 Welcome to the session, I am Deepika here. Let's discuss a question which says verify that the given function is the solution of the corresponding differential equation y is equal to x and x. xy dash is equal to y plus x into under root of x square minus y square, where x is not equal to 0 and x is greater than y or x is less than minus y. Now, we know that the given function is the solution of the corresponding differential equation if it satisfies the differential equation. That is when the function is substituted for the unknown y in the given differential equation, the left hand side becomes equal to right hand side. So, let's start the solution. Now, the given differential equation is xy dash is equal to y plus x into under root of x square minus y square, where x is not equal to 0 and x is greater than y or x is less than minus y. Let us give this as number one and the given function is y is equal to x sin x, let us give this as number two. Now, we will find the first derivative of this function and we will substitute the value of y and y dash in one to verify whether the given function is the solution of the given differential equation. Now, on differentiating both sides of equation two with respect to x, we get d y by dx is equal to, now here we will apply the product root x into derivative of sin x, which is cos x plus sin x into derivative of x, which is one of y dash is equal to x cos x plus sin x. Now, on substituting the value of y dash in the left hand side of equation one, we get, now the left hand side of equation one is x y dash and this is equal to x into y dash, which is x cos x plus sin x and this is equal to x square cos x plus x sin x and this is again equal to x square into, now cos x can be written as under root of one minus sin square x plus now from two because the given function is y is equal to x sin x, so we can write x sin x is equal to y and this is equal to x square into under root of one minus. Now, y is equal to x sin x, this implies sin x is equal to y over x, so this is under root of one minus y square over x square plus y and this is equal to x square into under root of x square minus y square over x square plus y. Now, since under root of x square minus y square is greater than zero, this implies x square is greater than y square, this implies x is greater than y and x is less than minus y, so the left hand side of one is equal to x square into under root of x square minus y square over x plus y and this is equal to y plus x into under root of x square minus y square, but this is our right hand side, hence left hand side is equal to right hand side, therefore the given function is a solution of the given differential equation, so this completes our session, I hope you have enjoyed the session, bye and take care.