 Hi, and welcome to the session I am Deepika here. Let's discuss the question which says point the general solution of the following differential equation. Cos square x d pi by d x plus y is equal to tan x, 0 is less than equal to x, x is less than pi by 2. Let's start the solution. The given differential equation is cos square x d y by d x plus y is equal to tan x or this can be written as dy by d x plus y over cos square x that is secant square x into y is equal to secant square x into tan x. Let us get this equation as number one. Now this is a linear differential equation of the time dy by d x plus py is equal to q where p and q are constants of functions of x only. Here p which is the coefficient of y is secant square x and q is equal to secant square x tan x. Therefore integrating factor is equal to e raise to power integral of p d x and this is equal to e raise to power integral of secant square x d x which is equal to e raise to power tan x on multiplying both sides of equation one by the integrating factor. We get dy by dx into e raise to power tan x plus y secant square x into e raise to power tan x is equal to e raise to power tan x into tan x into secant square x. Now left hand side of this equation is a differential of y into e raise to power tan x. So this equation can be written as dy d x of y into e raise to power tan x is equal to e raise to power tan x into tan x secant square x. Now integrating both sides with respect to x we get integral of dy dx of y into e raise to power tan x dx is equal to integral of e raise to power tan x into tan x secant square x dx or y into e raise to power tan x is equal to i plus c where i is equal to integral of e raise to power tan x into tan x secant square x dx. Now we will solve this integral for this let us put tan x is equal to t therefore secant square x dx is equal to dt therefore i is equal to integral of e raise to power t into t dt. Now we will integrate this integral by parts so this is equal to let us take t as a first function and e raise to power t as a second function so the integral of e raise to power t into t dt is equal to t into integral of e raise to power t dt minus integral of d over dt of t into integral of e raise to power t dt and this is equal to t into e raise to power t minus integral of 1 into e raise to power t dt and this is equal to t into e raise to power t minus e raise to power t and again this can be written as tan x into e raise to power tan x minus e raise to power tan x now substitute this value of integral in the equation y into e raise to power tan x is equal to i plus c let us get this equation as number 2 so on substituting the value of i in equation 2 we have 1 into e raise to power tan x is equal to e raise to power tan x into tan x minus e raise to power tan x plus c or this can be written as y is equal to tan x minus 1 plus c into e raise to power minus tan x hence the general solution of the given differential equation is y is equal to tan x minus 1 plus c into e raise to power minus tan x and this is the answer for the above question I hope the solution is clear to you and you have enjoyed the session bye and take care