00:01
The given differential equation we have is x plus 2y minus y dx plus 2x plus 4y minus 3 dy is equal to 0.
00:12
This will imply that 2x plus 4y minus 3, dy dy is equal to negative of x plus 2y minus 1 d x.
00:22
So we can say that dyy upon d x is equal to negative x plus 2y minus 1 and we can take 2 common from this part.
00:31
And we can write down x plus 2y minus 3 so this is our equation 1 now let x plus 2y be equal to let x plus 2y equal to v therefore we can say that 1 plus 2 d y upon d x is going to be equal to dv upon d x so here 2d y by d x is equal to dv by d x minus 1 from this we can easily say that d y upon d x is equal to half d v by d x minus half from one we have is half dv here we can say from one we can write down half dv upon dv x is equal to here we can have this negative half over here so negative half is equal to minus v minus one divided by 2 v minus 3 so when we do this this is going to be half dv by d x is equal to 2v minus 3 minus 2 v minus 1 divided by 2v minus 3 2v minus 3 therefore dv by d x is equal to negative 1 divided by 2v minus 3 this implies that 2v minus 3 dv is equal to negative d x then what we can do we can integrate this so on integrating on integrating what we can have is integration of 2v minus 3 dv is equal to negative integration of d x and when we solve the integration part of this so we are going to get the value as v square minus 3v plus x is equal to c from this putting the value of v that is equal to x plus 2y this is going to give x plus 2y the whole square minus 3 into x plus 2y plus x is equal to c so from this we conclude that x square plus 4 x y plus y square that is 4 y square this will become this will have 4 y square minus 2x minus 6y is equal to c is the required general solution general solution for the first of a part moving forward to the second part we have the differential equation as 2x plus 3y minus 1 d x plus 2x plus 3y minus 5 dy is equal to 0 then we can say that dy up on d x is equal to negative 2x plus 3y minus 1 divided by 2x plus 3y minus 5 this is equation number 1 now we'll say this then let 2x plus 3y let 2x plus 3y is equal to therefore we can say that 2 plus 3 d y upon d x is equal to d v by d x so this is going to be d y upon d x is equal to 1 by 3 times dv by d x minus 2 by 3 from 1 we can say that 1 upon 3 dv by d x minus 2 by 3 is equal to negative v minus 1 divided by v minus v minus now further solving what we can do is we can take this 2 by 3 on the right hand side and this will become 1 by 3 dv upon d x and after taking 2 by 3 and multiplying it so we are going to get 2 v minus 5 minus 3 v minus 1 divided by 3 v minus y after solving this we are going to get that d y by d x taking 3 over there 1 by 3 over there so we will get divide by d x is equal to minus v plus 7 divided by v minus 5.
04:50
This is v minus 5, which we are going to obtain.
04:55
What we can do? we can say that v plus 7 minus 12, or we can say there is one more step before that, that is going to be minus v minus 5 by v plus 7, dv is equal to negative d x.
05:14
We can say, write this as 5 as v plus 7 minus 12.
05:20
And this is v plus 7 this is dv is equal to negative d x so this can be written as v plus 7 v plus 7 dv minus 12 by v plus 7 dv is equal to minus d x here this is going to be dv minus 12 by v plus 7 dv minus 12 by v plus 7 dv is equal to negative d x now we can integrate on both sides so on integrating integration of dv minus integration of 12 by v plus 7 dv is equal to negative integration of d x when we solve this this is going to be v minus 12 log of v plus 7 is equal to minus x plus c from this we can conclude that 3x plus 3y minus 12 log of 2x plus 3y we have substituted the value of v over there so plus 7 is equal to c and where c is an arbitrary constant therefore the general required solution is this one so this will be our general required solution now moving forward to the second part of this question like the given differential equation which we have for this part is 2x minus 3y d x minus 3x plus 2y dy is equal to 0 this equation can be written as 2x minus 3y d x minus 3y d x plus negative 3x negative 2y dy is equal to 0 this is of the form m d x plus n dy is equal to 0 where value of m is equal to 2x minus 3y and value of n is equal to minus 3x minus 2y so from this we can conclude that dm we can see the partial derivative of this now d m dy is equal to negative 3 and d n d x is equal to negative 3 for m n values right so we can say that del m by del y is equal to del n here we have small n so we'll take small n only so small n d x is going to be equal to negative so the equation is an exact equation so it is an exact equation therefore integration of m d x is equal to integration of 2x minus 3y d x which is going to give the value as x square minus 3xy similarly integration of n dy is equal to integration of minus 3x minus 2y which is going to give the value as 3xy minus y square so the required general solution is therefore required general solution is x square minus 3xy x square minus 3xy minus y square is equal to c we are rejecting the repeated terms that is c is an arbitrary constant so here this is the required general solution i'm moving forward to the second part of the b question so the given differential equation we have now is xy d xy d xy minus x square plus 2y square d y d y is equal to 0 so we can write this as x y d x x x xx is equal to x square plus 2y square so next we can say that dx is going to be equal to, we can divide this by xy, so this becomes x divided by y plus 2 y by x, d y by x, next we can say that d y upon d x, d y upon d x is equal to 1 divided by 1 by x, 1 by y by x plus 2 y by x...