 Hi and welcome to the session. I am Shashi and I am going to help you with the following question. Question says, in a bank, principal increases continuously at the rate of 5% per year. An amount of Rs 1000 is deposited with this bank. How much will it worth after 10 years? Let us now start with the solution. Let us assume that the principal is equal to Rs P. So we can write, let principal be Rs P. Now we are given that principal increases at the rate of 5% per year. So here we can write principal increases at the rate equal to 5% per year. Now according to the question, dp upon dt is equal to 5 upon 100 multiplied by P. Here dp upon dt represents rate of change of principal and we know rate of change of principal is 5% of P. As principal is increasing every year at the rate of 5%, now separating the variables in this equation, we get dp upon P is equal to 5 upon 100 dt. Now integrating both the sides of this equation, we get integral of dp upon P is equal to 5 upon 100 multiplied by integral of dt. Now first of all we will find out this integral. Using this formula of integration, we get this integral is equal to log P. And using this formula of integration, we get this integral is equal to T. So here we can write 5 upon 100 multiplied by T plus C1, where C1 is the constant of integration. Now applying this law of logarithms on both the sides of this equation, we get P is equal to e raised to the power 5 upon 100 multiplied by T multiplied by e raised to the power C1. We know e raised to the power 5 upon 100 multiplied by T plus C1 is equal to e raised to the power 5 upon 100 multiplied by T multiplied by e raised to the power C1. Now we will substitute C for e raised to the power C1 and we get P is equal to C multiplied by e raised to the power 0.05T. We know 5 upon 100 can be written as 0.05. Now let us name this expression as 1. Now we are given P is equal to rupees 1000 when T is equal to 0. Here T represents the time in years. We know initially amount of rupees 1000 has been deposited with the bank. Now we will substitute P is equal to rupees 1000 and T is equal to 0 in this equation. That is equation 1, we get 1000 is equal to C multiplied by e raised to the power 0.05 multiplied by 0. Now this further implies 1000 is equal to C multiplied by 1. We know 0.05 multiplied by 0 is equal to 0 and e raised to the power 0 is equal to 1 only. Now we get value of C is equal to 1000. Now we will substitute this value of C in equation 1 and we get P is equal to 1000 multiplied by e raised to the power 0.05T. Now we have to find worth of rupees 1000 in the bank after 10 years. Now to find the required value of P after 10 years we will substitute 10 for T here. So we can write when T is equal to 10 years P is equal to 1000 multiplied by e raised to the power 0.05 multiplied by 10. Now this further implies P is equal to 1000 multiplied by e raised to the power 0.5. Now e raised to the power 0.5 is equal to 1.648. This is already given in the question. Now we can write this implies P is equal to 1000 multiplied by 1.648 which is further equal to 1648. So we get an amount of rupees 1000 worth rupees 1648 after 10 years. So this is our required answer. This completes the session. Hope you understood the solution. Take care and have a nice day.