Question

For the differential equation, find the general solution:

`cos^2 x dy/dx + y = tan x(0 <= x < pi/2)`

Answer 1

The given equation is

`cos^2 x dy/dx + y = tan x`

⇒ `dy/dx + (sec^2 x) y = tan x sec^2 x`

Which is a linear equation of the type

`dy/dx + Py = Q`

Here P = sec² x and Q =  tan sec² x

∴ `I.F. = e^(intsec^2 x  dx) = e^(tan x)`

∴ The solution is `y. (I.F.) = int Q. (I.F.) dx + C`

⇒ `y.e^(tan x) = int tan x sec^2 x e^(tan x)  dx + C = I + C`  ...(1)

Now, `I = int tan x sec^2 xe^(tan x)  dx`

Put tan x = t

⇒ sec²  x dx = dt

∴ `I = int t. e^t  dt = t. e^t - int (1) e^t  dt`        ....[Integrating by parts]

`= te^t - e^t = e^t (t - 1)`

`= e^(tan x) (tan x - 1)`

∴ From (1) `y.e^(tan x) = e^(tan x) (tan x - 1) + C`

⇒ ` y = (tan x - 1) + Ce^(-tan x),` Which is the required solution.