 Good morning friends, I am Purva and today we will discuss the following question. For the following differential equation, find a particular solution satisfying the given condition. And we are given x into x square minus 1 into dy by dx is equal to 1. And the given condition is y is equal to 0 when x is equal to 2. Let us now begin with the solution. Now, the given differential equation is x into x square minus 1 into dy by dx is equal to 1. And the condition is y is equal to 0 when x is equal to 2. Now this differential equation can be written as this implies dy by dx is equal to 1 upon x into x square minus 1. Now this further implies dy is equal to 1 upon x into x square minus 1 into dx. Now integrating both the sides we get integral of dy is equal to integral of 1 upon x into x square minus 1 dx. Now this implies integrating 1 with respect to y we get here y is equal to now we denote integral of 1 upon x into x square minus 1 dx as i. So we write i plus c where c is the constant of integration and we mark this as equation 1. So we have i is equal to integral of 1 upon x into x square minus 1 dx. Now we consider 1 upon x into x square minus 1. Now we can write this as this is equal to 1 upon x into x minus 1 into x plus 1 because we know that x minus 1 into x plus 1 gives x square minus 1. Now again 1 upon x into x minus 1 into x plus 1 is equal to a upon x plus b upon x minus 1 plus c upon x plus 1 by partial fractional method. Now this implies 1 is equal to a into x minus 1 into x plus 1 plus b into x into x plus 1 plus c into x into x minus 1. Now putting x is equal to 1 in the above equation we get 1 is equal to b into 1 into 2 and this implies b is equal to 1 upon 2. Putting x is equal to 0 we get 1 is equal to a into minus 1 and this implies a is equal to minus 1. And finally putting x is equal to minus 1 we get 1 is equal to c into minus 1 into minus 2 and this implies c is equal to 1 upon 2. Now putting the values of a b and c in this equation 2 we get 1 upon x into x minus 1 into x plus 1 is equal to minus 1 upon x plus 1 upon 2 into x minus 1 plus 1 upon 2 into x plus 1. That is we get 1 upon x into x square minus 1 is equal to minus 1 upon x plus 1 upon 2 into x minus 1 plus 1 upon 2 into x plus 1. Thus equation 1 becomes y is equal to minus integral 1 upon x dx plus integral 1 upon 2 into x minus 1 dx plus integral 1 upon 2 into x plus 1 dx plus c. Now this implies y is equal to minus now integrating 1 upon x we get log of mod x plus integrating 1 upon 2 into x minus 1 we get 1 upon 2 into log of mod x minus 1. Plus integrating 1 upon 2 into x plus 1 we get 1 upon 2 into log of mod x plus 1 plus c. Now this implies y is equal to minus log of mod x plus now taking out 1 upon 2 common from these both terms we get 1 upon 2 into log of mod x minus 1. Plus log of mod x plus 1 plus c. Now this implies y is equal to minus log of mod x plus 1 upon 2 into log x minus 1 into x plus 1 plus c. Since we know that log of m plus log of n is equal to log of m into n this implies y is equal to now we can also write right hand side as 1 upon 2 into log of x minus 1 into x plus 1 minus 2 into log x plus c. This implies y is equal to 1 upon 2 into log of now x minus 1 into x plus 1 is equal to x square minus 1 also 2 log x can be written as log x square plus c. Since we know that m log n is equal to log n to the power m this implies y is equal to 1 upon 2 into log x square minus 1 upon x square plus c. Since we know that log of m minus log of n is equal to log m upon n so using this formula here we get log of x square minus 1 upon x square. We mark this as equation 2. Now we are given a condition that y is equal to 0 when x is equal to 2 so putting x is equal to 2 and y is equal to 0 in equation 2 we get 0 is equal to 1 upon 2 into log of x square that is 4 minus 1 upon x square that is 2 square which is equal to 4 plus c this implies 0 is equal to 1 upon 2 into log of 3 upon 4 plus c and this implies c is equal to minus 1 upon 2 into log of 3 upon 4. Now putting this value of c in this equation 2 we get y is equal to 1 upon 2 into log of x square minus 1 upon x square minus 1 upon 2 into log of 3 upon 4. This is our required answer. Hence we write our answer as y is equal to 1 upon 2 into log of x square minus 1 upon x square minus 1 upon 2 into log of 3 upon 4. Hope you have understood the solution. Bye and take care.