00:01
Hello, i am your teacher for the question.
00:03
You need to solve out this ordinary differential equation, reducing it to the homogeneous equation.
00:12
So, dy by dx is given as x plus 3y plus 2 over 2x plus 4y minus of 2.
00:27
Now just see that this is an ordinary differential equation.
00:31
You can see this as that a1 upon a2 is 1 by 2 wherever b1 upon b2 is equals to 3 by 4 coefficient ratio of y's coefficient ratios of axis coefficients right so these are not going to be equal so hence this is an ordinary differential equation we need to convert it to homogeneous differential equations so what we will do we will be writing that x can be written as capital x plus h then y can be written as a written or replaced it with y plus k so i'll be using this substitution here so i'll be writing that d y by d x you can see here from here you can write that d x is going to be d x and then d of y can be written as d of y all right where h and k are coefficients we will be evaluating them so just see that we can write first of all d y by d x d y by d x as x plus 3y plus 2 so i'll be writing that capital x plus 3y plus h plus 3k this must be there not x plus h y plus k so hence h plus 3k and then plus of 2 plus of 2 over 2 h sorry 2 x 2 x 2 times x plus 4 4 y 2x plus 4y plus 2h plus 4k minus of 2 all right so just see that h plus 3k plus 2 and then 2h plus 4k minus 2 so i'll be writing this h and k as that h plus 3k plus 2 has to be equal to 0 and then an outer 2 h plus 4k minus 2 is equal to 0 h and k are in such a way.
02:36
So we will be solving this equation simultaneously, right? so what i'll be doing is that is that we are writing this as first and second equation.
02:45
So i'll be first multiplying this first equation with 2 and then we'll be writing here 2h plus 6k plus 2 twos of 4 and that is equal to 0.
02:56
Then we will be subtracting them both.
02:58
So minus minus and then minus.
03:01
So you'll be writing 4k minus 6k can be written as minus of 2k.
03:04
And then and after minus of 6 and that is equal to 0.
03:07
So i'll be writing this as minus 2k equal to 6.
03:11
So i'll be writing k can be written as minus of 3.
03:14
So value of k is minus 3.
03:16
Now i'll be writing that h plus 3 times minus 3, h plus 3 times minus 3 plus 2 equals to 0.
03:26
H plus 3 times minus 3 plus 2 equals to 0.
03:29
So i'll be writing h minus 9 plus 2 equals to 0.
03:32
So you'll be writing that value of h can be written as positive 7.
03:37
Positive 7 minus 9 plus 2 is negative 7.
03:40
So you'll be writing this as a positive 7 here.
03:43
So h is going to be positive 7.
03:45
Now, by replacing h and k's value here, you will be able to write this as 0.
03:51
So you will be writing only x plus 3y over 2x plus 4y.
03:55
So now, d, y by dx can be written as x plus 3y over...
