdx+c

∴

−1

y

−1

=−4(

2

x

2

)+c

∴

y

−1

=−2x

2

+c

Take x=0 and y=1

∴−1=−2(0)+c

∴c=−1

∴

y

−1

=−2x

2

−1

∴

y

1

=1+2x

2

∴y=(1+2x

2

)

−1