[SOLVED] How to answer this question

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September 24th 2009, 10:53 PM
mark1950

[SOLVED] How to answer this question

Show that $y\frac{d^2y}{dx^2} + (\frac{dy}{dx})^2 - 2y\frac{dy}{dx} = 4$

given that $y = (e^{2x} - 4x + 3)^{\frac{1}{2}}$
September 24th 2009, 10:55 PM
Prove It

Quote:

Originally Posted by mark1950

Show that $y\frac{d^2y}{dx^2} + (\frac{dy}{dx})^2 - 2y\frac{dy}{dx} = 4$

given that $y = (e^{2x} - 4x + 3)^{\frac{1}{2}}$

Can you work out $\frac{dy}{dx}$ ?

Or $\frac{d^2y}{dx^2}$ ?

Can you substitute these into the DE?
September 25th 2009, 12:34 AM
mark1950

Yes, I can but what I got was a very, very long equation.
September 25th 2009, 12:35 AM
Prove It

Quote:

Originally Posted by mark1950

Yes, I can but what I got was a very, very long equation.

Well that's how it's solved, so do it.
September 25th 2009, 12:57 AM
mr fantastic

Quote:

Originally Posted by mark1950

Yes, I can but what I got was a very, very long equation.

Please post your work so that it can be reviewed. Be sure to include your answer for $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$ .
September 25th 2009, 01:32 AM
mark1950

@Prove it

Dude, if I knew how to solve it, I would not have posted this question here.

@Mr. Fantastic

Thanks. After this step, I simply do not know how to continue. I tried substituting dy/dx into them but simply couldn't get rid of the exponential, e so as to prove that its right. You know what I mean, right?

$\frac{dy}{dx} <br /> = \frac{1}{2}(e^{2x} - 4x + 3)^{-\frac{1}{2}}(2e^{2x} - 4)$
$= \frac{e^{2x} - 2}{\sqrt{e^{2x} - 4x + 3}}$
$\frac{d^2y}{dx^2}<br /> = \frac{(e^{2x} - 4x + 3)(2e^{2x}) - (e^{2x} - 2)^2}{\sqrt{(e^{2x} - 4x + 3)^3}}$
September 25th 2009, 01:39 AM
Prove It

Quote:

Originally Posted by mark1950

Show that $y\frac{d^2y}{dx^2} + (\frac{dy}{dx})^2 - 2y\frac{dy}{dx} = 4$

given that $y = (e^{2x} - 4x + 3)^{\frac{1}{2}}$

So now plug everything you've found into

$y\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2 - 2y\frac{dy}{dx}$

and simplify.
September 25th 2009, 01:48 AM
mr fantastic

Quote:

Originally Posted by mark1950

@Prove it

Dude, if I knew how to solve it, I would not have posted this question here.

@Mr. Fantastic

Thanks. After this step, I simply do not know how to continue. I tried substituting dy/dx into them but simply couldn't get rid of the exponential, e so as to prove that its right. You know what I mean, right?

$\frac{dy}{dx} <br /> = \frac{1}{2}(e^{2x} - 4x + 3)^{-\frac{1}{2}}(2e^{2x} - 4)$
$= \frac{e^{2x} - 2}{\sqrt{e^{2x} - 4x + 3}}$
$\frac{d^2y}{dx^2}<br /> = \frac{(e^{2x} - 4x + 3)(2e^{2x}) - (e^{2x} - 2)^2}{\sqrt{(e^{2x} - 4x + 3)^3}}$

Your derivatives are OK. Now the unfortunate fact is that you need to substitute y and the derivatives into the left hand side of the differential equation. Then simplify it (hopefully to 4). This will take a fair bit of algebra but is not difficult - just very tedious (hence my use above of the word unfortunate).
September 25th 2009, 01:49 AM
Prove It

Quote:

Originally Posted by mr fantastic

Your derivatives are OK. Now the unfortunate fact is that you need to substitute y and the derivatives into the left hand side of the differential equation. Then simplify it (hopefully to 4). This will take a fair bit of algebra but is not difficult - just very tedious (hence my use above of the word unfortunate).

My CAS tells me the answer is, indeed, 4.
September 25th 2009, 01:53 AM
mr fantastic

Quote:

Originally Posted by Prove It

My CAS tells me the answer is, indeed, 4.

It's always good to know that what you're trying to do can be done (there's always a niggling suspicion that the given solution is wrong in problems like this ....)
September 25th 2009, 01:54 AM
Prove It

Quote:

Originally Posted by mr fantastic

It's always good to know that what you're trying to do can be done (there's always a niggling suspicion that the given solution is wrong in problems like this ....)

Indubidably.
September 25th 2009, 04:10 AM
mark1950

Oh darn...I was looking for a simpler way but nvm...there are some things that can't be simplified. Thanks, anyway for the assurance that my derivatives are correct.