I'm trying to solve a linear differential equation as in the following code:

DSolve[x'[t] == a + b E^(g t) + (c + d E^(-g t)) x[t], x[t], t]

which yields

enter image description here

K[1] in the integral term is mysterious. I looked at here and here, but wasn't able to find a solution. I couldn't find anything in my equation that makes it uncomputable.

Thanks for your help!

  • 1
    $\begingroup$ The 1 in K[1] just means it's the first integration dummy variable needed. If you had many you'd get K[1], K[2], ... $\endgroup$
    – b3m2a1
    Jul 22, 2019 at 20:36
  • 1
    $\begingroup$ Now that they've got the generated constant formerly known as C[1] nicely formatted, they should do something similar with the K[1]! $\endgroup$
    – Chris K
    Jul 22, 2019 at 20:50

1 Answer 1


This comes from integrating the ODE

The ODE to solve is \begin{align*} {\frac {\rm d}{{\rm d}t}}x \left( t \right) =a+b{{\rm e}^{gt}}+ \left( c+d{{\rm e}^{-gt}} \right) x \left( t \right) \end{align*}

In canonical form, the ODE is written as \begin{align*} x' &= F(t,x)\\ &= {{\rm e}^{-gt}}dx+b{{\rm e}^{gt}}+cx+a \end{align*}

The ODE is linear in $x$ and has the form $$ x' = x f(t) + g(t) $$ Where $f(t) = c+d{{\rm e}^{-gt}}$ and $g(t) = b{{\rm e}^{gt}}+a$.

Writing the ODE as \begin{align*} x' - \left( \left( c+d{{\rm e}^{-gt}} \right) x\right) &= b{{\rm e}^{gt}}+a\\ x' - \left( c+d{{\rm e}^{-gt}} \right) x &= b{{\rm e}^{gt}}+a \end{align*} Therefore the integrating factor $\mu$ is $$ \mu = e^{\int -c-d{{\rm e}^{-gt}}\mathop{\mathrm{d}t}} = {{\rm e}^{-ct+{\frac {d{{\rm e}^{-gt}}}{g}}}} $$ The ode becomes \begin{align*} \frac{\mathop{\mathrm{d}}}{ \mathop{\mathrm{d}t}} \mu x &= \mu \left(b{{\rm e}^{gt}}+a\right) \\ \frac{\mathop{\mathrm{d}}}{ \mathop{\mathrm{d}t}} \left(x{{\rm e}^{-ct+{\frac {d{{\rm e}^{-gt}}}{g}}}}\right) &= \left( b{{\rm e}^{gt}}+a \right) {{\rm e}^{-ct+{\frac {d{{\rm e}^{-gt}}}{g}}}}\\ \mathrm{d} \left(x{{\rm e}^{-ct+{\frac {d{{\rm e}^{-gt}}}{g}}}}\right) &= \left( \left( b{{\rm e}^{gt}}+a \right) {{\rm e}^{-ct+{\frac {d{{\rm e}^{-gt}}}{g}}}}\right) \mathrm{d} t \end{align*} Integrating both sides gives \begin{align*} x{{\rm e}^{-ct+{\frac {d{{\rm e}^{-gt}}}{g}}}} &= \int \! \left( b{{\rm e}^{gt}}+a \right) {{\rm e}^{-ct+{\frac {d{{\rm e}^{-gt}}}{g}}}}\,{\rm d}t + C_1 \end{align*} Dividing both sides by the integrating factor $\mu={{\rm e}^{-ct+{\frac {d{{\rm e}^{-gt}}}{g}}}}$ results in $$ x = {1\int \! \left( b{{\rm e}^{gt}}+a \right) {{\rm e}^{-ct+{\frac {d{{\rm e}^{-gt}}}{g}}}}\,{\rm d}t \left( {{\rm e}^{-ct+{\frac {d{{\rm e}^{-gt}}}{g}}}} \right) ^{-1}}+{{\it \_C1} \left( {{\rm e}^{-ct+{\frac {d{{\rm e}^{-gt}}}{g}}}} \right) ^{-1}} $$ Simplifying the solution gives $$ x={{\rm e}^{{\frac {ctg-d{{\rm e}^{-gt}}}{g}}}} \left( \int \! \left( b{{\rm e}^{gt}}+a \right) {{\rm e}^{-{\frac { \left( ctg{{\rm e}^{gt}}-d \right) {{\rm e}^{-gt}}}{g}}}}\,{\rm d}t+{\it \_C1} \right) $$

Now, instead of writing $\int \!b{\mathrm{e}^{{\frac{-ctg+{g}^{2}t+d{\mathrm{% e}^{-gt}}}{g}}}}+{\mathrm{e}^{{\frac{-ctg+d{\mathrm{e}^{-gt}}}{g}}}}a\,% \mathrm{d}t$, Mathematica wrties as \begin{equation*} \int_{1}^{t}b{\mathrm{e}^{{\frac{-cKg+{g}^{2}K+d{\mathrm{e}^{-gK}}}{g}}}}+% {\mathrm{e}^{{\frac{-cKg+d{\mathrm{e}^{-gK}}}{g}}}}a\,\mathrm{d}K \end{equation*}

i.e. it uses $K[n]$ as dummy variable of integration inside the integral. That is all. The choice of 1 as lower limit vs. 0 is arbitrary. Mathematica always uses 1 for lower limit in such cases.

  • $\begingroup$ Thanks so much! So, when I Plot the solution along with parameter values assigned, I should get concrete value for $t=1,2,3,\cdots$, right? But Mathematica doesn't compute the integral for $t=1,2,3,\cdots$ but simply reproduce the integral with $t=1,2,3,\cdots$ in the output. For example, I tried x[1] and I don't get any specific value. (Of course, I assigned values for the other parameters. $\endgroup$
    – ppp
    Jul 22, 2019 at 20:54
  • $\begingroup$ @ppp you can't plot the solution as is, since there is an unknown constant of integration in it. You need to have an initial condition specified to get specific solution to plot. $\endgroup$
    – Nasser
    Jul 22, 2019 at 21:03
  • $\begingroup$ Yes, as just mentioned, I assigned specific values for all unknown constants of integration, i.e. $a$, $b$, $c$, $d$, $g$, and $C_1$. Then I should be able to plot the solution, right? But it doesn't work... $\endgroup$
    – ppp
    Jul 22, 2019 at 21:08
  • $\begingroup$ @ppp In this case, yes, it should work. I do not know why. May you are making a mistake somewhere. May be you can post a separate question on this if you want. $\endgroup$
    – Nasser
    Jul 22, 2019 at 21:11
  • 2
    $\begingroup$ @user10478 K[1] means it is indexed symbol. Mathematica uses C[1] and C[2] and K[1] and K[2] etc... as reserved for internal constants of integration and other purposes where it needs to make a constant. It is not an actual function, just an indexed variable. One is the index. That is why it is not good idea to use C or K for own variables. Only use lower case symbols. $\endgroup$
    – Nasser
    Jun 21, 2021 at 0:23

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