00:01
In the problem, we have to solve the differential equation, that is, d .y upon d x, that equals minus y into 1 plus 7x to the 7x into y to the power 3.
00:12
So, d .y upon dx equals to minus y, minus 7x, y to power 4.
00:19
Now, this can be written as d .y upon dx plus y that equals minus 7x, y to the power 4.
00:25
So here, this is of the linear differential equation form.
00:29
Now, we can write in this form, that is.
00:32
Is 1 upon y to the power 4 d y upon x plus y upon y to the power 4 that equals to minus 7 x so here we'll assume y to per minus 3 that equals u so this becoming minus 3 y to power minus 4 to y upon d by x that equals to u upon d x minus 3 upon d u upon d x minus 3 to up on d x minus 3 upon d up on d x or here it is 1 upon y to power 4 d y to y by by by x that equals 1 upon 3 t u upon d x so we will put the very dev.
01:03
Values this one in this equation.
01:07
So after putting the value, we have this equals to minus 1 upon 3, d u x plus u, that equals to minus 7x.
01:16
And now, d u upon d x minus 3 u that equals to 27x.
01:23
So this is of the form that is d u upon d x plus 1 minus n px u that equals to 1 minus n qx.
01:31
So from these we obtained the value of n which is equal to 4...