00:02
In the given question, we have to solve the differential equations based on reducible separable form, based on reducible, separable form.
00:18
So we have given the first differential equation is, dy over d x is equal to y minus x over y plus x.
00:28
We will simplify it by substituting x plus y is equal to.
00:33
This gives us y is equal to v minus x so we get if we differentiate it with respect to y we get 1 plus dyy over d x is equal to dv over d x therefore we have d y over d x is equal to dv over d x minus 1 so this gives us d .y over dv over d x minus 1 is equal to we can rewrite it as v minus x minus x over v.
01:08
So, simplifying it, we have dv over dx is equal to v minus 2 times x over v minus 1.
01:22
So we have dv over d x is equal to v minus 2x minus v over v which is equal to negative 2 times x over v.
01:32
So we have v times dv is equal to negative 2 times x over d x.
01:39
Now if we integrate it from the both side, we get the value of integrally v square over 2 is equal to negative x square plus c.
01:49
So we have v square is equal to negative 2 times x squared plus c1 since we denote c as a 2 times sorry, we denote c1 as a 2 times c please substitute the value of v we get the solution is x plus y whole square is equal to negative 2 times x square plus c1 which is a required solution for the first differential equation now we have given the second differential equation is x square plus 3 times y square d x 3 times y square d x minus 2 times x y is equal to 0 this gives us d y over d x is equal to x squared sorry x square plus 3y square over 2 times x y again we will simplify it by substituting y is equal to v times x so differentiating it we get d y over d x is equal to v plus x times dv over d x so this gives us v plus x times dv over d x is equals to x square plus three times x square plus three times v square x square over two times vx that is we how v plus x times t v over d x is equals to x squared times 1 plus 3 times v square over 2 times x squared times v here we cancel out the x square from numerator and denominator so we get this value is 1 plus 3 times v square over 2 times sorry 2 times v that is we get x times x times we get x times dv over d x is equals to 1 plus 3 times v squared over 2 times v minus v so we have x times d v is equal to 1 plus 3 times v square minus 2 times v square over 2 times v which is equal to 1 plus v square over sorry 1 plus 1 plus v square over 2 times v so, changing the terms to v over 1 plus v squared times dv is equal to dx over x.
04:27
Now if we integrate this obtained equation from both the side, we get the value of integrally's log of 1 plus v square is equal to log x plus constant of integration log c.
04:40
That is we have 1 plus v square is equals to x times c.
04:46
By using logarithmic rule.
04:48
So we get v square is equal to x times c minus 1.
04:52
The substitute the value of v, we get y square over x square is equal to c times x minus 1...