I want to solve the initial value problem $$x''(t)+3x'(t)+2x(t)=\frac{1}{1+e^t}, \quad x(0)=2\ln2,~x'(0)=1-3\ln 2$$

considering the homogeneous couterpart $$x''(t)+3x'(t)+2x(t)=0$$ with the characteristic polynomial $$ \chi(\lambda)=\lambda^2+3\lambda+2=(\lambda+1)(\lambda+2)$$ the general solution to the homogeneous equation is of the form $$x(t)=c_1e^{-1t}+c_2e^{-2t}$$ using my initial conditions $$x(0)=c_1+c_2=2\ln 2, \quad x'(0)=-c_1-2c_2=1-3\ln 2$$ I have $$c_1=1+\ln2,~c_2=-1+\ln 2,\quad x(t)=(1+\ln 2)e^{-t}+(-1+\ln 2)e^{-2t}$$

Solution using @Karn Watcharasupat's method: $u := x+x', \quad u'=x'+x''$ the equation becomes a first order linear differential equation, with $\mu$ integration factor such that $\mu 2=\mu'\Rightarrow \mu = e^{2t}$ $$u'+2u=\frac{1}{1+e^t}$$ $$(\mu u)'=\mu \frac{1}{1+e^t}\iff \frac{d\left(e^{2t}u\right)}{dt}=\frac{e^{2t}}{1+e^t}$$ $$ e^{2t}u=\int \frac{e^{2t}}{1+e^t}dt=\int \frac{v-1}{v} dv=\int 1 dv - \int \frac{1}{v}dv=v-\ln |v|=1+e^t-\ln|1+e^t|+C_1$$ for $v:=1+e^t$. And then $$u=\frac{e^t-\ln|1+e^t|+C_2}{e^{2t}}=e^{-t}-e^{-2 t}\ln|1+e^t|+e^{-2t}C_2$$ Plugging back in $$x+x'=C_2e^{-2t}+e^{-t}-e^{-2 t}\ln(1+e^t)$$ which is similarly a first order ODE with $\mu=\mu'\Rightarrow \mu=e^t$ $$\left(x\mu\right)'=\mu\cdot \left(C_2e^{-2t}+e^{-t}-e^{-2 t}\ln(1+e^t)\right)=e^t\left(C_2e^{-2t}+e^{-t}-e^{-2 t}\ln(1+e^t)\right)$$ $$\frac{d\left(xe^t\right)}{dt}=\left(C_2e^{-t}+1-e^{- t}\ln(1+e^t)\right)$$ \begin{align}xe^t&=\int \left(C_2e^{-t}+1-e^{- t}\ln(1+e^t)\right) dt\\&=\int C_2e^{-t} dt+\int 1 dt -\int e^{- t}\ln(1+e^t) dt\\&=-C_2e^{-t}+t-\left[-e^{-t}\ln \left(e^t+1\right)+t-\ln \left|e^t+1\right|+C\right]\\&=-C_2e^{-t}+t+e^{-t}\ln \left(e^t+1\right)-t+\ln \left(e^t+1\right)+C\\&=C_2e^{-t}+e^{-t}\ln \left(e^t+1\right)+\ln \left(e^t+1\right)+C\end{align} And finally $$x=\frac{C_2e^{-t}+e^{-t}\ln \left(e^t+1\right)+\ln \left(e^t+1\right)+C}{e^t}=C_2e^{-2t}+e^{-2t}\ln \left(e^t+1\right)+e^{-t}\ln \left(e^t+1\right)+Ce^{-t}$$ $$x(t)=C_1e^{-t}+C_2e^{-2t}+e^{-t}\ln \left(e^t+1\right)+e^{-2t}\ln \left(e^t+1\right)$$ $$x'(t)=-C_1e^{-t}-2C_2e^{-2t}-e^{-t}\ln \left(e^t+1\right)+\frac{1}{e^t+1}-2e^{-2t}\ln \left(e^t+1\right)+\frac{e^{-t}}{e^t+1}$$ Finding $C_1,~C_2$ $$x(0)=C_1+C_2+2\ln(2)=2\ln(2)\Rightarrow C_1=-C_2$$ $$x'(0)=-C_1-2C_2-\ln 2+\frac{1}{2}-2\ln 2+\frac{1}{2}=1-3\ln 2 \Rightarrow C_1=-2C_2$$ So $$C_1=C_2=0$$ and the solution to the IVP is $$x(t)=e^{-t}\ln \left(e^t+1\right)+e^{-2t}\ln \left(e^t+1\right)$$

  • $\begingroup$ Wolfram|Alpha gives $e^{-2t}\ln(e^t+1)$ and $e^{-t}\ln(e^t+1)$ as the particular solutions. I'm trying to figure out how to get these... $\endgroup$ – Karn Watcharasupat Dec 22 '17 at 8:04

I will first use the substitution $$u=x+x'$$

So we have $$x''(t)+3x'(t)+2x(t)=\frac{1}{1+e^t}$$ transforming into $$u'+2u=\frac{1}{1+e^t}$$

So we have the IF $$e^{\int 2 \ dt}=e^{2t}$$

So \begin{align} \frac{d}{dt}(e^{2t}u)&=\frac{e^{2t}}{1+e^t}\\ e^{2t}u&=\int \frac{e^{2t}}{1+e^t}\ dt\\ &=e^t - \ln(1 + e^t)+C_1\\ u&=C_1e^{-2t}+e^{-t}-e^{-2t}\ln(1 + e^t)\\ \end{align} Substituting $u=x+x'$ back in, \begin{align} x+x'&=C_1e^{-2t}+e^{-t}-e^{-2t}\ln(1 + e^t)\\ \end{align}

This is a first order LODE again with IF $=e^t$.

\begin{align} \frac{d}{dt}(e^{t}x)&=(e^t)(C_1e^{-2t}+e^{-t}-e^{-2t}\ln(1 + e^t))\\ &=C_1e^{-t}+1 - e^{-t} \ln(1 + e^t)\\ e^{t}x&=\int C_1e^{-t}+ 1 - e^{-t} \ln(1 + e^t) \ dt\\ &=-C_1e^{-t}+ \ln(1 + e^{-t}) + e^{-t} \ln(1 + e^t)+C_2\\ x&=-C_1e^{-2t}+C_2e^{-t}+ e^{-t}\ln(1 + e^{-t}) + e^{-2t} \ln(1 + e^t) \end{align}

  • $\begingroup$ Thank you for answer, it will take some time to process. $\endgroup$ – badatmath Dec 22 '17 at 8:27
  • $\begingroup$ would you say such a substitution is possible for 2nd order linear inhomogeneous equations (simple ones) in general? or just in this case? (the little mistake is in the integration, the $t$ cancles out) $\endgroup$ – badatmath Dec 22 '17 at 9:10
  • $\begingroup$ @Noob It just happens to be nice in this case. $\endgroup$ – Karn Watcharasupat Dec 22 '17 at 9:15
  • 1
    $\begingroup$ You can use this process in all cases where you have constant coefficients. You are transforming $(D-λ_1)(D-λ_2)...(D-λ_n)x=f$ into $(D-λ_1)u_1=f$, $(D-λ_2)u_2=u_1$, ..., $(D-λ_n)u_n=u_{n-1}$ and $x=u_n$. Sometimes this gives a faster solution than the more general variation of constants. $\endgroup$ – LutzL Dec 22 '17 at 11:08
  • $\begingroup$ @LutzL I didn't know how to proceed with your answer. Would appreciate a link to where that method is explained. $\endgroup$ – badatmath Dec 22 '17 at 12:13

You can only apply the guessing process if the left side has constant coefficients and the right side only terms of the form polynomial times exponential.

As your right side is not of this form, you will need to apply the variation of constants method. $$ y(t)=c_1(t)y_1(t)+c_2(t)y_2(t) $$ where $$ c_1'(t)y_1(t)+c_2'(t)y_2(t)=0\\ c_1'(t)y_1'(t)+c_2'(t)y_2'(t)=f(t) $$ so that $$ c_1'(t)=-\frac{f(t)y_2(t)}{W(t)},\\ c_2'(t)=\frac{f(t)y_1(t)}{W(t)},\\ W(t)=y_1(t)y_2'(t)-y_1'(t)y_2(t), $$ As $W(t)=\det\pmatrix{e^{-t}&e^{-2t}\\-e^{-t}&-2e^{-2t}}=-e^{-3t}$. This gives the coefficient functions as This involves computing the integrals $$ c_1(t)=-\int\frac{e^t}{1+e^t}dt\text{ and }c_2(t)=\int\frac{e^{2t}}{1+e^t}dt $$ which both are symbolically solvable.

  • $\begingroup$ I can't upvote yet but thanks for the general solution! $\endgroup$ – badatmath Dec 23 '17 at 18:09

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