Year 12s: what will be the first way you research your future options? >>

Differential Equations

me stuck

how does one solve the following such that it satisfies the condition lim(x->0) y(x) = 0

2(x^2)(dy/dx) = (x^2) + (y^2)

well, first i think it were normal. made the substitution y= vx and rearranged it to:

2x dv/dx = (v-1)^2

then proceeded to intemagrate (prolly were i went wrong)

Int (2 / (v-1)^2 dv = Int (1/x) dx

2(v-1)ln(v-1) = lnAx

(v-1)ln(v-1) = lnBx

here on in me no likey

i tried

ln[(v-1)^(v-1)] = lnBX

(v-1)^(v-1) = Bx

sub back in and...

(y/x -1)^(y/x - 1) = Bx

then what?

cheers for any replies :smile:

Reply 1

greeniev

How did you manage to get

El Chueco

Int (2 / (v-1)^2 dv = Int (1/x) dx

2(v-1)ln(v-1) = lnAx

because after the former step I got:
ln |(v-1)^2| = ln x + C

Might be wrong though... :confused:

Reply 2

m:)ckel

Surely the integral of 2/(v-1)^2 is -2/(v-1)

??

Reply 3

El Stevo

mockelinho

Surely the integral of 2/(v-1)^2 is -2/(v-1)

??

i knew it was somethign daft like that - damn a level maths for persuading me to logify everything...

Reply 4

El Stevo

greeniev

How did you manage to get

because after the former step I got:
ln |(v-1)^2| = ln x + C

Might be wrong though... :confused:

cheers for the try, but Int f'(x)/f(x) dx = ln(f(x)) + C, and the derivative of the bottom row is 2v-2 not 1, so your workign doesnt hold/

[C can be written as lnA, and use laws of logs to get ln(Af(x))]

Reply 5

greeniev