Question

d^2y/dx^2 + dy/dx +y = cos^2x

Answer 1

Answer:

Given y=cos(sinx)

cos

−1

y=sinx

Differentiating wrt x both sides we get ,

1−y

2

−1

dx

dy

=cosx

dx

dy

=−

1−y

2

cosx

Differentiating wrt x both sides we get ,

dx

2

d

2

y

=

1−y

2

sinx+cosx

1−y

2

y

dx

dy

dx

2

d

2

y

=sinx

1−y

2

−cosx

1−y

2

y

1−y

2

cosx

dx

2

d

2

y

=

1−y

2

sinx−ycos

2

x

L.H.S =

dx

2

d

2

y

+tanx

dx

dy

+ycos

2

x=

1−y

2

sinx−ycos

2

x+

cosx

sinx

(−

1−y

2

cosx)+ycos

2

x

dx

2

d

2

y

+tanx

dx

dy

+ycos

2

x=0

= R.H.S