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FP2: 2nd order differential equations - Substitution

tumblr_m8fhc6QMya1rww3jzo1_500.png
I've found the complementary integral and got y=e^-x(Acosx + Bsinx) but I wasn't sure what to use to find my particular integral! Would it be y=lamdaE^mx or y=lamdaE^-x
Original post by creativebuzz
tumblr_m8fhc6QMya1rww3jzo1_500.png
I've found the complementary integral and got y=e^-x(Acosx + Bsinx) but I wasn't sure what to use to find my particular integral! Would it be y=lamdaE^mx or y=lamdaE^-x


You can't use y=λe−xy= \lambda e^{-x} because it's part of the C.I, so what would you use instead?
Reply 2
Avatar for Gome44
Gome44
10
Original post by Jordan\
You can't use y=λe−xy= \lambda e^{-x} because it's part of the C.I, so what would you use instead?


λe^-x is not part of the C.F?
(edited 8 years ago)
Original post by Gome44
λe^-x is not part of the C.I?

Oh I read it wrong, I just saw it in the C.F :getmecoat: It would be fine to just use λe^-x then wouldn't it?
Original post by Gome44
λe^-x is not part of the C.F?


Yup, I've got the answer now thanks!

Would you mind giving me a hand on this question
tumblr_mls07tVxpg1qd7o7ao1_500.jpg

I've got as far as dz/dx = cosx

dy/dx = cosxdy/dx

and I tried to use the product rule to find d^y/dx^2 and I got -sinxdy/dx + cosxd^2y/dx^2 or did I go wrong somewhere?
Reply 5
Avatar for Gome44
Gome44
10
Original post by creativebuzz
Yup, I've got the answer now thanks!

Would you mind giving me a hand on this question
tumblr_mls07tVxpg1qd7o7ao1_500.jpg

I've got as far as dz/dx = cosx

dy/dx = cosxdy/dx

and I tried to use the product rule to find d^y/dx^2 and I got -sinxdy/dx + cosxd^2y/dx^2 or did I go wrong somewhere?


This part in bold is incorrect. dydx=dzdxdydz \dfrac{dy}{dx} = \dfrac{dz}{dx} \dfrac{dy}{dz} , so what you should end up with is dydx=cos⁡(x)dydz \dfrac{dy}{dx} = \cos (x) \dfrac{dy}{dz}

You then need to use implicit differentiation
(edited 8 years ago)

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