FP2 Second Order Differential Equation Q

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GPODT
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#1
Report Thread starter 7 years ago
#1
The question:

http://i.imgur.com/8K7p32n.png

I let the PI = x2ex but when I substitute this into the differential equation I obtain a invalid solution

http://i.imgur.com/HXOXFgF.jpg

Why is this?
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lazy_fish
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#2
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Check the \displaystyle \frac {d^2y}{dx^2} of partial integral.

Hope that helps.
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GPODT
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#3
Report Thread starter 7 years ago
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(Original post by lazy_fish)
Check your second derivative of partial integral.

Hope that helps.
Just corrected my mistake. However, I still get an invalid solution (albeit a different one); 2ex = ex ?

Thanks
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lazy_fish
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#4
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That is because your particular integral is not correct. It may be clear through inspection but a more systematic approach would be through:

Set a trial y_t (x) = Ax^2e^x.

Do same thing, that is find \displaystyle \frac {dy_t}{dx} and \displaystyle \frac {d^2y_t}{dx^2} before using in original differential equation, and you'll end up solving for A.

Substitute this value back into trial to get a particular integral.

Good luck.
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GPODT
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Report Thread starter 7 years ago
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(Original post by lazy_fish)
That is because your particular integral is not correct. It may be clear through inspection but a more systematic approach would be through:

Set a trial y_t (x) = Ax^2e^x.

Do same thing, that is find \displaystyle \frac {dy_t}{dx} and \displaystyle \frac {d^2y_t}{dx^2} before using in original differential equation, and you'll end up solving for A.

Substitute this value back into trial to get a particular integral.

Good luck.
I was actually provided with the PI in a subtle manner (I didn't notice it earlier lol). The question says ''show that y = 0.5x2ex is a solution of the differential equation''. This may be quite obvious but how do I know that y = 0.5x2ex represents the particular integral?
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lazy_fish
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#6
Report 7 years ago
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A particular integral is another term for particular solution so just plug it into and thus show it satisfies the original differential equation.
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user2020user
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#7
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(Original post by lazy_fish)
...
(Original post by GPODT)
...
Did you guys get this as your solution at the given boundary conditions?

y = \left( \dfrac{x^2}{2} + x + 1 \right) e^{x}
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lazy_fish
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#8
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Yes.
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