 Hello and welcome to this session. In this session we discussed the following question which says solve the following differential equation cos square x dy by dx plus y equal to tan x. Let's move on to the solution. The given differential equation is cos square x dy by dx plus y is equal to tan x. Now on dividing both sides by cos square x we get dy by dx plus y upon cos square x is equal to tan x over cos square x. This can also be written as dy by dx plus y sex square x since we know that 1 over cos square x is sex square x. This is equal to tan x into sex square x since 1 over cos square x is sex square x. Let this be equation one. Now you can see that this is a linear differential equation of the form dy by dx plus py equal to q. So on comparing these two equations we get that p is equal to sex square x and q is equal to tan x into sex square x. Now to solve the linear differential equation we need to find the integrating factor that is is and this is equal to e to the power integral p dx that is this is equal to e to the power integral sex square x dx. We know that integral sex square x dx is tan x. So this is equal to e to the power tan x is the integrating factor. Now the solution is given by y into the integrating factor equal to integral of q into integrating factor dx plus c that is we put the values for integrating factor and q. So this gives us y into e to the power tan x equal to integral of tan x into sex square x into e to the power tan x dx plus c. That is we have y into e to the power tan x is equal to integral of e to the power tan x into sex square x into tan x dx plus c. Now let's try and solve this integral for this we put tan x equal to p. Now differentiating both the sides we get sex square x dx is equal to dt. So on substituting sex square x dx equal to dt and tan x equal to t we get y into e to the power tan x equal to integral e to the power t into t dt plus c. Now this integral is solved by by parts method by taking this as the first function and this as the second function. So this would be equal to the first function that is t into integral of the second function that is integral of e to the power t which would be e to the power t minus integral of differential of the first function that is differential of p would be 1 into integral of the second function which would be e to the power t dt plus c. So we get y into e to the power tan x is equal to t into e to the power t minus integral 1 into e to the the power t dt plus c. This gives y into e to the power tan x is equal to t into e to the power t minus e to the power t plus c that is we get y into e to the power tan x is equal to e to the power t into t minus 1 plus c. Now substituting t equal to tan x in this we get y into e to the power tan x is equal to e to the power tan x into tan x minus 1 plus c. Now dividing both sides by e to the power tan x we get y equal to tan x minus 1 plus c into e to the power minus tan x. So, this is the required solution. Hence our final answer is y is equal to tan x minus 1 plus c into e to the power minus tan x. So, this completes the session. Hope you have understood the solution for this question.