Solve for, and then substitute.

I posted before, but I am not getting the answers I am looking for. This is a simplified version of what I want to do.

say, f(t)=y*t, and we are given an initial condition f(1)=10. We can use this information to solve for y..

Now that we know, y=10, solve for f(2)=?, obviously f(2) will equal 20.

How can I write this code in MATLAB?

[Kyle Langford]

I am trying to solve this:

given

y(1.2)=80

, and the equation

y(t)==K*A + (y0-K*A)*exp(-t/T)
K=1
A=100
y0=0
y_t=80 %y(1.2)=80
t=1.2 
eq1=y_t==K*A + (y0-K*A)*exp(-t/T)
vpasolve(eq1,T)

This gives me that T=0.7456.

Now using knowing T, solve for y(1.5)

[Kyle Langford]

I can do this, but i was trying but I am trying to find a more creative way to code this.

clear;clc;
syms x
syms
K=1
A=100
y0=0
y_t=80 %y(1.2)=80
t=1.2 %subs for t=1.5
eq1=y_t==K*A + (y0-K*A)*exp(-t/x)
T=vpasolve(eq1,x)
clear eq1
y=@(t) K*A + (y0-K*A)*exp(-t/T)
y(1.5)

Answer 1

[Walter Roberson]

K=1

K = 1

A=100

A = 100

y0=0

y0 = 0

syms y(t) T
y(t) = K*A + (y0-K*A)*exp(-t/T)

y(t) = 

y_known = 80 %y(1.2)=80

y_known = 80

t_known = 1.2 

t_known = 1.2000

eq1 = y(t_known) == y_known

eq1 = 

sol_T = solve(eq1, T)

sol_T = 

y(t) = subs(y(t), T, sol_T)

y(t) = 

y(1.5)

ans = 

vpa(ans)

ans = 

86.625193900471559519935338534827

[Kyle Langford]

@Walter Roberson as always, I appreciate you!

Answer 2

[Arif Hoq]

you can use anonymous function:

y=10;
f=@(t) y*t;
f(1)

ans = 10

f(2)

ans = 20

[Kyle Langford]

Although that does work, that is not what I asked. In this specific scenario, it is easy to figure out y. In a more complex scenario, it would not be so easy.

I think you are on to something, but how could you solve for y given the initial condition f(1)=10, and then additionally solve for f(2) using code?