I want to find values for $a$, $b$, and $c$ that will make $y(x)$ a 'more linear' equation from $x=0$ to $x=100$ where $y(0)=0$. How do I find values for $a$, $b$, and $c$ that satisfy the two conditions below?
Values should satisfy the following two conditions:
1) $y(0)=0$
2) $y(50)=(1/2)y(100)$
Equation I wish to 'linearize':
$y(x)=5(ab-d(x)c)/(b(a+c))$
$d(x)=10000exp(3900(1/(x+273.15)+1/298))$
My attempt:
$y(0)=0$ is satisfied when $d(0)/a=b/c$
$d(0)=32891$ therefore $a=32891c/b$
Substituting $a=32891c/b$ into the general equation $y(50)=(1/2)y(100)$ yields:
$c/b=0.08552$
Substituting $c/b=0.08552$ into the general equation $d(0)/a=b/c$ yields:
$a=384600$
I don't really know where to go from here to find $b$ and $c$ (or even if I correctly solved for a). The $d(x)$ in the equation makes this difficult for me to set up a system of equations.