 Hi, and welcome to the session Let us discuss the following question question says find the particular solution of the differential equation x minus y multiplied by dx plus dy is equal to dx minus dy Given that y is equal to minus 1 when x is equal to 0 Let us now start with the solution First of all, let us rewrite the given differential equation That is x minus y multiplied by dx plus dy Is equal to dx minus dy Now this equation can be further written as x minus y minus 1 multiplied by dx x plus x minus y plus 1 multiplied by dy is equal to 0 Adding dy minus dx on both the sides of this equation we get this equation Now this further implies x minus y plus 1 multiplied by dy is equal to minus x minus y minus 1 multiplied by dx Now dividing both the sides of this equation by this bracket we get dy is equal to minus x minus y minus 1 dx upon x minus y plus 1 Now dividing both the sides of this equation by dx we get dy upon dx is equal to minus x minus y minus 1 upon x minus y plus 1 Now clearly we can see x minus y is present in numerator as well as denominator here So we will substitute x minus y is equal to t So here we can write put x minus y is equal to t Now differentiating both the sides with respect to x we get 1 minus dy upon dx is equal to dt upon dx Now subtracting 1 from both the sides of this equation we get minus dy upon dx is equal to dt upon dx minus 1 Multiplying both the sides of this equation by minus 1 we get dy upon dx is equal to 1 minus dt upon dx Now let us name this equation as equation 1 Now in this equation 1 we will substitute t for x minus y and 1 minus dt upon dx for dy upon dx Now we can write substituting x minus y is equal to t and dy upon dx is equal to 1 minus dt upon dx in equation 1 we get 1 minus dt upon dx is equal to minus t minus 1 upon t plus 1 Now this further implies 1 minus dt upon dx is equal to 1 minus t upon t plus 1 Multiplying this minus sign by this bracket we get 1 minus t Now subtracting 1 from both the sides of this equation we get minus dt upon dx is equal to 1 minus t upon t plus 1 minus 1 Now multiplying both the sides by minus 1 we get dt upon dx is equal to 1 minus 1 minus t upon t plus 1 Now this further implies dt upon dx is equal to t plus 1 minus 1 plus t upon t plus 1 Now we will cancel plus 1 minus 1 we know plus 1 minus 1 is equal to 0 So we get dt upon dx is equal to 2t upon t plus 1 t plus t is equal to 2t Now separating the variables in this equation we get t plus 1 upon 2t dt is equal to dx Now multiplying both the sides of this equation by 2 we get t plus 1 upon t dt is equal to 2dx Now integrating both the sides of this equation we get integral of t plus 1 upon t dt is equal to 2 multiplied by integral of dx Now this integral can be further written as integral of dt plus integral of dt upon t And here we will write is equal to sign as it is And we will write 2 multiplied by integral of dx as it is Now using this formula of integration we can find these two integrals And using this formula of integration we can find this integral Now we get integral of dt is equal to t Integral of dt upon t is equal to log t Is equal to 2 multiplied by We know integral of dx is equal to x So here we will write 2 multiplied by x plus c Where c represents constant of integration Now replacing t by x minus y in this expression we get x minus y plus log of x minus y Is equal to 2x plus c Now adding y minus x on both the sides of this equation we get log of x minus y is equal to 2x Plus y minus x plus c Now we know 2x minus x is equal to x So we can write log of x minus y is equal to x plus y plus c Now we are given that for the given differential equation y is equal to minus 1 when x is equal to 0 Now we will substitute y is equal to minus 1 And x is equal to 0 in this equation Let us name this equation as equation 2 So here we can write substituting y is equal to minus 1 and x is equal to 0 In equation 2 we get log of 0 minus minus 1 is equal to 0 plus minus 1 plus c Now this further implies log 1 is equal to minus 1 plus c Now we know log 1 is equal to 0 So here we will substitute 0 for log 1 and we get 0 is equal to minus 1 plus c Now adding 1 on both the sides of this equation we get 1 is equal to c or we can simply write c is equal to 1 Now substituting value of c is equal to 1 in equation 2 We get log of x minus y is equal to x plus y plus 1 So this is the required particular solution of the given differential equation So this is our required answer This completes the session. Hope you understood the solution. Take care. Have a nice day