# Accumulation and Differential Equations

Accumulation 6: Differential equations

When students first learn about antiderivatives they are given simple initial value problems to solve such as $\frac{dy}{dx}=3-2x,\quad y\left( 1 \right)=-5$. They are instructed to find the antiderivative, then tack on a +C , then substitute in the initial condition, then solve for C and finally write the particular solution. So we hope to see:

$\frac{dy}{dx}=3-2x$

$y=3x-{{x}^{2}}+C$

$-5=3(1)-{{1}^{2}}+C$

$-7=C$

$y=3x-{{x}^{2}}-7$

But thinking of it as an accumulation problem you can write

$\displaystyle y=-5+\int_{1}^{x}{3-2t\,dt}$

$y=-5+\left. \left( 3t-{{t}^{2}} \right) \right|_{1}^{x}$

$y=-5+\left( 3x-{{x}^{2}} \right)-\left( 3(1)-{{1}^{2}} \right)$

$y=3x-{{x}^{2}}-7$

Now I’ll admit there is not too much difference in the amount of work involved there, but once the first line is on the paper there is not much to remember in what to do from there on. Common student mistakes in  the first method include entirely forgetting the +C and not using the initial condition correctly.

A differential equation in which dy/dx can be written as a function of x only, $\frac{dy}{dx}={f}'\left( x \right)$, with an initial condition $\left( a,f\left( a \right) \right)$ has the solution

$\displaystyle f\left( x \right)=f\left( a \right)+\int_{a}^{x}{{f}'\left( t \right)dt}$

Look familiar? Yes, that’s my favorite accumulation equation.

Furthermore, if all you need is some function values, this expression can be entered into a graphing calculator and the values calculated without finding an antiderivative.

In my first post on accumulation I discussed the AP exam question 2000 AB 4. The solution to part (a) looks like this

$\displaystyle \int_{0}^{3}{\sqrt{t+1}dt}=\left. \tfrac{2}{3}{{\left( t+1 \right)}^{3/2}} \right|_{0}^{3}=\tfrac{2}{3}\left( {{4}^{3/2}}-{{0}^{3/2}} \right)=\tfrac{14}{3}\text{ gallons}$

The scoring standard included a second method that went this way:

$L(t)=\text{ gallons leaked in first }t\text{ minutes}$

$\frac{dL}{dt}=\sqrt{t+1};\quad L\left( t \right)=\tfrac{2}{3}{{\left( t+1 \right)}^{3/2}}+C$

$L(0)=0;\quad C=-\tfrac{2}{3}$

$L(t)=\tfrac{2}{3}{{\left( t+1 \right)}^{3/2}}-\tfrac{2}{3};\quad L(3)=\tfrac{14}{3}$

You decide which is easier. If you’re still not sure compare the two methods shown for part (c) on the scoring standard.