- #1
chaotixmonjuish
- 287
- 0
this is the given equation
y'=(4y-3x)/(2x-y)
and here is all the work I've done so far:
(4v-3)/(2-v)=v+x*dv/dx
i moved v over and came up with this
(-3+2v+v^2)/(2-v)=x*dv/dx
did a flip
(2-v)/(-3+2v+v^2)dv=dx/x
by partial fractions I got a=-3/2 and b=1/2
so -3/2*ln|v-1|+1/2*ln|v+3|+c=ln|x|
after making the -3/2 and 1/2 powers to ln and throwing e in
(v-1)^(-3/2)+(v+3)^(1/2)+c=x
or
((y+x)/x)^(-3/2)+((y+3x)/x)^(1/2)+c=x
that's the part I'm stuck at.
y'=(4y-3x)/(2x-y)
and here is all the work I've done so far:
(4v-3)/(2-v)=v+x*dv/dx
i moved v over and came up with this
(-3+2v+v^2)/(2-v)=x*dv/dx
did a flip
(2-v)/(-3+2v+v^2)dv=dx/x
by partial fractions I got a=-3/2 and b=1/2
so -3/2*ln|v-1|+1/2*ln|v+3|+c=ln|x|
after making the -3/2 and 1/2 powers to ln and throwing e in
(v-1)^(-3/2)+(v+3)^(1/2)+c=x
or
((y+x)/x)^(-3/2)+((y+3x)/x)^(1/2)+c=x
that's the part I'm stuck at.