Cannot perform Laplace transformation to integrate ODE correctly

Question

Given

• two state (time-dependent) parameters: $C,A\ge0$
• these fixed parameters: $R,k_0,V,k_1,k_2\gt0$

and the following two differential equations

$\begin{cases} \frac{dC}{dt}=\frac{R}{V}-(k_1+k_0)C+\frac{k_2A}{V}\\ \frac{dA}{dt}=k_1CV-k_2A \end{cases}$

I would like to integrate the above and obtain the analytical solution for $C(t)$ for any set of non-negative initial conditions: $C(0)$ and $A(0)$.

I adopted the technique of Laplace transformation as introduced in this article.

The followings are the general step that I have followed:

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Applying Laplace transformation to domain $s$ for $A$:

$s{\cdot}A(s)-A(0)=k_1VC(s)-k_2A(s)$

$A(s)=\frac{k_1VC(s)+A(0)}{s+k_2}$ --------------------------------------- (1)

Applying Laplace transformation to domain $s$ for $C$:

$s{\cdot}C(s)-C(0)=\frac{R}{V}-(k_1+k_0)C(s)+\frac{k_2A(s)}{V}$

By substituting equation (1), followed by some simplification, we have:

$C(s)=\frac{R+VC(0)}{V}{\cdot}\frac{s+k_2+\frac{k_2A(0)}{R+VC(0)}}{s^2+s{\cdot}k_1+s{\cdot}k_0+s{\cdot}k_2+k_2{\cdot}k_0}$

Then, the denominator can be simplified with the quadratic formula:

$\begin{cases} C(s)=\frac{R+VC(0)}{V}{\cdot}\frac{s+k_2+\frac{k_2A(0)}{R+VC(0)}}{(s+\alpha)(s+\beta)}\\ \alpha,\beta=\frac{k_1+k_2+k_0\pm\sqrt{(k_1+k_2+k_0)^2-4k_2{\cdot}k_0}}{2} \end{cases}$

This form contains a constant term times a proper polynomial in $s$ (i.e. degree in the numerator is smaller than the denominator), so it can be converted back to the time domain by inverse Laplace transform:

$C(t)=\frac{R+VC(0)}{V}\left(\frac{-\alpha+k_2+\frac{k_2A(0)}{R+VC(0)}}{\beta-\alpha}e^{-\alpha t}+\frac{-\beta+k_2+\frac{k_2A(0)}{R+VC(0)}}{\alpha-\beta}e^{-\beta t}\right)$

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The problem is, I verified my solution against a differential equation solver and they gave very different numbers... So I suspect that there are some errors with my integration, but I cannot identify the error... Can anyone please help check my answer?

Answer 1

The Laplace transform of $R/V_p$ is $R/(sV_p)$. Does that fix your problem?

Answer 2

This all seems too complicated. Take the first equation:

$$\frac{dC}{dt}=\frac{R}{V}-(k_1+k_0)C+\frac{k_2A}{V}$$ and differentiate it with respect to $t,$ to get

$$\ddot{C} = \frac{R}{V}-(k_1+k_0)\dot{C}+\frac{k_2\dot{A}}{V}.$$

Plug in the second equation for $\dot{A}$ and get

$$\ddot{C} = \frac{R}{V}-(k_1+k_0)\dot{C}+\frac{k_2k_1CV-k_2A}{V}.$$

Now, there is still a pesky $A$ but this can be solved from from the first equation:

$$A = V(\dot{C} - \frac{R}{V} - (k_1 + k_0)C).$$ Substitute this above, and you have a second order linear equation with constant coefficients, which, I assume, you know how to solve.