Verify that $y=\frac{1}{x}-\log x$ is a solution of the differential equation $x^2\frac{d^2y}{dx^2}+x\frac{dy}{dx}-y=\log x$

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> [!TLDR]
> Computing $y'$ and $y''$ for $y=	frac1x-\log x$ and substituting reduces the LHS to exactly $\log x$.

> [!FORMULA]
> Substitute $y,\;y'=\dfrac{dy}{dx},\;y''=\dfrac{d^2y}{dx^2}$ into $x^2y''+xy'-y$ and check it equals $\log x$. Recall $\dfrac{d}{dx}x^{-1}=-x^{-2}$ and $\dfrac{d}{dx}\log x=\dfrac1x$.

**Given:** $y=\dfrac1x-\log x=x^{-1}-\log x$.

1. First derivative: $y'=-x^{-2}-\dfrac1x=-\dfrac{1}{x^2}-\dfrac1x$.
2. Second derivative: $y''=2x^{-3}+x^{-2}=\dfrac{2}{x^3}+\dfrac{1}{x^2}$.
3. $x^2y''=x^2\!\left(\dfrac{2}{x^3}+\dfrac{1}{x^2}ight)=\dfrac{2}{x}+1$.
4. $xy'=x\!\left(-\dfrac{1}{x^2}-\dfrac1xight)=-\dfrac1x-1$.
5. $-y=-\dfrac1x+\log x$. …

Q.Verify that $y=\frac{1}{x}-\log x$ is a solution of the differential equation $x^2\frac{d^2y}{dx^2}+x\frac{dy}{dx}-y=\log x$

Concept understanding — Solution Verification

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Q.Verify that y=1x−log⁡xy=\frac{1}{x}-\log xy=x1​−logx is a solution of the differential equation x2d2ydx2+xdydx−y=log⁡xx^2\frac{d^2y}{dx^2}+x\frac{dy}{dx}-y=\log xx2dx2d2y​+xdxdy​−y=logx

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Q.Verify that $y=\frac{1}{x}-\log x$ is a solution of the differential equation $x^2\frac{d^2y}{dx^2}+x\frac{dy}{dx}-y=\log x$

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