Question

28. Find the particular solution of the differential equation
dy /dx =1+y^2/1+x^2 given that y(1) = 1?​

Answer 1

Answer:

(x+1)

dx

dy

=2e

−y

−1⟶eqn.1

dx

dy

+

1+x

1

=

1+x

2e

−y

Eqn.1 can be written as:

2e

−y

−1

1

dy=

1+x

1

dx

⇒

e

y

2

−1

1

dy=

1+x

1

dx

⇒

2−e

y

−(−e

y

)

dy=

1+x

1

dx

⇒(−log∣2−e

y

∣)=log∣x+1∣+C

y=0,x=0

⇒−log2=log1+C

⇒C=−log2

Answer: log(x+1)−log2=−log(2−e

−y

)

⇒log(

2

(x+1)(2−e

−y

)

)=0