Question

3. First order ordinary differential equation) A solution containing 90% by volume of alcohol (in water) runs at 1 liter per
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Answer #1

a)

= 0.9-0.01N dt

dNdt 0 0.9-0.01N, Jo

log(0.9 - 0.01N)N -0.01

log(0.9 0.01N-0.01t

log(0.9-0.01N) _ log(0.9)--0.012

.9-0.01N0.01t log( 0.9

9-0.01N -0.01t 0.9

0.9-0.01N-0.9e-0.01t

0.01N0.9(1 001t)

-0.01t N = 90(1-e

b)

= 0.9-0.01N dt

By FDM

\small \frac{N(t+h)-N(t)}{h} = 0.9 - 0.01N(t+1)

take : h=Δι=1

0.9 _ 0.01x(t + 1)

\small N(t+1)-N(t) = 0.9 - 0.01N(t+1)

1.01 * Mt + 1) = 0.9 + N(t)

This formula can be used to iterate from t = 0 to 200 seconds.

c)

Matlab Code:

t = 0:1:200;


ya = 90*(1-exp(-0.01.*t));

yb = zeros(201,1);

for i = 2:201
yb(i) = (0.9 + yb(i-1))/1.01;
end

[t,yc] = ode45(@deriv,t,0.9);
plot(t,ya);
hold on;
plot(t,yb);
hold on;
plot(t,yc);
hold on;
legend('a','b','c');

function dndt = deriv(t,N)

dndt = 0.9 - 0.01*N;
  
end

80 70 60 40 20 10 0 20 140 200 40 60 80 100 160 180 120

d)

as \small t\rightarrow \infty

-0.01t N = 90(1-e

\small e^{-0.01t} \rightarrow 0

So, N = 90.

Since all three method closely follow each other, they end up with same result.

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