The Student Room Group
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>

FP2 Second Order Differential Equation Q

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?
Check the d2ydx2\displaystyle \frac {d^2y}{dx^2} of partial integral.

Hope that helps. :smile:
(edited 9 years ago)
Reply 2
Avatar for GPODT
GPODT
OP
16
Original post by lazy_fish
Check your second derivative of partial integral.

Hope that helps. :smile:


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

Thanks
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 yt(x)=Ax2exy_t (x) = Ax^2e^x.

Do same thing, that is find dytdx\displaystyle \frac {dy_t}{dx} and d2ytdx2\displaystyle \frac {d^2y_t}{dx^2} before using in original differential equation, and you'll end up solving for AA.

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

Good luck. :smile:
(edited 9 years ago)
Reply 4
Avatar for GPODT
GPODT
OP
16
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 yt(x)=Ax2exy_t (x) = Ax^2e^x.

Do same thing, that is find dytdx\displaystyle \frac {dy_t}{dx} and d2ytdx2\displaystyle \frac {d^2y_t}{dx^2} before using in original differential equation, and you'll end up solving for AA.

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

Good luck. :smile:


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?
A particular integral is another term for particular solution so just plug it into and thus show it satisfies the original differential equation. :smile:
(edited 9 years ago)
Original post by lazy_fish
...

Original post by GPODT
...


Did you guys get this as your solution at the given boundary conditions?

y=(x22+x+1)exy = \left( \dfrac{x^2}{2} + x + 1 \right) e^{x}
Yes. :smile:

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you