Question

xy' - y = x^3
in the method of variation of parameters
how can i solve?

Answer 1

oh this one is not as fun haha. Anyways do you know Kramers rule?

Answer 2

yeah i know

Answer 3

alright so first find your complementary using an auxilary equation ( putting the equation in terms of m)

Answer 4

i learnt not to much but in the examples it does not have x in the LHS

Answer 5

actually in this case you will have to use another way other than auxilary

Answer 6

divide by x and you'll hvae a linear equation

Answer 7

y' - y/x = x^2 :/ nothing changes

Answer 8

solve the homogenous equation

\[y'-\frac{1}{x}y=0\]

Answer 9

are you sure about that it is a part of variation of parameters method

Answer 10

i do not see like that

Answer 11

well the first part in my book uses an auxilary equation however since you do not have the x variable you need to find it out another way.... variation of parameter has to deal with the particular solution if i'm not mistaken

Answer 12

hmm i guess you are right i do not have much information about that

Answer 13

yeah you have to have a way to find the complementary function

Answer 14

y = (x^3/2) + cx

Answer 15

one way that you might have learn is to use linear to find one solution and then using that solution to find a y_2

Answer 16

@Outkast3r09 i found the solution of h.eq but is it right ?

Answer 17

hold on i find this weird because usually variation of parameters is ued int higher order equations

Answer 18

yeah you are right

Answer 19

are you sure that's what your teacher wrote?

Answer 20

in the book :)

Answer 21

if my eyes are not damaged , it is true :D

Answer 22

since this is a first order DE, just find the integrating factor

Answer 23

it is 1/x

Answer 24

yes, multiply 1/x to every term

Answer 25

that's what i'm saying dumbcow? you'll find the general solution

Answer 26

yeah i do that and it was y = (x^3/2) + cx

Answer 27

if you do the non homogenous however reduction formula is used to find another general solution but you cannot use it... if you solve the nonhomog linear you'l get

y=yc+yp which you do not need to use variation of parameters on it

Answer 28

yeah it is easy but book says the variation of parameter :(

Answer 29

is it a darker prime... i'm looking at another site and it has a darker prime mark to show y''

Answer 30

personally i believe this is a typo i'd ask my teacher about it

Answer 31

o-oo i do not think so i even do not know what darker prime is

Answer 32

hmm i have to give it tomorrow :/