# Differential equation with Euler's method

Unfortunately from online classes i missed this lesson and now have an assignment question that has to be solved however im struggling to work this one out! Any help and answer would be appreciated thankyou. Im very willing to learn how to solve this problem as its going to be part of my job. This forum is my last hope of learning and working this out. It would be very much appreciated.

Regards!

The differential equation of a circuit is: $$0.045\frac{dv}{dt} = 15t^2 - v,$$ where $v$ is the voltage. When $t = 0\; v = 1$.

(a) Use Euler's method to obtain a numerical solution for the range $0 >\leq t \leq 0.5$ with intervals of $0.01$.

(b) Plot the graph of $v$ against $t$ for $0 \leq t \leq 0.5$. Briefly explain the graph.

• Your enthusiasm is welcome, but please show it in a more concrete way by giving us some details about what you've attempted. Have you tried reading about Euler's method? – epimorphic May 31 '15 at 13:27
• do you have a program that implement the eludes method? – abel May 31 '15 at 13:45

euler's method for the initial value problem $$\frac{dv}{dt} = f(t, v) = \frac1{0.045} (15t^2 - v), v = v_0=1 \text{ at } t = 0.$$ is
$$v_{n+1} = v_n+hf(nh, v_n) \text{ where }v_n \simeq v(nh), v(0) = v_0, h = 0.01$$
for example, $$v_1 = 1+0.01 \times \frac1{0.045}(0-1)=0.7777$$
i did this on my ti-$83.$ here is what i got:
$\begin{array}{|c|c|c|c|c|c|}\hline t & 0.00 &0.01 & 0.02 & 0.03 &0.04 &0.05\\ \hline v & 1.00 &0.7777 &0.6053 & 0.4721 &0.3702 &0.2536 \\ \hline\end{array}$