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Could someone help me solve this differential equation? $$y''-\frac{1}{x\ln x}y'=12x^2\ln x$$ I tried doing $y'=z$ and that leads me to $$z'-\frac{1}{x\ln x}z=12x^2\ln x$$ and then I do $z=t\cdot x\:\:and\:z'=t'x+t$ which will lead me to $$t'x+t=12x^2\ln x+\frac{1}{x\ln x}\cdot t\cdot x$$ but trying to solve this(by parts) yields me $$t=x^{6x^2}-e^{3x^2}+\ln x+\frac{1}{x}$$ and to find $y$ I will have to integrate a pretty nasty equation, which leads me to believe I'm doing something wrong.

I'm pretty sure I need to use substitution to get a first order linear equation but perhaps I am doing something wrong? Could someone take a look at this and point me in the right direction?

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  • $\begingroup$ I don't think I'm supposed to solve this by using an integrating factor, I must use substitution and get an equation that can be solved by parts. $\endgroup$ – MikhaelM Jul 1 '17 at 19:08
  • $\begingroup$ Then replace $x=e^t$ with $t$ as new independent variable, $u(t)=y(e^t)$, $u'(t)=y'(e^t)e^t$ etc. $\endgroup$ – Lutz Lehmann Jul 1 '17 at 19:09
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It's easier if you start by solving the homogeneous equation: $$(x\ln x) z'= z$$ Or: $$\frac{dz}{z} = \frac{dx}{x\ln x}$$ $$\ln z + C= \ln\ln(x)\rightarrow z =C\ln x$$ Now you have to find a particular solution. We can guess the general form $z_0 = a x^n \ln^m x$:

$$(x\ln x) z'- z = 12 x^3 \ln^2 x$$ $$a x^n \ln^m x (m -1+ n\ln x) = 12 x^3 \ln^2 x$$ This can only be solved by: $$m=1,\ n=3, a =4$$ So you get a particular solution: $$z_0 = 4 x^3 \ln x$$, And a final solution of: $$z = (4 x^3 + C)\ln x$$

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  • $\begingroup$ I get $z_p=lnx\cdot C\left(x\right)$ and $z_p'=\frac{1}{x}C\left(x\right)+lnx\cdot C'\left(x\right)$. After replacing in the original equation I get $lnx\cdot C\left(x\right)+xln^2\left(x\right)\cdot C'\left(x\right)=lnx\cdot C\left(x\right)$ but that leaves me with the constant being 0? Hence the particular solution is 0? What am I missing here? $\endgroup$ – MikhaelM Jul 1 '17 at 19:17
  • $\begingroup$ Nevermind, I forgot the right side T_T $\endgroup$ – MikhaelM Jul 1 '17 at 19:23
  • $\begingroup$ Ok, solved it fully! End result was $C_1xlnx\:-\:C_1x\:+\:x^4lnx\:-\:\frac{x^4}{x}+C_2$, wolfram confirms it! Thank you for all your help :) $\endgroup$ – MikhaelM Jul 1 '17 at 19:28

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