 Hi and welcome to the session. Let us discuss the following question. Question says, for 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. First of all, let us consider the given function. Given function is pi is equal to e raised to the power x multiplied by a cos x plus b sin x. First of all, let us name this expression as 1. Now differentiating both sides of 1 with respect to x, we get dy upon dx is equal to e raised to the power x multiplied by minus a sin x plus b cos x plus a multiplied by cos x plus b sin x e raised to the power x. We know derivative of phi with respect to x is dy upon dx. Now we can find derivative of this term by using product rule. Now this further implies dy upon dx is equal to e raised to the power x multiplied by minus a sin x plus b cos x plus y. We know e raised to the power x multiplied by a cos x plus b sin x is equal to y. So we can substitute for this term y. Now let us name this equation as equation 2. Now we will differentiate equation 2 with respect to x again to find the value of d square y upon dx square. After differentiation, equation 2 becomes d square y upon dx square is equal to e raised to the power x minus a sin x plus b cos x plus e raised to the power x minus a cos x minus b sin x plus dy upon dx. We know derivative of dy upon dx is d square y upon dx square. We can find derivative of this term by using product rule and derivative of y is dy upon dx. Now we can replace this term by minus y. Clearly we can see this term is equal to minus e raised to the power x multiplied by a cos x plus b sin x and e raised to the power x multiplied by a cos x plus b sin x is equal to y only. So we will write this minus sign as it is and we will replace this term by y. So we can write d square y upon dx square is equal to e raised to the power x minus a sin x plus b cos x minus y plus dy upon dx. Also from equation 2 we get dy upon dx minus y is equal to e raised to the power x minus a sin x plus b cos x. So we can substitute dy upon dx minus y for this term. Clearly we can see this term is right hand side of this equation. Now let us name this equation as 3. Using 3 we get d square y upon dx square is equal to dy upon dx minus y minus y plus dy upon dx. Now this further implies d square y upon dx square is equal to 2 dy upon dx minus 2y. Now adding 2y minus 2 dy upon dx on both the sides we get d square y upon dx square plus 2y minus 2 dy upon dx is equal to 0. Now this equation is same as the given differential equation. So the given function is the solution of the given differential equation. Hence verified this completes the session. Hope you understood the solution. Take care and have a nice day.