 Hello and welcome to the session. In this session we will discuss the following question which says solve the following differential equation dy by dx minus 1 is equal to e to the power of x plus y. Consider the equation of the form f of x into dx plus phi of y into dy is equal to zero. Such equations in which the coefficient of dx is only a function of x and the coefficient of dy is only a function of y are the differential equations in which variables are separable. So for the equations of this kind we separate the variables of x and y and then we integrate both sides adding an arbitrary constant on one side. This is the key idea that we would use in this question. Let's proceed with the solution now. The given differential equation is dy by dx minus 1 is equal to e to the power of x plus y. As you can see that this given differential equation is not of this form in which the variables are separable. But let us try to reduce this equation to variables separable. For this we substitute x plus y equal to z. Now differentiating both sides with respect to x we get 1 plus dy by dx is equal to dz by dx and from here we get that dy by dx is equal to dz by dx minus 1. If the given equation that is the given differential equation be given number 1. Now substituting the values of dy by dx so obtained x plus y. In the equation 1 we get dz upon dx minus 1 that is in place of dy by dx we put dz upon dx minus 1. This minus 1 is equal to e to the power of z. So from here we have dz upon dx minus 2 is equal to e to the power of z. Further we get dz upon dx is equal to 2 plus e to the power of z. Now we can easily separate the variables z and x that is we take the coefficient of dz as a function of z only. So we can have here 1 upon 2 plus e to the power of z into dz is equal to dx. So coefficient of dz is a function of z only and the coefficient of dx is 1 which is a constant. Now further we have dz upon now 2 plus e to the power of z can be written as e to the power of z into 2 upon e to the power of z plus 1 the whole is equal to dx. Now further we have e to the power of minus z into dz upon 2 into e to the power of minus z plus 1 is equal to dx. Now putting 2 into e to the power of minus z plus 1 as t and differentiating both sides with respect to t we get minus 2 into e to the power of minus z dz is equal to dt. And from here we have e to the power of minus z dz is equal to minus 1 upon 2 dt. So now considering this equation to be number 2. So further substituting the values in equation 2 that is the value of e to the power of minus z dz and 2 into e to the power of minus z plus 1. So substituting these values we get minus 1 upon 2 dt upon t is equal to dx. Next we integrate both sides. So integrating both sides we have minus 1 upon 2 into integral dt upon t is equal to integral dx. So further we get minus 1 upon 2 log modulus t is equal to x plus your arbitrary constant c. But we have taken the value of t as 2 into e to the power of minus z plus 1. So substituting these values of t we get minus 1 upon 2 into log modulus 2 into e to the power of minus z plus 1 is equal to x plus c that is the constant of integration. Also we have taken the value of z as x plus y. So substituting the value of z as x plus y we get minus 1 upon 2 log modulus 2 into e to the power of minus of x plus y plus 1 is equal to x plus c where the c is some arbitrary constant. Or we can also write this as x plus 1 upon 2 into log modulus 2 into e to the power of minus x plus y plus 1 plus c that is the arbitrary constant is equal to 0. And this is the required solution. So this is our final answer. This completes the session. Hope you have understood the solution of this question.