Question

xy' - y = x^3
using the method of variation of parameter ???

Answer 1

Do you know something about the variation of parameter method?

Answer 2

it is one order diff equ i know the second order and higher so i could not do this

Answer 3

@nitz and also your attachment is full of second order bro

Answer 4

ok wait

Answer 5

i m waiting

Answer 6

Nitz is not bro she is a sis..

Answer 7

@waterineyes good observation :)

Answer 8

Thanks bro..

Answer 9

@nitz any improvements sis

Answer 10

?

Answer 11

can you please wait for 5 min i am solving someone other problem

Answer 12

xy' - y = x^3

let y = xu

then y' = u + u'x

x(u + u'x) - ux = x^3

hence u + u'x - u = x^2

hence u'x = x^2

hence u' = x

hence u = x^2/2 + C

hence y/x = x^2/2 + C

hence y = x^3/2 + Cx