Question

s

Answer 1

@lgbasallote @EulerGroupie

Answer 2

ok..i admit it...this is a good one

Answer 3

im getting a tingly feeling something's wrong in this question =_=

Answer 4

Haha, I dont think so..

Answer 5

you'd need to change the variable , use \[\frac{y}{x} = t\]differentiate this with respect to x and change the equation so that all y/x is replaced everywhere by t, also in the square root part use \[x^2-y^2=(x+y)(x-y)\]

Answer 6

lol nice. didnt see that =))

Answer 7

but dy/dx changes too wont it?

Answer 8

dy/dx = t + sqrt( 1-t^2)

Answer 9

i think you do something to the dy/dx side dont know what thouhg

Answer 10

\[\frac{y}{x}= t\]So,\[y= tx\]\[\frac{dy}{dx}=\frac{dt}{dx} +1\]

Answer 11

you have to do product rule there

Answer 12

so you would get t + x(dt/dx)

Answer 13

Sorry, made a typo in the previous post\[\frac{dy}{dx} =\frac{dt}{dx} +t\] so the new equation becomes \[\frac{dt}{dx} +t = t+\sqrt{1-t^2}\]simplify and it reduces to form you'd be able to solve

Answer 14

dt/dx =sqrt(1-t^2) then what?

Answer 15

wait there hsould be an x in front of the dy/dx
xdt/dx =sqrt(1-t^2)

Answer 16

then you move the dt and dx's to the right side and integrate to get sin(t)^-1 +c = ln(x) + c. right?

Answer 17

\[\frac{dt}{\sqrt{1-t^2}} =dx\]integrate both sides, you'd get \[sin^{-1}t= x +C\] now plug \[\frac{y}{x}=t\]\[Sin^{-1}\frac{y}{x} = x+C\] Use, y(1)=0\[Sin^{-1}0= 1+C\]hence, C= -1, plug this back into\[Sin^{-1}\frac{y}{x} = x+C\]\[Sin^{-1}\frac{y}{x} = x-1\]So we get, \[y = x Sin(x-1)\] this is the particular solution of the given differential equation.

Answer 18

alright thanks

Answer 19

You are welcome, it was fun. Good day.

Answer 20

wait quick question where did the x go

Answer 21

Becuase you have to do the product rule to get t + x(dt/dx)

Answer 22

I am having that issue too. It gave me a natural log down the road.

Answer 23

yeah same. do you think he made that mistake?

Answer 24

Yeah, I finally found my book. That is how the substitution usually goes. I am scanning my work... brb

Answer 25

Alright. Thanks man

Answer 26

rhs=(1/x)dx
which leads to ln x +c
instead of x+c

Answer 27

Yeah I got the same stuff @EulerGroupie. @hartnn what is rhs

Answer 28

right hand side

Answer 29

oh. so we got it right?

Answer 30

your work is correct @EulerGroupie

Answer 31

Thank you, couldn't have done it without @akitav though.

Answer 32

thanks again for the help guy s