 Hi and welcome to the session. Let us discuss the following question. Question says, for each of the exercise given below, verify that the given function is a solution of the corresponding differential equation. This is the given function and this is the corresponding differential equation. Let us now start with the solution. Now given function is x square is equal to 2y square log y. Let us name this equation as equation 1. Now differentiating both the sides of equation 1 with respect to x, we get 2x is equal to 2 multiplied by 2y log y. dy upon dx plus y square multiplied by 1 upon y dy upon dx. Using this formula, we can find derivative of x square using product rule. We can find derivative of 2y square log y. Now dividing both sides by 2, we get x is equal to 2y log y dy upon dx plus here we will cancel y from numerator and denominator both and we get y dy upon dx. Now multiplying both the sides of this equation by y, we get xy is equal to 2y square log y dy upon dx plus y square dy upon dx. Now this further implies xy is equal to dy upon dx multiplied by 2y square log y plus y square. Clearly we can see dy upon dx is common in both the terms. Now from equation 1, we know x square is equal to 2y square log y. Now we will substitute x square for 2y square log y in this equation and we get xy is equal to dy upon dx multiplied by x square plus y square. Now subtracting xy from both the sides of this equation, we get 0 is equal to dy upon dx multiplied by x square plus y square minus xy or we can simply write dy upon dx multiplied by x square plus y square minus xy is equal to 0. Now clearly we can see this equation is same as given differential equation. So given function is a solution of the given differential equation. So we can write the above equation is same as the given differential equation. Therefore given function is a solution of given differential equation. Hence verified this completes the session. Hope you understood the solution. Take care and have a nice day.