Question

Let y'(x) + y(x)g'(x) = g(x)g'(x), y(0) = 0, x Î R, where ƒ'(x) denotes
dƒ(x)/dx
and g(x) is a given non-constant
differentiable function on R with g(0) = g(2) = 0. Then the value of y(2) is

Answer 1

we need to multiply both sides by g(x)

Answer 2

(gy)'=g^2g'

Answer 3

hmm i don't know let me think some more

Answer 4

hmm i don't know let me think some more

Answer 5

i thing i got divide both sides by g'
so we have
y'/g'+y=g
so we want
(1/g')'=1
=> 1/g'=x
=>g'=1/x
so then we have
(xy)'=g
=>
xy=g+C
y=g/x+C/x

Answer 6

if g'=1/x
then g=lnx+C

Answer 7

but this isn't defined at 0
so again i don't know lol

Answer 8

i made a mistake above
we have
(xy)'=g right!
=>(xy)'=1/x
xy=ln(x)+C
=>
y=ln(x)/x+C/x
but still this doesn't help

Answer 9

try determining c

Answer 10

oh we are given y(0)=0
but still we can't plug in
ln(0) doesn't exit
1/0 also doesn't exist

Answer 11

wait...

Answer 12

i think i got it
so we have y'+yg'=gg'
if we multiply both sides by v we get
vy'+vg'y=vgg'

so we want this v'=vg'
so we have dv/dx=v*dg/dx
dv=v*dg
1/v dv=dg
integrating both sides gives
ln|v|=g+k
let k=0
so
v=e^g

so we have
e^g*y'+e^g*g'y=e^g*g'
which we can write as
(e^gy)'=(e^g)'
integrate both sides
e^gy=e^g+C

y=1+C/e^g
....

Answer 13

i have no clue :(

Answer 14

i have no clue :(