How do you integrate this differential equation then solve it? - Equation included

Question

Assumption x=0 when v=0

1057 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-08T17:17:41

Given: $v \frac{d v}{d x} = 9.8 - 0.004 v^{2}, v (0) = 0$ $v(dv)/dx = 9.8 - 0.004v^2, v(0) = 0$

Use the separation of variables method:

$\frac{v}{9.8 - 0.004 v^{2}} d v = d x$ $v/(9.8 - 0.004v^2) dv = dx$

Integrate both sides:

$\int \frac{v}{9.8 - 0.004 v^{2}} d v = \int d x$ $intv/(9.8 - 0.004v^2)dv = intdx$

Multiply by 1 in the form of $\frac{- 250}{- 250} :$ $(-250)/-250:$

$\int (\frac{- 250}{- 250}) \frac{v}{9.8 - 0.004 v^{2}} d v = \int d x$ $int((-250)/-250)v/(9.8 - 0.004v^2) dv = intdx$

$- 125 \int \frac{2 v}{v^{2} - 2450} d v = \int d x$ $-125int(2v)/(v^2-2450)dv = intdx$

Let $u = v^{2}$ $u = v^2$ then $d u = 2 d v$ $du = 2dv$ :

$- 125 \int \frac{1}{u - 2450} d u = \int d x$ $-125int1/(u-2450)du = intdx$

$ln (u - 2450) = \frac{x}{- 125} + C$ $ln(u-2450) = x/-125 + C$

Reverse the substitution:

$ln (v^{2} - 2450) = \frac{x}{- 125} + C$ $ln(v^2-2450) = x/-125 + C$

$v^{2} - 2450 = e^{- 0.008 x + C}$ $v^2-2450 = e^(-0.008x + C)$

$v^{2} = 2450 + C e^{- 0.008 x}$ $v^2= 2450 + Ce^(-0.008x)$

Use the boundary condition:

$0^{2} = 2450 + C$ $0^2 = 2450+ C$

$C = - 2450$ $C = -2450$

$v^{2} = 2450 - 2450 e^{- 0.008 x}$ $v^2= 2450 - 2450e^(-0.008x)$

$v^{2} = 2450 (1 - e^{- 0.008 x})$ $v^2= 2450(1 - e^(-0.008x))$ Q.E.D.