Differential Equation

This is a separable equation. It means that x and v can be separated to the right and the left respectively.

So you can put x on one side and v on another side and integrate.

x dv/dx = (-3v² - 8v + 3)/(3v + 4)
(3v + 4)/(-3v² - 8v + 3) dv = 1/x dx
∫ (3v + 4)/(-3v² - 8v + 3) dv = ∫ 1/x dx
-1/2 ∫ 1/(-3v² - 8v + 3) d(-3v² - 8v + 3) = ln |x|
-1/2 ln |-3v² - 8v + 3| + C1 = ln |x|
ln (1/√|-3v² - 8v + 3| * e^C1) = ln |x|
|x| = e^C1/√|-3v² - 8v + 3|
|x| = C2/√|-3v² - 8v + 3|, where C2 = e^C1


Hope it helps! ^^

xdv/dx=(-3v^2-8v+3)/(3v+4)
(3v+4)dv/(-3v^2-8v+3) = dx/x
Integrate [(3v+4)dv/(-3v^2-8v+3)] = Integrate (dx/x)
-Integrate [(3v+4)dv/(3v^2+8v-3)] = Integrate (dx/x)
(1/2)Integrate [d(3v^2+8v-3)/(3v^2+8v-3)] = Integrate (dx/x)
(1/2) ln |3v^2+8v-3| + C = ln |x| where C is an arbitrary constant (for indefinite integration)

Taking exponent on both sides:
exp [(1/2) ln |3v^2+8v-3| + C] = exp [ln |x|]
C' {exp [ln |3v^2+8v-3|]}^(1/2) = |x| where C' is another arbitrary constant equal to exp (C)
C' [|3v^2+8v-3|]^(1/2) = |x|

Since both C' = exp(C) and [|3v^2+8v-3|]^(1/2) must be positive values, the absolute sign for x may be removed.

Thus,
x = C' [|3v^2+8v-3|]^(1/2)

The value of C' (or C) is determined based on the initial value given in the question, if any.

xdv/dx=(-3v^2-8v+3)/(3v+4)
(3v+4) / (-3v^2-8v+3)dv/dx=1/x
∫(3v+4) / (-3v^2-8v+3)dv = ∫1/x dx
-1/2 ∫1 / (-3v^2-8v+3)d(-3v^2-8v+3) = ∫1/x dx
∫1 / (-3v^2-8v+3)d(-3v^2-8v+3) = -2∫1/x dx
ln((-3v^2-8v+3)) = -2lnx + C
-3v^2-8v+3 = kx^-2
x^2(-3v^2-8v+3) = k
where c,k are constants