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Solving partial differential equations

du/dt = Qcosnt + P (d^2)u/dy^2

Q and P are constant

using u = re {f(y) e^(int)}

how does the equation simplify in steps?

the derivatives are partials
Reply 1
Original post by Calculator878


using u = re {f(y) e^(int)}


Well, what's ut\frac{\partial u}{\partial t}?
Original post by Zacken
Well, what's ut\frac{\partial u}{\partial t}?


Re {f(Y) in e^int}
and the other one i got as Re{f'' e^iwt}

dont know how to simplify
Reply 4
Original post by Calculator878
Re {f(Y) in e^int}


Huh? What's the partial derivative w.r.t t of (f(y)eint)\Re ( f(y) e^{int})?
Original post by Zacken
Huh? What's the partial derivative w.r.t t of (f(y)eint)\Re ( f(y) e^{int})?


is it not

1.

Re {f(Y) in e^int}

Reply 6
Original post by Calculator878
is it not

1.

Re {f(Y) in e^int}



Yep!
Original post by Zacken
Yep!


then for the 2nd partial derivative of y i got:

Re{f'' e^iwt}


but after substituting it in I didn't know how to simplify it to whats its supposed to
(something like f'' = in f, with maybe a Q somewhere)
Reply 8
Original post by Calculator878
then for the 2nd partial derivative of y i got:

Re{f'' e^iwt}


but after substituting it in I didn't know how to simplify it to whats its supposed to
(something like f'' = in f, with maybe a Q somewhere)


Why don't you think about writing e^{i...} in cosines and sines and extracting the real part from that? :smile:
Original post by Zacken
Why don't you think about writing e^{i...} in cosines and sines and extracting the real part from that? :smile:


not sure that would help. f(y) is a complex valued function

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