 Hi friends, I am Purva and today we will work out the following question. Find the equation of a curve passing through the origin given that the slope of the tangent to the curve at any point x,y is equal to the sum of the coordinates of the point. Now, the slope of the tangent to the curve at the point x,y is given by dy by dx. This is the key idea behind our question. Let us now begin with the solution. Now, we are given that the slope of the tangent to the curve at the point x,y is equal to the sum of the coordinates of the point. Therefore, according to the given condition we have, dy by dx is equal to x plus y or we can write this as dy by dx minus y is equal to x. Let us mark this as equation 1. Now, equation 1 is a linear differential equation of the form dy by 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 which is equal to e raised to the power integral. Now, here p is equal to minus 1. So, we have minus 1 dx and this is further equal to e raised to the power now integral of minus 1 with respect to x is minus x. So, we get the integrating factor as e raised to the power minus x. Now, multiplying both the sides of equation 1 by e raised to the power minus x we get e raised to the power minus x into dy by dx minus y into e raised to the power minus x is equal to x into e raised to the power minus x. And this implies now we can write e raised to the power minus x into dy by dx minus y into e raised to the power minus x as d upon dx of y into e raised to the power minus x. And this is equal to x into e raised to the power minus x. Now, integrating both the sides with respect to x we get integral d upon dx of y into e raised to the power minus x dx is equal to integral x into e raised to the power minus x dx. And this implies now integrating left hand side we get y into e raised to the power minus x and this is equal to now on right hand side we applied by path method. So, we take x as first term and e raised to the power minus x as second term. So, we get x into integral of e raised to the power minus x dx minus integral derivative of x with respect to x into integral e raised to the power minus x dx dx plus c. And this implies y into e raised to the power minus x is equal to x into now integral of e raised to the power minus x is minus of e raised to the power minus x minus integral of minus e raised to the power minus x dx plus c. Because derivative of x with respect to x gives 1 so we have 1 and 2 and integral of e raised to the power minus x is minus e raised to the power minus x. So, we get integral of minus e raised to the power minus x dx. And this further implies y into e raised to the power minus x is equal to minus x into e raised to the power minus x minus now integral of minus e raised to the power minus x is e raised to the power minus x plus c. And this implies now dividing both the sides by e raised to the power minus x we get y is equal to minus x minus 1 plus c into e raised to the power x. And this further implies x plus y plus 1 is equal to c into e raised to the power x and we mark this as equation 2. Now in the question we are given that find the equation of a curve passing through the origin. Therefore, since this curve passes through 0 comma 0 therefore we have putting x as 0 and y as 0 we get 0 plus 0 plus 1 is equal to c into e raised to the power 0. And this implies 1 is equal to c because e raised to the power 0 is equal to 1 therefore we get c is equal to 1. Now substituting the value of c in equation 2 we get x plus y plus 1 is equal to e raised to the power x and this is the required equation of the curve. Hence the required curve is x plus y plus 1 is equal to e raised to the power x. This is our answer. Hope you have understood the solution. Bye and take care.