How to solve $\frac{dy}{dx}\ln(x)+\frac{2y}{x}=1$?

Question

$$\frac{dy}{dx}\ln(x)+\frac{2y}{x}=1$$

The solution has evaded me thus far. I've tried pursuing the integrating factor technique (writing the equation as a differential and trying to find an integrating factor that will make it a full differential) but with no success. Any help is appreciated.

Answer 1

Write your equation in the form $$y'(x)+\frac{2y(x)}{x\ln(x)}=\frac{1}{\ln(x)}$$ and multiply the equation by $$\mu=e^{\int\frac{2}{x\log(x)}dx}=\log^2(x)$$ and now note that $$\frac{2\log(x)}{x}=\frac{d}{dx}\log^2(x)$$ and we get $$\int\frac{d}{dx}(\log^2(x)y(x))dx=\int\log(x)dx$$ Can you finish?

Answer 2

The integrating factor technique can always solve these types of problems, and will reveal if the solution can only be written as an integral. We divide both sides by $ln(x)$ to get

$$\frac{dy}{dx}+\frac{2y}{x\ln(x)}=\frac{1}{\ln(x)}$$

The integrating factor is then

$$\mu(x)=e^{\int\frac{2}{x\ln(x)}dx}=e^{2\ln(\ln(x))}=\ln^2(x)$$

Multiplying by this and rewriting the left side gives

$$\frac{d}{dx}\left(\ln^2(x)y\right)=\ln(x)$$

This equation can be integrated, where the right side requires integration by parts.

Answer 3

You can almost apply the product rule to the LHS, but the factor of $2$ on the second term ruins this opportunity. If only we had a square somewhere, that gives us this factor of $2$ when differentiated.

To that end, you can multiply both sides of the equation by $ln(x)$ to get that:

$$y'(x)\cdot ln^2(x) + y(x) \cdot2ln(x) \cdot \frac{1}{x} =(ln^2(x)y(x))'= ln(x) \Leftrightarrow ln^2(x)y(x) = xlnx-x+C $$

Can you continue from here?

Answer 4

The equation is in the linear form. You first solve the homogeneous equation, which is separable:

$$\frac{y'}y+\frac{2}{x\ln(x)}=0$$

giving

$$\ln(y)+2\ln(\ln(x))=C$$ or $$y=\frac C{\ln^2(x)}.$$

Now by variation of the constant,

$$\frac{C'}{\ln(x)}-\frac{2C}{x\ln^2(x)}+\frac{2C}{x\ln^2(x)}=1$$

and

$$C=x\ln(x)-x+K.$$

Finally,

$$y=\frac{x\ln(x)-x+K}{\ln^2(x)}.$$

---

You obtain this result faster with the integrand factor $\ln(x)$, turning the equation to

$$y'\ln^2(x)+y\frac{2\ln(x)}x=(y\ln^2(x))'=\ln(x).$$