# Solve x^3y''' + x^2y'' - 2xy' + 2y = x^3lnx

Solve ${x}^{3}y+{x}^{2}y-2xy+2y={x}^{3}\mathrm{ln}x$
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Jeffrey Jordon

Given: ${x}^{3}y+{x}^{2}y-2xy+2y={x}^{3}\mathrm{ln}x$

Let $\mathrm{ln}x=z$ then

$\frac{dy}{dx}=\frac{dy}{dz}\frac{dz}{dx}=\frac{1}{x}\frac{dy}{dx}$

Similarly:

${x}^{2}\frac{{d}^{2}y}{d{x}^{2}}=\frac{{d}^{2}y}{d{z}^{2}}-\frac{dy}{dz}$ and ${x}^{3}\frac{{d}^{3}y}{d{x}^{3}}=\frac{{d}^{3}y}{d{z}^{3}}-3\frac{{d}^{2}y}{d{z}^{2}}+2\frac{dy}{dx}$

Using above three expressions, the original differential equation reduces to:

$\left(\frac{{d}^{3}y}{d{z}^{3}}-3\frac{{d}^{2}y}{d{z}^{2}}+2\frac{dy}{dz}\right)+\left(\frac{{d}^{2}y}{dz}-\frac{dy}{dz}\right)-2\frac{dy}{dz}+2y=z{e}^{3z}$

The auxilary equation of the differential equation

$\frac{{d}^{3}y}{d{z}^{3}}-2\frac{{d}^{2}y}{d{z}^{2}}-\frac{dy}{dz}+y=0$ becomes

${m}^{3}-2{m}^{2}-m+2=0$

$⇒m=2,1,-1$

Therefore the complimentary function becomes:

${y}_{c}\left(x\right)=a{e}^{2z}+b{e}^{z}+c{e}^{-z}$

The potential integral is given by:

${y}_{p}\left(x\right)=\frac{1}{{D}^{3}-2{D}^{2}-D+2}z{e}^{3z}$

$={e}^{3z}\frac{1}{\left(D+3{\right)}^{3}-2\left(D+3{\right)}^{2}-\left(D+3\right)+2}z$

$={e}^{3z}\frac{1}{{D}^{3}+7{D}^{2}+14D+4}z$

$={e}^{3z}\frac{1}{4}\left(1+\frac{{D}^{3}+7{D}^{2}+14D}{4}{\right)}^{-1}z$

$=\frac{{e}^{3z}}{4}\left(1-\frac{{D}^{3}+7{D}^{2}+14D}{4}+...\right)z$

$=\frac{{e}^{3z}}{4}\left(z-\frac{14}{4}\right)$

$=\frac{{x}^{3}}{4}\left(\mathrm{ln}x-\frac{14}{4}\right)$

Therefore the solution is

$y\left(x\right)={y}_{c}\left(x\right)+{y}_{p}\left(x\right)=a{x}^{2}+bx+\frac{c}{x}+\frac{{x}^{3}}{4}\left(\mathrm{ln}x-\frac{14}{4}\right)$

where a,b,c are arbitrary constants