Question

Solve the following Linear differential equation:

`cos x  ("d"y)/("d"x) + y sin x ` = 1

Answer 1

The given differential equation can be written as

`cosx/cosx ("d"y)/("d"x) + y sinx/cosx = 1/cosx`

`("d"y)/("d"x) + (sinx/cosx)y` = sec x

`("d"y)/("d"x) + (tan x)y` = sec x

This is the form `("d"y)/("d"x) + "p"y` = Q

Where P = tan x

Q = sec x

Thus, the given differential equation is linear.

I.F = `"e"^(int "pd"x)`

= `"e"^(int tan x  "d"x)`

= `"e"^(log (sec x)`

= sec x

So, the required solution is given by

[y × I.F] = `int ["Q" xx "I.F"]  "d"x + "c"`

y × sec x = `int sec x xx sec x  "d"x + "c"`

y sec x = `int sec^2x  "d"x + "c"`

y sec x = tan x + c

`÷ sec x, ysecx/secx = tanx/secx + "c"/secx`

y = `sinx/cosx xx 1/secx + "c"/secx`

y = `sinx/cosx xx cos x + "c" cos x`

= sin x + c cos x

y = sin x + c cos x is the required solution.