 Hello and welcome to the session I am Deepika here. Let's discuss a question which says find the general solution of the differential equation T by by Tx plus Ck xy is equal to tan x, 0 is less than equal to x, x is less than pi by 2. So let's start the solution. The given differential equation is D by by Tx plus Ck xy is equal to tan x, 0 is less than equal to x, x is less than pi by 2. Let us get this equation as number one. Now this is a first order linear differential equation with P is equal to Ck x and Q is equal to tan x. Therefore, the integrating factor is given by e raised to power integral of P ds and this is equal to e raised to power integral of Ck x dx and we know that integral of Ck x dx is equal to log of mod Ck x plus tan x. Now e raised to power log of mod Ck x plus tan x is equal to Ck x plus tan x. So this is the integrating factor and this is defined for 0 is less than equal to x, x is less than pi by 2. Now on multiplying equation one by integrating factor Vg over dx into Ck x plus tan x plus Ck x plus tan x y is equal to tan x into Ck x plus tan x. Left hand side is a differential of the function y into Ck x plus tan x. So this equation can be written as y into Ck x plus tan x Ck x plus tan x integrating both sides with respect to x. We have integral of d over dx of y into Ck x plus tan x integral of tan x Ck x plus tan square x. This equation can be written as y into Ck x plus tan square x integral of Ck square integral of 1 dx because tan square x can be written as Ck square x. This equation can be written as y into Ck x plus tan x. Now integral of x dx the general solution of the given differential equation is y into Ck x plus tan x is equal to Ck x plus tan x minus x plus Ck for the above question. I hope the solution is clear to you and you have enjoyed the session. Bye and take care.