How can I handle exp(ln|x|) to solve 1st order linear DE?RHS and LHS are same.exp⁡(ln⁡(x))=exp⁡(ln⁡(x))&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(1)Taking log.ln⁡(exp⁡(ln⁡(x)))=ln⁡(exp⁡(ln⁡(x)))&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(2)ln⁡(x)=ln⁡(exp⁡(ln⁡(x)))&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(3)∴≈x=exp⁡(ln⁡(x))&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(4)Then what about ln|x| ?This problem is a sub-problem of below ODE.dydx-2xy=x6&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(5)My thoughts are as below.P(x)&nbsp;=−2x&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(6)Q(x)&nbsp;=x6&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(7)dydx+P(x)y=Q(x)⏟1st order linear DE&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(8)Integrating factor=exp(∫P(x)dx)&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(9)exp(∫-2xdx)&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(10)=exp⁡(−2∫1xdx)&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(11)=exp(-2lnx+const)&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(12)=exp(-2lnx)exp(const)&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(13)=exp(lnx)-2exp(const)&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(14)exp(lnx)=exp(lnx)&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(15)I got the following general solution as I forcefully assumed~&nbsp;x=exp(lnx)&nbsp;~y=15x7+const1x2&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(16)

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How can I handle exp(ln|x|) to solve 1st order linear DE?RHS and LHS are same.exp⁡(ln⁡(x))=exp⁡(ln⁡(x))&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;(1)Taking log.ln⁡(exp⁡(ln⁡(x)))=ln⁡(exp⁡(ln⁡(x)))&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;(2)ln⁡(x)=ln⁡(exp⁡(ln⁡(x)))&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;(3)∴≈x=exp⁡(ln⁡(x))&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;(4)Then what about ln|x| ?This problem is a sub-problem of below ODE.dydx-2xy=x6&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;(5)My thoughts are as below.P(x)&amp;nbsp;=−2x&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;(6)Q(x)&amp;nbsp;=x6&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;(7)dydx+P(x)y=Q(x)⏟1st order linear DE&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;(8)Integrating factor=exp(∫P(x)dx)&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;(9)exp(∫-2xdx)&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;(10)=exp⁡(−2∫1xdx)&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;(11)=exp(-2lnx+const)&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;(12)=exp(-2lnx)exp(const)&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;(13)=exp(lnx)-2exp(const)&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;(14)exp(lnx)=exp(lnx)&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;(15)I got the following general solution as I forcefully assumed~&amp;nbsp;x=exp(lnx)&amp;nbsp;~y=15x7+const1x2&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;(16)

catcher1307ieh · Accepted Answer

Explanation:exp⁡(ln|x|)−2&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;(1)=1exp⁡(ln|x|)2&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;(2)=1exp⁡(2ln|x|)&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;(3)=1exp⁡(ln⁡(|x|2))&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;(4)=1exp⁡(ln⁡(x2))&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;(5)=1exp⁡(2ln⁡(x))&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;(6)=1exp⁡(ln⁡(x))2&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;(7)=1(x)2&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;(8)

pecanjima9vo · Accepted Answer

Firstly, on your equations for exp and ln,x=exp⁡[ln⁡(x)]only holds for&amp;nbsp;x&amp;gt;0. Secondly, the integrating factor for the equationy′(x)−2xy(x)=x6is given by the functionμ(x)={C0exp⁡[−2ln⁡(−x)]x&amp;lt;0C1exp⁡[−2ln⁡(x)]x&amp;gt;0={−C0x2x&amp;lt;0C1x2x&amp;gt;0with&amp;nbsp;C0,C1&amp;gt;0. What this means is that iff(x)=1x2,then&amp;nbsp;(fy)′(x)=f(x)x6=x4.Since 0 is a singularity, we have thatf(x)y(x)=x55+C0=y(x)x2&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;x&amp;lt;0f(x)y(x)=x55+C1=y(x)x2&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;x&amp;gt;0.What this implies is thaty(x)={x75+C0x&amp;lt;0x75+C1x&amp;gt;0.

How can I handle exp(ln|x|) to solve 1st order linear

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