00:01
Hello student, we have to solve given differential equation by variable separable method, right? given differential equation is, right, first question, d .y, divide by d x, right, plus, x plus 2 is multiplied by y cube is equal to 0.
00:21
Now, this one can be written as d .y divide by d x is equal to minus x plus 2.
00:30
Write x plus 2 is multiplied by y3 right y cube right now what we will get by variable separable method we can write d y divide divide by y cube is equal to minus right x plus 2 multiplied by d x now i am going to take integration on both side what we will get now this one can be written as integration now y power minus 3 is multiplied by d .y is equal to right integration of x plus 2.
01:07
Right, here minus is there, x squared, divide by 2 plus 2x plus 6 plus c is the integration constant.
01:16
Here integration of this one is right, y power minus 3 plus 1.
01:23
Divide by minus 3 plus 1 is equal to minus x square divide by x squared, divide by 2 ,000, divide by 2 plus 2 x right plus c here what we will get 1 divide by right 1 divide by y square here minus 2 is there right minus minus get cancelled what we will get 2 divide by right 1 divide by 2 is equal to write x square divide by 2 plus 2 x plus c this is the solution this is the solution of the given differential equation right and the second problem is second problem right a d -y divide by d x plus right plus x square plus four divide by right divide by x multiplied by y square plus four y square plus four is equal to zero now this one can be written as d .y, divide by d x, is equals to minus of x square plus 4, divide by x times y square plus 4...