 Hello friends and how are you all today? The question says solve the following differential equation. Here we have the differential equation as dx square plus xy into dy is equal to x square plus y square into dx. That is, x square plus xy into dy is equal to x square plus y square into dx. Or we can write it as dy by dx is equal to x square plus y square upon x square plus xy. Now here let us put y equal to v into x. So this implies dy by dx is equal to v plus x dy by dx. So on substituting these two in here we have v plus x dv by dx square plus vx the whole square upon x square plus x into v. Or we have v plus x x square plus v square x square upon x square plus vx common from the numerator and the denominator x square 1 plus v square in the bracket upon x square 1 plus v. Or v plus x dv by dx is equal to 1 plus v square upon 1 plus v. Now on subtracting v from both the sides we get v plus x dv by dx minus v equal to 1 plus v square upon 1 plus v minus v. So we are left with x dv by dx equal to 1 plus v square minus v minus v square upon 1 plus v. x dv by dx is equal to 1 minus v upon 1 plus. Now further we can write it as here we have dv so taking these terms left inside we have 1 plus v upon 1 minus v dv x by x. Integrating both the sides let us integrate it as 2 minus 1 minus v upon 1 minus v dv equal to log A1. Or we can write it as 2 upon 1 minus v dv minus 1 minus v upon 1 minus v dv equal to log x plus c1. Solving it we have this v minus substituting back the value of v or before doing that let us solve it we have minus v. Now here let us take our constant also as log so as to simplify. So we have 2 log 1 minus v minus log c1 equal to v it can be written as log v whole square minus log x minus log c1 is equal to v. So we have log 1 minus v square 1 minus v the whole square into x into c is equal to v. So we have log x into c 1 minus v the whole square equal to v or we can write it as cx 1 minus v the whole square is equal to e raised to the power minus v. Now putting back the value of v as y upon x we have 1 minus y upon x the whole square equal to x minus y upon we have cx x minus v the whole square upon x square equal to e raised to the power minus y by x. Simplifying it a little more we have minus y the whole square upon x equal to e raised to the power minus y upon x or we have now the answer as c into x minus y the whole square equal to x e raised to the power minus y by x where x is not equal to 0. Right so this completes the session hope you understood it well enjoyed it to have a nice day.