tibukooinm

2022-03-26

How can I handle $exp\left(\mathrm{ln}|x|\right)$ to solve 1st order linear DE?
RHS and LHS are same.

Taking log.

This problem is a sub-problem of below ODE.

My thoughts are as below.

I got the following general solution as I forcefully assumed
$~x=\mathrm{exp}\left(\mathrm{ln}|x|\right)~$

Aidyn Wall

Explanation:

pastuh7vka

Firstly, on your equations for exp and ln,
$x=\mathrm{exp}\left[\mathrm{ln}\left(x\right)\right]$
only holds for $x>0$. Secondly, the integrating factor for the equation
${y}^{\prime }\left(x\right)-\frac{2}{x}y\left(x\right)={x}^{6}$
is given by the function

$\mu \left(x\right)=\left\{\begin{array}{ll}{C}_{0}\mathrm{exp}\left[-2\mathrm{ln}\left(-x\right)\right]& x<0\\ {C}_{1}\mathrm{exp}\left[-2\mathrm{ln}\left(x\right)\right]& x>0\end{array}=\left\{\begin{array}{ll}-\frac{{C}_{0}}{{x}^{2}}& x<0\\ \frac{{C}_{1}}{{x}^{2}}& x>0\end{array}$

with ${C}_{0},{C}_{1}>0$. What this means is that if
$f\left(x\right)=\frac{1}{{x}^{2}}$,
then ${\left(fy\right)}^{\prime }\left(x\right)=f\left(x\right){x}^{6}={x}^{4}$.
Since 0 is a singularity, we have that
$f\left(x\right)y\left(x\right)=\frac{{x}^{5}}{5}+{C}_{0}=\frac{y\left(x\right)}{{x}^{2}},;;;;x<0$
$f\left(x\right)y\left(x\right)=\frac{{x}^{5}}{5}+{C}_{1}=\frac{y\left(x\right)}{{x}^{2}},;;;;x>0.$
What this implies is that

$y\left(x\right)=\left\{\begin{array}{ll}\frac{{x}^{7}}{5}+{C}_{0}& x<0\\ \frac{{x}^{7}}{5}+{C}_{1}& x>0\end{array}.$

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