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After not doing serious math since I finished college I've decided to try to teach myself Differential Equations. I'm using the Tenenbaum and Pollard text, but occasionally run into things I can't figure out. Since it's been almost 20 years since my last math class, I can't always tell if the problem is my not understanding the new topic, or just a failure in doing calculus (or even algebra).

In this case, it's problem 6 in the exercises at the end of section 6, separable equations.

The problem asks to find a 1-parameter family of solutions, including intervals where it's valid, plus any particular solutions which are not members of the family of solutions.

The equation to solve is: $$yx^2dy-y^3dx = 2x^2dy$$

My attempted solution is as follows:

Assuming $y\neq0$ and $x\neq0$, divide both sides by $y^3x^2$ to get: $$\frac1 y dy-\frac1 {x^2} dx=\frac2 {y^3} dy$$ Rearrange to get: $$x^{-2}dx=y^{-1}dy - 2y^{-3}dy$$ Integrating both sides, we get: $$\int x^{-2}dx=\int y^{-1}dy - 2\int y^{-3}dy$$ $$-\frac1 x +C = \ln y + \frac1 {y^2}$$ Multiplying both sides by $y^2x$ and rearranging gets me to: $$-y^2 + Cy^2x=y^2x\ln y +x$$ $$(Cx-1)y^2=(y^2\ln y+1)x, x\neq0, y\neq0$$

However, the solution in the books isn't the same. Instead, they offer: $$(cx+1)y^2=(y-1)x, x\neq0, y\neq0$$

For the life of me, I can't find either an error or how to get rid of the $\ln y$ term.

Additionally, the book lists $y=0$ as a particular solution not part of the family of solutions. Why isn't $x=0$ also a particular solution?

Help much appreciated.

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    $\begingroup$ The first term in the equation after dividing by $y^3x^2$ should be $\dfrac{1}{y^2}\,dy$. $\endgroup$ Oct 29, 2017 at 15:41

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Assuming $y\neq0$ and $x\neq0$, divide both sides by $y^3x^2$ to get $$\frac1 y dy-\frac1 {x^2} dx=\frac2 {y^3} dy$$

You made a mistake here. It should be $$\frac1 {y^2} dy-\frac1 {x^2} dx=\frac2 {y^3} dy$$

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  • $\begingroup$ Aha! I was afraid it was something silly like that. Thanks! $\endgroup$
    – jgalak
    Oct 29, 2017 at 17:08

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