 Hi and welcome to the session. Let us discuss the following question. Question says verify that the given function is a solution of the corresponding differential equation. Now the given function is y is equal to square root of a square minus x square where x belongs to open interval minus a a and given differential equation is x plus y dy by dx is equal to 0. Let us now start with the solution. Now the given function is y is equal to square root of a square minus x square. Now differentiating both sides of this equation with respect to x we get dy upon dx is equal to 1 upon 2 multiplied by square root of a square minus x square multiplied by minus 2x. Now this is further equal to minus x upon square root of a square minus x square. Now the given differential equation is x plus y dy upon dx is equal to 0. Now substituting values of y and dy upon dx in this given differential equation we get LHS as x plus square root of a square minus x square multiplied by minus x upon square root of a square minus x square. Now we will cancel common factor square root of a square minus x square in numerator and denominator both. Now LHS is further equal to x minus x. Now we know x minus x is equal to 0. Now clearly we can see 0 is equal to right hand side of the given differential equation. So we get LHS is equal to RHS. Now we can write the given function is a solution of given differential equation. Hence verified this completes the session. Hope you understood the solution. Take care and keep smiling.