 Hello and welcome to the session. In this session, we discuss the following question with face, find the particular solution of the differential equation, satisfy the given conditions, where we have derived by dx equal to y into tan x, given that y equal to 1 when x is equal to 0. Now, we know that integral of 1 upon x dx is equal to log of modulus x plus the constant of integration c, then integral of tan x dx is equal to log of modulus of sec x plus c, which is the constant of integration. This is the key idea that we use in this question. Let's go on to the solution. We have given the differential equation as dy by dx equal to y into tan x. This can be written as dy upon y is equal to tan x into dx. Now integrating both sides we get integral of dy upon y is equal to integral of tan x dx. Now integral of 1 upon x dx is log of modulus x plus c. So this means we have log of modulus y is equal to integral of tan x dx, which is log of modulus sec x plus c. So this is log of modulus sec x plus log c where the c is the constant of integration. So this means we have log of modulus y equal to log c into modulus sec x. This gives us modulus y is equal to c into modulus sec x. Or you can say y is equal to c into sec x. Let this be equation 1. Now the condition given to us is when x is equal to 0 we get y equal to 1. So putting x equal to 0 y equal to 1 in equation 1 we get 1 is equal to c into sec 0. Now we know that cos 0 is 1. So this means sec 0 which is 1 upon cos 0 would also be 1. So we have 1 is equal to c. So we have got the value of c as 1. So now substituting c equal to 1 in equation 1 we get y is equal to sec x. So therefore y equal to sec x is the equivalent solution given differential equation. So this is our final answer. This completes the session. Hope you understood the solution of this question.