# Solve the differential equation: $x\dfrac{dy}{dx}=\dfrac{1}{y^3}$

Would anybody be able to help me work through this problem? I am having some trouble with it.

$$x\dfrac{dy}{dx}=\dfrac{1}{y^3}$$

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The keyword here is separable. What have you tried, and what are you having trouble with? Have you seen any separable equations before? – user61527 Jul 6 '14 at 20:29
Would I start it with – Kell Jul 6 '14 at 20:33

From

$$x\dfrac{dy}{dx}=\dfrac{1}{y^3}$$

we have

$$\int y^3dy=\int\frac{dx}{x}$$

So

$$\frac{1}{4}y^4=\ln{|x|}+C\Rightarrow y=\pm\sqrt[4]{4\ln{|x|}+C}$$

You can find a specific solution from an initial condition.

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Rearrange your equation and you have

$y^3\frac{dy}{dx}=\frac{1}{x}$ (this method is called seperation of variables)

integrating with respect to $x$

$\int y^3\frac{dy}{dx}\,dx=\int \frac{1}{x}\,dx$

$\int y^3\,dy=\ln x +c$

$\frac{1}{4}y^4 =\ln x +c$, so

$y=(4\ln x+4c)^{\frac{1}{4}}$

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Recall the definition of a separable differential equation. Can you see why $x\dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac{1}{y^3}$ is an example of such an ODE?

You may try rearranging the equation and integrating: $$x\dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac{1}{y^3}\\ \implies {y^3}{\mathrm{d}y}=\dfrac{\mathrm{d}x}{x}\\ \implies \int{y^3}{\mathrm{d}y}=\int\dfrac{\mathrm{d}x}{x}$$ Do not forget the integration constant.

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