# How would I integrate this ODE representing an production and degradation biochemical reaction?

I'm a biologist trying to model a production and degradation reaction of species x which is dependent on species y over time. I'm used to using software to integrate my differential equations numerically (because most ODEs I deal with generally do not have analytical solutions). However, I'm particurarly interested in seeing whether the following equation has an analytical solution and how to find it if there is one. I do not however have enough time to go back to the basics of calculus as I'm on a limited time frame. My question is therefore, can this equation be integrated analytically and how do I do it if it can?

$$\frac{d[x]}{dt} = k1 \cdot y_{t} - k2 \cdot x_{t}$$

In this equation, x and y are variables that change over time t and k1 and k2 are constants which represent the rate of reaction.

Also, the square brackets is simply chemistry notation for concentration of

• Does the $[x]$ mean some operation on $x$ or how is it to be interpreted? – LutzL Jun 2 '17 at 12:57
• Hi, the square brackets just indicate that its the change in the concentration of chemical species x – CiaranWelsh Jun 2 '17 at 13:01
• Is this then the second derivative if $[x]$ indicates the change, which is the first derivative? My problem is simply that on both sides of the equation should be the same quantities, $[x]$ and $x_t$ should be the same quantity, or their more complicated relation should be explained. – LutzL Jun 2 '17 at 13:04
• I don't believe so no, I think its just my lack of experience with writing equations. You are correct in saying that the x and [x_t] are the quantity. Thanks for answering – CiaranWelsh Jun 2 '17 at 13:37

The equation $$\dot x=k_1y-k_2x$$ can be integrated via integrating factor as $$\frac{d}{dt}(e^{k_2t}x(t))=e^{k_2t}(\dot x+k_2x)=e^{k_2t}k_1y(t)$$ which can be integrated to $$e^{k_2t}x(t)-x(0)=k_1\int _0^te^{k_2s}y(s)\,ds$$ As nothing is known about $y$, the remaining integral can not be further reduced at this point.