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I have worked the problem, but I keep getting the wrong answer. I've also looked at some other similar questions from here, but I still don't understand what I'm doing incorrectly.

Attempt:

First, the derivations I got:

$y=C_1 x^{2}+C_2 x^{3}$

$y'=2C_1x+3C_2x^2$

$y''=2C_1+6C_2x$

$\Longrightarrow C_2=\dfrac{y''-2C_1}{6x}$

Then, I started substituting:

$y'=2C_1 x+3\left(\dfrac{y''-2C_1}{6x}\right) x^2=2C_1 x+\left(\dfrac{1}{2}\right) xy''-C_1 x$

$y'=C_1x+\left(\dfrac{1}{2}\right) xy''$

$\Longrightarrow C_1=\dfrac{y'}{x}-\dfrac{y''}{2}$

When I plugged $C_1$ and $C_2$ back into $y=C_1 x^{2}+C_2 x^{3}$:

$y=\left(\dfrac{y'}{x}-\dfrac{y''}{2}\right) x^2+\left(\dfrac{y''-2\left(\dfrac{y'}{x}-\dfrac{y''}{2}\right)}{6x}\right) x^3$

And when I simplify, I end up with:

$y=\dfrac{2xy'}{3}-\dfrac{x^2 y''}{2}$

But I'm being told this is incorrect. Is there a different approach to this problem and/or have I made a mistake somewhere?

Thanks in advance!

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    $\begingroup$ Your simplification at the end is incorrect. You missed $\frac{1}{3}x^2y''$, hence it should be $\displaystyle y=\frac{2xy'}{3}-\frac{x^2y''}{6}$. Rather than that your solution is perfect. $\endgroup$ – Galc127 Aug 31 '16 at 4:20
  • $\begingroup$ Thank you! I saw where I made a mistake. $\endgroup$ – nexicon Aug 31 '16 at 4:46
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You want to perform differentiations that remove the constants directly. So I would probably start with dividing everything by $x^2$ and differentiating the result, like thus:

$\begin{eqnarray}y & = & C_1 x^2 + C_2 x^3 \\ \frac{y}{x^2} & = & C_1 + C_2 x \\ \frac{d}{dx}\left(yx^{-2}\right) & = & \frac{d}{dx}\left(C_1 + C_2 x\right) \\ y'x^{-2} - 2yx^{-3} & = & C_2 \\ \frac{d}{dx}\left(y'x^{-2} - 2yx^{-3}\right) & = & \frac{d}{dx} C_2 \\ y''x^{-2} - 4y'x^{-3} + 6yx^{-4} & = & 0 \\ y'' - 4y'x^{-1} + 6yx^{-2} & = & 0\\ y'' & = & \frac{4y'}{x} - \frac{6y}{x^2} \end{eqnarray}$

Which is, I believe, equivalent to the corrected answer @Galc127 has suggested, but where I've focused on the highest order derivative of $y$.

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  • $\begingroup$ Thanks, I never thought about doing it that way! $\endgroup$ – nexicon Aug 31 '16 at 4:47
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It is true that $y=C_1x^2+C_2x^3\Rightarrow y=\frac{2xy'}{3}-\frac{x^2y''}{2}$. But conversely, it doesn't hold. Maybe because of this reason, your answer is not incorrect. The correct answer should be such that $y=C_1x^2+C_2x^3\Leftrightarrow$correct answer

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  • $\begingroup$ No, it does not hold. If $y=C_1x^2+C_2x^3$, then $y'=2C_1x+3C_2x^2,y''=2C_1+6C_2x$ and then $$\frac{2xy'}{3}-\frac{x^2y''}{2}=\frac{2x(2C_1x+3C_2x^2)}{3}-\frac{x^2(2C_1+6C_2x)}{2}=\frac{1}{3}C_1x^2-C_2x^3\neq y$$ $\endgroup$ – Galc127 Aug 31 '16 at 4:24

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