I'm blocked in a step of getting to the equation $(7)$, more precisely in the step between $(5)$ and $(6)$. I show my way to solve it down, but I don't get the same as $(6)$, so I wanted to know if I was wrong and an explanation of the correct aproach. $$\ce{A ->[k_\mathrm{1}]I->[k_\mathrm{2}]P}$$

$$\frac{\mathrm d[\ce{A}]}{\mathrm dt} = -k_\mathrm{1}[\ce{A}] \tag{1}$$ $$\frac{\mathrm d[\ce{I}]}{\mathrm dt} = k_\mathrm{1}[\ce{A}] -k_\mathrm{2}[\ce{I}] \tag{2}$$ $$\frac{\mathrm d[\ce{P}]}{\mathrm dt} = k_\mathrm{2}[\ce{I}] \tag{3}$$

$$[\ce{A}]_\mathrm{0}\neq0 ; [\ce{I}]_\mathrm{0}=0; [\ce{P}]_\mathrm{0}=0$$ $$[\ce{A}]=[\ce{A}]_\mathrm{0}e^{-k_\mathrm{1}t} \tag{4}$$

$$\frac{\mathrm d[\ce{I}]}{\mathrm dt} = k_\mathrm{1}[\ce{A}]_\mathrm{0}e^{-k_\mathrm{1}t} -k_\mathrm{2}[\ce{I}] \tag{5}$$

$$[\ce{I}]=\frac{k_\mathrm{1}}{k_\mathrm{2}-k_\mathrm{1}}(e^{-k_\mathrm{1}t}-e^{-k_\mathrm{2}t})[\ce{A}]_\mathrm{0} \tag{6}$$

$$[\ce{A}]_\mathrm{0}=[\ce{A}]-[\ce{I}]-[\ce{P}]$$ $$[\ce{P}]=[\ce{A}]-[\ce{A}]_\mathrm{0}-[\ce{I}]$$

$$[\ce{P}]=(\frac{k_\mathrm{1}e^{-k_\mathrm{2}t}-k_\mathrm{2}e^{-k_\mathrm{1}t}}{k_\mathrm{2}-k_\mathrm{1}}+1)[\ce{A}]_\mathrm{0} \tag{7}$$

And here is what I've done:

$$\int_{0}^{[\ce{I}]} \, \mathrm{d}[\ce{I}]=k_\mathrm{1}[\ce{A}]_\mathrm{0}\int_{0}^{t}e^{-k_\mathrm{1}t} \, \mathrm{d}t-k_\mathrm{2}[\ce{I}]\int_{0}^{t} \, \mathrm{d}t \tag{A}$$

$${[\ce{I}]}=-e^{-k_\mathrm{1}t}[\ce{A}]_\mathrm{0}-k_\mathrm{2}t[\ce{I}] \tag{B}$$

$${[\ce{I}]}=-[\ce{A}]_\mathrm{0}(\frac{e^{-k_\mathrm{1}t}}{k_\mathrm{2}t+1}) \tag{C}$$

Anyone could help me? Thanks :)

  • $\begingroup$ This symbolic integrator to resolve both infinite and finite integrals may help you: integral-calculator.com // It is very powerful. $\endgroup$
    – Poutnik
    Aug 16, 2022 at 9:21
  • $\begingroup$ Thanks a lot, but I actually know how to solve thos integrals and I've just checked them. The question was how to solve the differential equation to reach equation (6). $\endgroup$ Aug 16, 2022 at 9:27
  • $\begingroup$ It looks to me you need to separate terms to integral equation for I and t on respective sides. $\endgroup$
    – Poutnik
    Aug 16, 2022 at 13:43
  • $\begingroup$ I end up with the same because I can't pass "dt" to the right without affecting "k2[I]". $\endgroup$ Aug 16, 2022 at 13:51
  • 1
    $\begingroup$ @JorgeBonifaz I claimed it was related. I did not claim it was a duplicate. $\endgroup$ Aug 18, 2022 at 9:00

1 Answer 1


This is a form of a linear first-order differential equation, and you can solve it like this (or just plug it into wolfram alpha).

We have

$$\frac{d[I]}{dt} = k_1[A]_0e^{-k_1t} - k_2[I]$$ $$\frac{d[I]}{dt} + k_2[I] = k_1[A]_0e^{-k_1t}$$

If we let $P(t) = k_2$, $Q(t) = k_1[A]_0e^{-k_1t}$, and $I(t) = e^{\int P(t)dt} = e^{k_2t}$ the solution to this sort of differential equation is

$$[I] = \frac{1}{I(t)} \left( \int I(t)Q(t)dt + C \right)$$

where C is a constant. Plugging in the values set for this problem, we have

$$[I] = \frac{1}{e^{k_2t}} \left( \int e^{k_2t}\cdot k_1[A]_0e^{-k_1t} dt + C\right)$$

$$[I] = Ce^{-k_2t} + k_1[A]_0\cdot\frac{1}{k_2-k_1}e^{(k_2-k_1)t}\cdot e^{-k_2t}$$

$$[I] = Ce^{-k_2t} + k_1[A]_0\cdot \frac{1}{k_2-k_1}e^{-k_1t}$$

Now plugging in $t = 0$, we know $[I]_0 = 0$. So:

$$C = -\frac{k_1[A]_0}{k_2-k_1}$$

Reinputting C to the previous equation gives:

$$[I] = -\frac{k_1[A]_0}{k_2-k_1} e^{-k_2t} + \frac{k_1[A]_0}{k_2-k_1} e^{-k_1t}$$

$$[I] = \frac{k_1}{k_2-k_1}(e^{-k_1t} - e^{k_2t})[A]_0$$

We have to do this because [I] depends on t as well, so you can't just take it out of the integral like you did.

  • 2
    $\begingroup$ Thanks a lot. I'm not in university yet, so I didn't know how to solve differential equations, but now I understand it. $\endgroup$ Aug 17, 2022 at 13:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.