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Which of the following is a solution to the separable differentiable equation:

$$\frac{dy}{dx}=\frac{xy}{\ln y }$$

$A.\ \displaystyle e^{|x|}$

$B.\ \displaystyle e^{\sqrt{\frac{x^2}2}}$

$C.\ \displaystyle \frac12$$e^{\sqrt{x^2+1}}$

$D.\ \displaystyle e^{-x}$

What I did was separate the functions and get:

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

Then integrating I get:

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

But that doesn't match any of my answers. Help will be appreciated.

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  • $\begingroup$ The statement of the problem is a bit weak -- it should ask you for an explicit function $ \ y \ = \ f(x) \ $ that would solve the differential equation. So you want to solve your result for $ \ y \ $ as some function of $ \ x \ $ on the right-hand side; keep in mind that " a solution of the differential equation" means that the arbitrary constant $ \ C \ $ can take on a specific value. $\endgroup$ – colormegone Jul 29 '14 at 0:24
  • $\begingroup$ Yeah I think the problem is simply asking for the equation without C in it although it should really be in there. Like they would normally give " for a value F(3) = 5" or something like that. But I think this is asking for the samething just without C in it., $\endgroup$ – Panthy Jul 29 '14 at 0:25
  • $\begingroup$ Any function with the correct form and some value for $ \ C \ $ would be a solution, so the choice should be a function with the correct form. Let me know if I edited any of your choices incorrectly. $\endgroup$ – colormegone Jul 29 '14 at 0:28
  • $\begingroup$ Yeah I just don't know how to do this problem or if its even do-able. But yes you did edit them properly. Sorry for my poor mathjax skills :p $\endgroup$ – Panthy Jul 29 '14 at 0:29
  • $\begingroup$ What B.S. shows is correct: multiply your result by 2 on both sides, take the square-root of both sides, and then "exponentiate" both sides. That gives the general solution to the differential equation. Do any of your choices resemble that (with a numerical value for $ \ C \ $ inserted)? $\endgroup$ – colormegone Jul 29 '14 at 0:31
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You were right. We have:

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

Now multiply both sides by $2$:

$$\ln^2 y = x^2+C$$

NOTE: $C$ is any arbitrary constant, so we'll just make $2C=C$ because it doesn't really matter (still some constant).

Now you want to isolate for $y$, that's how you'll get the solution. Hence, first take the square root of both sides so you'll be 1 step closer:

$$\ln y = \sqrt{x^2+C}$$

And remember the definition of the natural logarithm. This can be turned into:

$$y=e^\sqrt{x^2+C}$$

Remember that $C$ can be any number, so you can figure it out now.

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$$\ln^2(y)=x^2+C\to \ln(y)=\pm\sqrt{x^2+C}\to y=\exp(\pm\sqrt{x^2+C})$$

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  • $\begingroup$ that made no sense to me... :( $\endgroup$ – Panthy Jul 29 '14 at 0:16
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    $\begingroup$ Makes a whole lot of sense! $\endgroup$ – Namaste Jul 29 '14 at 11:18
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Here this might help:

Okay, so the first thing you do is:

Separate variables on either sides of the equation:

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

If we integrate notice that if we make $u = \ln y$

$$udu = xdx$$

And it becomes an easy integral: $$\frac{\ln^2 y}{2} = \frac{x^2}{2} + C$$

Multiply both sides by 2:

$$\ln^2 y = x^2 + 2C$$

It doesn't matter what C is, so $C = 2C$ in this instance.

Continuing:

$$\ln y = \sqrt{x^2+C}$$ $$y = e^{\sqrt{x^2+C}}$$

EDIT

The answer can be either A or D, depending on how you look at it.

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  • $\begingroup$ How does the constant term escape ending up under the radical? $\endgroup$ – colormegone Jul 29 '14 at 0:36
  • $\begingroup$ Your last two equations have incorrect exponents $\endgroup$ – ClassicStyle Jul 29 '14 at 0:36
  • $\begingroup$ @RecklessReckoner oops didn't mean for that too happen, forget formatting :D $\endgroup$ – Varun Iyer Jul 29 '14 at 0:36
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    $\begingroup$ @Panthy The answer is D. You end up having $y=e^\sqrt{x^2+C}$. If $C=0$, then you have $y=e^\sqrt{x^2}$. $\sqrt{x^2}=\pm x$. $\endgroup$ – Shahar Jul 29 '14 at 0:40
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    $\begingroup$ @Shahar but NOTE THAT $|x| = \sqrt{x^2}$ as well. So answer can be A too. $\endgroup$ – Varun Iyer Jul 29 '14 at 0:41

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