 Hello and welcome to the session. The given question says solve the following differential equation and the given equation is sin x into dy divided by dx plus cos x into y is equal to cos x into sin x square. Now let's start with the solution and the given first order linear differential equation is sin x into dy divided by dx plus cos x into y is equal to cos x into sin square x. Now divided both the sides by sin x we have divided by dx plus cos x divided by sin x into y is equal to cos x into sin x. Here we have divided both the sides by sin x. Now this is a linear first order differential equation of the form dy by dx plus dx into y is equal to qx and for this type of first order linear differential equation its integrating factor is given by exponential raised to the power integral px into dx and its solution curves are given by y into the integrating factor is equal to integral qx into the integrating factor into dx plus c. So here we have px is equal to cos x divided by sin x and qx is equal to cos x into sin x. Now let us find the integrating factor is given by exponential raised to the power integral px into dx that is exponential raised to the power integral cos x divided by sin x into dx. Now first let us solve this integral let sin x is equal to t. So we have on differentiating both sides with respect to x we have cos x into dx is equal to dt. So here we have exponential raised to the power integral dt divided by t which is equal to exponential raised to the power log t this is further equal to t and t is sin x so integrating factor is sin x. Now let us be equation number one so multiplying both sides of equation one by the integrating factor which is sin x we have into dy divided by dx plus cos x into y is equal to cos x into sin square x. Now the left hand side can be written as dx of sin x into y that is derivative of sin x into y with respect to x is equal to cos x into sin square x. Now integrating both the sides on the left hand side we have sin x into y and on the right hand side we have integral cos x into sin square x into dx. Now let us take u is equal to sin x so this implies du is equal to cos x into dx. So this can further be written as integral u square into du which is further equal to u cube divided by 3 plus a constant. And this is further equal to sin cube x divided by 3 plus a constant or y can be written as dividing both sides by sin x we have sin square x divided by 3 plus c divided by sin x. Hence on solving the given differential equation we have y is equal to 1 by 3 sin square x plus c divided by sin x where c is a constant. So this completes the session by intake q.