 Hi friends, I am Purva and today we will work out the following question for the following differential equation finding a particular solution satisfying the given condition dy upon dx plus 2y into tan x is equal to sin x and we have y is equal to 0 when x is equal to pi by 3 Now the solution of the linear differential equation of the form dy upon dx plus dy is equal to q is given by y into e raised to the power integral p dx is equal to integral q into e raised to the power integral p dx plus c and here e raised to the power integral p dx is the integrating factor of the given equation and this c can be determined by the given conditions so we have where c can be determined by the given conditions so this is the key idea behind our question let us now begin with the solution so the given differential equation is dy upon dx plus 2 into y into tan x is equal to sin x let us mark this as equation 1 and here we have y is equal to 0 when x is equal to pi by 3 this is the given condition the above differential equation is a linear differential equation of the type dy upon dx plus py is equal to q therefore the integrating factor of this equation is given by e raised to the power integral p dx and this is equal to e raised to the power integral now comparing this equation 1 by the equation in a key idea we can clearly see that p is equal to 2 tan x so we have here e raised to the power integral 2 tan x dx and this is equal to e raised to the power now integrating 2 tan x with respect to x we get 2 log mod of sec x so we get e raised to the power 2 log sec x and this is equal to e raised to the power now we can write 2 log modulus of sec x as log sec square x so we get e raised to the power log sec square x and this is equal to sec square x because e raised to the power log sec square x is equal to sec square x so we get the integrating factor of this equation as sec square x now multiplying each term of equation 1 by sec square x we get sec square x into dy upon dx plus 2y tan x into sec square x is equal to sin x into sec square x this implies now we can write sec square x into dy upon dx plus 2y tan x into sec square x as d upon dx of y into sec square x because on differentiating y into sec square x with respect to x we get sec square x into dy upon dx plus 2y tan x into sec square x and this is equal to now we can write sin x into sec square x as tan x into sec x write now integrating both sides with respect to x we get integral d upon dx of y into sec square x dx is equal to integral tan x into sec x dx and this implies now integrating d upon dx of y into sec square x we get y into sec square x is equal to now integrating tan x into sec x we get sec x plus c plus c is a constant now dividing both the sides by sec square x we get y is equal to sec x upon sec square x plus c upon sec square x or we can write this as y is equal to 1 upon sec x plus c into cos square x or we can write this as y is equal to cos x plus c cos square x just mark this as equation 2 now to find the value of c we use the given condition that is y is equal to 0 when x is equal to pi by 3 this is the condition given to us so we put the value of y and x in equation 2 and we find the value of c therefore we have 0 is equal to cos pi by 3 plus c into cos square pi by 3 and this implies 0 is equal to now cos pi by 3 is equal to 1 upon 2 so we have 1 upon 2 plus c into 1 upon 4 because cos pi by 3 is equal to 1 upon 2 and whole square of 1 upon 2 gives 1 upon 4 and this implies c upon 4 is equal to minus 1 upon 2 and this further implies c is equal to minus 2 so we get the value of c as minus 2 now substituting this value of c in equation 2 we get y is equal to cos x minus 2 cos square x hence the required solution is y is equal to cos x minus 2 cos square x this is our answer hope you have understood the solution bye and take care