Can anyone walk me through calculating the differential equation...

Question

I need help solving this equation, i've attempted using numerous methods. But I'm give choices with square roots as an exponent of e, and I haven't been able to match any of them. $$ \frac{1}{6x}\frac{dy}{dx}=y\sqrt{3x^2-1} $$ The answers I'm given all have $$e^{\frac{2}{3}\sqrt{(3x^2-1)^3}}$$ and no method I've isolated has managed that answer.

Answer 1

The most straightforward way to approach this problem is by the method of separation of variables. This works as follows.

You try to rewrite the original equation such that on the left hand side, only terms with $y$ and $\frac{\text{d} y}{\text{d} x}$ are present, while the right hand side is a function of $x$ only. In your case, this yields \begin{equation} \frac{1}{y} \frac{\text{d} y}{\text{d} x} = 6 x \sqrt{3 x^2 -1}, \end{equation} by multiplying both sides with $\frac{6x}{y}$.

The following step is to integrate both sides with respect to $x$, i.e. \begin{equation} \int^x \frac{1}{y(x)} \frac{\text{d} y(x)}{\text{d} x}\,\text{d} x = \int^x 6 x \sqrt{3x^2-1} \,\text{d} x. \end{equation} You can recognise the integrand of the left integral as the $x$-derivative of $\log y(x)$, i.e. \begin{equation} \int^x \frac{1}{y(x)} \frac{\text{d} y(x)}{\text{d} x}\,\text{d} x = \int^x \frac{\text{d}}{\text{d} x}\left[\log y(x)\right]\,\text{d} x = \log y(x). \end{equation} Therefore, you can express the solution $y(x)$ as \begin{equation} y(x) = \exp \Big[ \int^x 6 x \sqrt{3x^2 -1}\,\text{d} x \Big] \end{equation} I leave it to you to calculate this integral.