Question

lfxy< l.tm1· 1 x + tan·y = _____

Answer 1

Answer:

Given differential equation is

dx

dy

+

1+x

tany

=(1+x)e

x

secy

⇒

secy

1

dx

dy

+

(1+x)secy

tany

=(1+x)e

x

⇒cosy

dx

dy

+

1+x

siny

=(1+x)e

x

Substitute siny=v⇒cosydy=dv

∴

dx

dv

+

1+x

v

=(1+x)e

x

...(1)

Here, P=

1+x

1

⇒∫Pdx=∫

1+x

1

dx=log(1+x)

∴I.F.=e

log(1+x)

=1+x

Multiplying 1 by I.F., we get (1+x)

dx

dv

+v=(1+x)

2

.e

x

Integrating both sides

(1+x)v=∫(1+x)

2

.e

x

dx+C

⇒siny=(1+x)(e

x

+C)