Solving this differential equation

[Mark44]

From post #14:

on integration,i have $ln x= -1.5 ln(3v+1)-0.5 ln (v-3)$

It took me a while to find it, but you have a mistake in the line above. Check your work for this integral, which produces the first term on the right side, above.
$$\int \frac{-3/2}{3v + 1}dv$$
To do this, use the substitution u = 3v + 1, so du = 3dv.
I think you'll find that you don't get $-1.5 \ln(3v + 1)$.

 

[chwala]

so integrating that yields,
$-0.5 ln (3v+1)$
therefore, we now have
$ln x + 0.5 ln (3v+1)+0.5 ln (v-3)=ln C$
substituting $y=vx$ and simplifying leads to,
$(3y+x)(y-3x)=c$
$3y^2-9xy+xy-3x^2=c$
$3y^2-8xy-3x^2=c$
on multiplying both sides by $-1$ and dividing by $2$ yields,

$-\frac {3y^2}{2}$+$4xy$+ $\frac {3x^2}{2}$

$\frac {3x^2}{2}$+$4xy$ $-\frac {3y^2}{2}$ the desired result.

 

[Mark44]

$3y^2-8xy-3x^2=c$
on multiplying both sides by $-1$ and dividing by $2$ yields,

$-\frac {3y^2}{2}$+$4xy$+ $\frac {3x^2}{2}$

$\frac {3x^2}{2}$+$4xy$ $-\frac {3y^2}{2}$ the desired result.

No, not quite. The "desired result" should be an equation,, as I wrote in post #27. $\frac {3x^2}{2}$+$4xy$ $-\frac {3y^2}{2} = C$
The constant C that I wrote is equal to the constant that you wrote, c, multiplied by -1/2.

 

[chwala]

aaaaargh left out the constant again...nice day mark