 Hi and welcome to the session I am Deepika here. Let's discuss the question which says find the general solution of the following differential equation dy by dx plus 2y is equal to sin x. Let us first understand how to find the solution of the differential equation of the form dy by dx plus py is equal to q where p and q are constants or functions of x only. This is known as the first order linear differential equation. To solve this equation we will first find out the integrating factor which is given by e raised to power integral of p dx and the solution of the given differential equation is given by y into integral of q into integrating factor dx plus c that is y into e raised to power integral of p dx is equal to integral of q into e raised to power p dx plus c where c is the constant of integration. So this is the key idea behind our question. We will take the help of this key idea to solve the above question. So let's start the solution. The given differential equation is dy by dx plus 2y is equal to sin x. Let us get this equation as number one. Now this equation is a linear differential equation of the form dy by dx plus py is equal to q. Here p is equal to 2 and q is equal to sin x. Therefore according to our key idea integrating factor is given by e raised to power integral of p dx which is equal to e raised to power integral of 2 dx and this is equal to e raised to power 2x. Now we will multiply both sides of equation one by the integrating factor because by this way the left hand side of the differential equation one will become exact differential of some function. So on multiplying both sides of equation one by e raised to power 2x we get e raised to power 2x into dy by dx plus 2y is equal to e raised to power 2x into sin x or e raised to power 2x into dy by dx plus e raised to power 2x into 2y that is 2 into e raised to power 2x into y is equal to e raised to power 2x into sin x. Now left hand side of this equation has become the differential of some function. Now it is the differential of y into e raised to power 2x. So we can write this equation as d over dx of y into e raised to power 2x is equal to e raised to power 2x into sin x. On integrating both sides of the above equation with respect to x we get integral of d over dx of y into e raised to power 2x dx is equal to integral of e raised to power 2x into sin x dx or y into e raised to power 2x is equal to integral of e raised to power 2x sin x dx plus let i is equal to integral of e raised to power 2x sin x dx. Now we will integrate this by parts let us take sin x as a first function and e raised to power 2x as a second function. So integral of e raised to power 2x sin x dx is equal to first function that is sin x into integral of second function that is e raised to power 2x dx minus integral of differential coefficient of the first function is d over dx of sin x into integral of the second function. So this is equal to sin x into e raised to power 2x over 2 minus integral of cos x into e raised to power 2x over 2 dx this is equal to 1 over 2 e raised to power 2x sin x minus 1 over 2 integral of e raised to power 2x cos x dx. Now this is equal to 1 over 2 e raised to power 2x sin x minus 1 by 2 now again we will integrate this function by parts let us take cos x as the first function and e raised to power 2x as the second function. So integral of e raised to power 2x cos x dx is equal to cos x into e raised to power 2x over 2 minus integral of minus sin x into e raised to power 2x over 2 dx. Now this is again equal to 1 over 2 e raised to power 2x sin x minus 1 over 4 e raised to power 2x cos x minus 1 over 4 integral of e raised to power 2x sin x dx. Now we have assumed i is equal to integral of e raised to power 2x sin x dx so this is equal to 1 over 2 e raised to power 2x sin x minus 1 over 4 e raised to power 2x cos x minus 1 over 4 i so this implies i plus 1 by 4 i is equal to 1 over 2 e raised to power 2x sin x minus 1 over 4 e raised to power 2x cos x so this implies 5 over 4 i is equal to let us take 1 over 4 e raised to power 2x common from these two terms so we have 5 over 4 i is equal to 1 over 4 e raised to power 2x into 2 sin x minus cos x this implies i is equal to 1 over 4 into 4 over 5 e raised to power 2x into 2 sin x minus cos x so this implies i is equal to 1 over 5 e raised to power 2x into 2 sin x minus cos x hence the integral of e raised to power 2x sin x dx is equal to 1 over 5 e raised to power 2x into 2 sin x minus cos x now we have 1 into e raised to power 2x is equal to integral of e raised to power 2x sin x dx plus c and the integral of e raised to power 2x sin x dx is 1 over 5 e raised to power 2x into 2 sin x minus cos x so we have y into e raised to power 2x is equal to 1 over 5 e raised to power 2x into 2 sin x minus cos x plus c or y is equal to 1 over 5 into 2 sin x minus cos x plus c into e raised to power minus 2x hence the general solution of the Vivian differential equation is y is equal to 1 over 5 into 2 sin x minus cos x plus c into e raised to power minus 2x so this is our answer for the above question I hope the solution is clear to you and you have enjoyed this session bye and take care.