How to plot the graphs by fixing constants of your choice
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Ibrahim Bin Umair
on 24 May 2021
Commented: Ibrahim Bin Umair
on 24 May 2021
I have this expression as output
C1*exp(-x*(9/(4*(37/8 - 10^(1/2))^(1/3)) + (37/8 - 10^(1/2))^(1/3) + 3/2)) + C2*exp(x*(9/(8*(37/8 - 10^(1/2))^(1/3)) + (37/8 - 10^(1/2))^(1/3)/2 - 3/2))*cos((3^(1/2)*x*(9/(4*(37/8 - 10^(1/2))^(1/3)) - (37/8 - 10^(1/2))^(1/3)))/2) - C3*exp(x*(9/(8*(37/8 - 10^(1/2))^(1/3)) + (37/8 - 10^(1/2))^(1/3)/2 - 3/2))*sin((3^(1/2)*x*(9/(4*(37/8 - 10^(1/2))^(1/3)) - (37/8 - 10^(1/2))^(1/3)))/2)
How can i plot the graph by fixing constants of my choice
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Accepted Answer
DGM
on 24 May 2021
Edited: DGM
on 24 May 2021
C = [1 2 3] % pick some constants
x = linspace(0,50,100); % pick an interval
y = C(1)*exp(-x*(9/(4*(37/8 - 10^(1/2))^(1/3)) + (37/8 - 10^(1/2))^(1/3) + 3/2)) ...
+ C(2)*exp(x*(9/(8*(37/8 - 10^(1/2))^(1/3)) + (37/8 - 10^(1/2))^(1/3)/2 - 3/2)) ...
.*cos((3^(1/2)*x*(9/(4*(37/8 - 10^(1/2))^(1/3)) - (37/8 - 10^(1/2))^(1/3)))/2) ...
- C(3)*exp(x*(9/(8*(37/8 - 10^(1/2))^(1/3)) + (37/8 - 10^(1/2))^(1/3)/2 - 3/2)) ...
.*sin((3^(1/2)*x*(9/(4*(37/8 - 10^(1/2))^(1/3)) - (37/8 - 10^(1/2))^(1/3)))/2);
plot(x,y)
or you could do the same thing using symbolic tools
syms x C1 C2 C3
y = C1*exp(-x*(9/(4*(37/8 - 10^(1/2))^(1/3)) + (37/8 - 10^(1/2))^(1/3) + 3/2)) ...
+ C2*exp(x*(9/(8*(37/8 - 10^(1/2))^(1/3)) + (37/8 - 10^(1/2))^(1/3)/2 - 3/2)) ...
*cos((3^(1/2)*x*(9/(4*(37/8 - 10^(1/2))^(1/3)) - (37/8 - 10^(1/2))^(1/3)))/2) ...
- C3*exp(x*(9/(8*(37/8 - 10^(1/2))^(1/3)) + (37/8 - 10^(1/2))^(1/3)/2 - 3/2)) ...
*sin((3^(1/2)*x*(9/(4*(37/8 - 10^(1/2))^(1/3)) - (37/8 - 10^(1/2))^(1/3)))/2);
y = subs(y,[C1 C2 C3],[1 2 3]); % specify the constants
fplot(y,[0 50]) % specify the interval
More Answers (1)
KSSV
on 24 May 2021
You can substitute your value using subs. Read about this function.
Also you can plot for range values using fplot. Read about it.
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