Question

find general solution to y prime +(4/x-1)y = 1/(x-1)^5 + sin(x)/(x-1)^4

Answer 1

\[y \prime + \frac{ 4 }{ (x-1) }y = \frac{ 1 }{ (x-1)^{5} }+\frac{ \sin(x) }{ (x-1)^{4} }\]

Answer 2

I'm totally lost and I don't know how to even go about this

Answer 3

me neither

Answer 4

ughhhhhhhhh math

Answer 5

multiply throught out by (x-1)^4 and see something magic happens on left hand side

Answer 6

\[(x-1)^{4} y \prime + f(x-1)^{3} y\] ?

Answer 7

does y prime cancel out?

Answer 8

i have no idea, but wolfram does
looks like @ganeshie8 had the right idea

Answer 9

I still don't understand the procedure :(

Answer 10

the trick is to multiply something both sides so that the left hand gets into form : \[\large (y*g)'\]

Answer 11

after that, integrating both sides simplifies the left hand side to \(\large y*g\) because the integral and derivative eat eachother out

Answer 12

that special factor to be multiplied both sides hapens to be (x+1)^4 here

Answer 13

\[\large y' + \frac{ 4 }{ (x-1) }y = \frac{ 1 }{ (x-1)^{5} }+\frac{ \sin(x) }{ (x-1)^{4} } \]

Multiply (x-1)^4 both sides : (there is a standard way to find this factor)

\[ \large (x-1)^4y' + 4(x-1)^3y = \frac{ 1 }{ x-1 }+\sin x \]

Answer 14

stare at left hand side,
does it look familiar : \( \large fg' + f'g\)

Answer 15

product rule

Answer 16

@ganeshie8 i have a question
was this cooked up so this gimmick works for this question, or is there something in the method that forces it?

Answer 17

yes use it to your advantage here,
write left hand side as (fg)'

Answer 18

satellite, it works in general for equations of form :
\[\large y' + f(x) y = g(x)\]

Answer 19

we can always find such a factor,
multiply it both sides,
and write the left hand side as derivative of product of two functions : (yg)'

Answer 20

and then integrate?

Answer 21

yes integrating isolates the required function "y"
thats ur goal function when solving a differential equation

Answer 22

\[\large (x-1)^4y' + 4(x-1)^3y = \frac{ 1 }{ x-1 }+\sin x \]

rewriting left hand side using reverse product rule you get

\[\large \left((x-1)^4y\right)' = \frac{ 1 }{ x-1 }+\sin x \]

Answer 23

integrate both sides and solve y

Answer 24

okay thanks I'm going to try to solve it

Answer 25

rest of the problem is straightforward :

\[\large \int \left((x-1)^4y\right)' dy= \int \frac{ 1 }{ x-1 }+\sin x dx \]

\[\large (x-1)^4y= \int \frac{ 1 }{ x-1 }+\sin x dx \]


evaluate the right hand side integral and you're done!

Answer 26

So just to verify, \[y=\frac{ -1 }{ 4 }-\frac{ \cos(x) }{ (x-1)^4 }\]