# Solving an equation with first degree polynomial and an exponential

For my Math C assignment I have to solve this function in regards to a falling mass with resistance 'kv' where $k=0.005$ and $g=9.8$. I have anti-derived the function from an acceleration to a velocity to finally a displacement. I would like to know how I can solve this algebraically without using graphing software, I have solved to 27.0811.

$$2400=196x+3920e^{-0.05x}-3920$$

• this can be solved by a numerical method or the LambertW function – Dr. Sonnhard Graubner May 8 '17 at 11:25
• @Dr.SonnhardGraubner are you able to link any papers to solving this using the specified method. Thanks. – Carlos Bray May 8 '17 at 11:32
• yes i have see here mathworld.wolfram.com/NewtonsMethod.html – Dr. Sonnhard Graubner May 8 '17 at 11:33
• @JeanMarie thankyou – Carlos Bray May 8 '17 at 11:57

Your equation can be written as the equation verified by the abscissas of the intersection points of the 2 curves with equations:

$$y=3920\exp(-0.05x) \ \ \ \text{and} \ \ \ y=-196x+6320.$$

(see figure below).

In fact, there are two roots, the one you have already found, and a negative one around $-18.7$, maybe without physical significance.

It is easy to device a fixed point iteration method for finding an accurate value of the positive root ; we rewrite the equation as

$$x=f(x) \ \ \ \text{with} \ \ \ f(x):=(6320-3920\exp(-0.05x))/196$$

and build recurrence relationship

$$x_{n+1}=(6320-3920\exp(-0.05x_n))/196 \ \ \ \text{with} \ \ \ x_0=20.$$

which converges to $L=27.081065708166108\cdots$ (the convergence is ensured by the fact that $|f'(x)|=\tfrac{39.2}{196}exp(-0.05x) < 1$ for all $x>0$.) In black, the curve with equation $y=3920\exp(-0.05x)$; in red, the straight line with equation $y=-196x+6320.$

Using a similar method, the other root is found to be $-18.700406404095428\cdots$.

As mentioned, there is the Lambert W method. The Lambert W function is defined so that $y=x e^x$ is equivalent to $x = W(y)$.

So in our problem $$2400=196x+3920e^{-x/20}-3920$$ rearrange to get $$\frac{79}{49} - \frac{x}{20} = e^{-x/20} \\ e^{79/49}\left(\frac{79}{49} - \frac{x}{20}\right) = e^{79/49 -x/20}$$ So that if $y = \frac{x}{20} - \frac{79}{49}$ our equation becomes $$-e^{79/49} y = e^{-y} \\ ye^y = -e^{-79/49} \\ y = W\left(-e^{-79/49}\right) \\ \frac{x}{20} - \frac{79}{49} = W\left(-e^{-79/49}\right) \\ x = 20 \;W\left(-e^{-79/49}\right) + \frac{1580}{49}$$

In fact $W$ is multi-valued, and we get all complex solutions of your equation by taking all the branches of $W$. The two real solutions are $$20 \;W_0\left(-e^{-79/49}\right) + \frac{1580}{49} \approx 27.081065 \\ 20 \;W_{-1}\left(-e^{-79/49}\right) + \frac{1580}{49} \approx -18.700406$$

• [+1] Interesting ! – Jean Marie May 8 '17 at 12:34