How to use initial conditions with an equation?
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Luis
le 18 Juin 2014
Réponse apportée : Star Strider
le 18 Juin 2014
Hello I'm wondering how can I use the given initial conditions to an equation e.g. x=0 and y=1 to be used on an equation e.g. Y=2*x^3+6*x^2-10*x+4 and its derivative dydx=6*x^2+12x-10 when I try to plot Y and dydx from 0 to 6 with 0.1 steps. Do I need a for loop or something of that matter?
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Star Strider
le 18 Juin 2014
The easiest way is to use polyval:
% Y=2*x^3+6*x^2-10*x+4
% dydx=6*x^2+12x-10
Y = [2 6 -10 4]; % EXPRESS COEFFICIENTS AS A VECTOR
dydx = [6 12 -10];
x = 0:0.1:6;
Y_x = polyval(Y,x);
dydx_x = polyval(dydx,x);
figure(1)
plot(x, Y_x, x,dydx_x)
grid
legend('Y', '^{dy}/_{dx}', 'Location', 'NW')
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Geoff Hayes
le 18 Juin 2014
Luis - no you don't need to use a for loop, but you will have to adjust your equations so that they can compute elementwise operations (i.e. handle vector input rather than just a scalar input).
To do this, any operation that affects your vector must be preceded by a period. Your first equation
Y=2*x^3+6*x^2-10*x+4
becomes
Y=2*x.^3+6*x.^2-10*x+4
Note that it was only necessary to precede the exponent operator with a period; all other operations are handled normally.
Then with
x=0:0.1:6
you can calculate Y with the above equation. You can do the same sort of thing for the derivative.
Note that you mention that your initial conditions are x=0 and y=1. Why is y=1, since if x=0 then your equation evaluates to 4?
Try the above and see what happens!
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