DSolve yields a strange character K[1]

Question

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

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!

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.