# dA/dt=0.5*A*(1−A/100)−10 with A(0)=70 and we want to use Euler's method to get an approximate value for A(10), with a step size of 1.

$\frac{dA}{dt}=0.5×A×\left(1-\frac{A}{100}\right)-10$
with $A\left(0\right)=70$ and we want to use Euler's method to get an approximate value for $A\left(10\right)$, with a step size of 1.
So the answer sheet says you basically have to use $\text{Ans}+0.5×\text{Ans}×\left(1-\frac{\text{Ans}}{100}\right)-10$ with the first Ans being 70, and then of course repeat 10 times.
But I'm wondering, doesn't this actually give you $\frac{dA\left(10\right)}{dt}$? How is this a correct method?
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Ricky Lamb
The Euler method does not give you $\frac{dA}{dt}$. You give it a formula for $\frac{dA}{dt}$, such as the one in your question. Then from any given point, like your start of (0,70) it puts a straight line through the point with slope $\frac{dA}{dt}$ of that point. From your expression, $\frac{dA}{dt}{|}_{\left(0,70\right)}=0.5$ so we step one unit in t at a slope of 0.5, giving the A value of the next point as 70+0.5⋅1=70.5. Now we are at (70.5,1), we calculate $\frac{dA}{dt}$ at this point and take another step along the t axis, and so on until we get to t=10
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tonan6e
Your formula is wrong. If the differential equation is $\frac{dA}{dt}=f\left(A\right)$ and your step size is $h$, the formula is ${A}_{n+1}={A}_{n}+hf\left({A}_{n}\right)$.So in this case, ${A}_{n}+0.5{A}_{n}\left(1-{A}_{n}/100\right)-10$.