$x\frac {dy}{dx}=2y$

Now of course we can rewrite as $\frac{1}{2y}dy=\frac1xdx$

and end on LHS after integrating with $\ln(2y)$.

But couldn't we think of LHS as $\frac12 \times \frac 1y$

Then when we integrate we just get $\frac12 \int \frac1y dy$ which is just $\ln (y^{\frac12})$

Thanks for any explanation! Rob

  • $\begingroup$ Your third line is incorrect. The fourth line is what you should do. $\endgroup$ – JohnD May 8 '14 at 20:29

The derivative of $\ln{2y}$ is $\displaystyle 2\cdot \ln'{2y}=2 \cdot \frac{1}{2y}=\frac{1}{y}$. An other way to see it is $\ln{2y}=\ln{2} + \ln{y}$ so it has the same derivative than $\ln $ and the primitive of $\displaystyle \frac{1}{2y}$ is not $\ln{2y}$.

Hence your first integration is wrong, you need to add a factor to integrate and this is exactly what you're doing in the second method, which is correct.

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  • $\begingroup$ Thanks! That's clear. $\endgroup$ – Saxobob May 8 '14 at 20:42

The integral $\displaystyle{\int \frac{1}{2y}~\mathrm{d}y}$ is not $\ln 2y$. You are right to say that $$\int \frac{1}{2y}~\mathrm{d}y = \frac{1}{2}\int \frac{1}{y}~\mathrm{d}y = \frac{1}{2}\ln|y| + C$$

For the integral to be an exact log, you need the numerator to be the derivative of the denominator:

$$\int \frac{\mathrm{f}'(x)}{\mathrm{f}(x)}~\mathrm{d}x = \ln|\mathrm{f}(x)| + C$$

To continue with your solution: from $\displaystyle{\frac{1}{2y}~\mathrm{d}y = \frac{1}{x}~\mathrm{d}x}$ we get:

$$\int \frac{1}{2y}~\mathrm{d}y = \int \frac{1}{x}~\mathrm{d}x$$

$$\frac{1}{2}\ln|y| = \ln|x|+C$$

$$\ln|y| = 2\ln|x| + D$$

$$\ln |y| = \ln x^2 + D$$

$$|y| = \mathrm{e}^{\ln x^2 + D}=\mathrm{e}^{\ln x^2}\cdot\mathrm{e}^D$$

$$|y| = Ex^2 \ \ \text{where} \ \ E=\mathrm{e}^D >0$$

$$y = Fx^2 \ \ \text{where} \ \ F \in \mathbb{R}$$

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  • $\begingroup$ Thanks for the complete explanation $\endgroup$ – Saxobob May 8 '14 at 21:49

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