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# Solving partial differential equations Watch

1. 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
2. (Original post by Calculator878)

using u = re {f(y) e^(int)}
Well, what's ?
3. (Original post by Zacken)
Well, what's ?
Re {f(Y) in e^int}
4. and the other one i got as Re{f'' e^iwt}

dont know how to simplify
5. (Original post by Calculator878)
Re {f(Y) in e^int}
Huh? What's the partial derivative w.r.t t of ?
6. (Original post by Zacken)
Huh? What's the partial derivative w.r.t t of ?
is it not
1. Re {f(Y) in e^int}
7. (Original post by Calculator878)
is it not
1. Re {f(Y) in e^int}
Yep!
8. (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)
9. (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?
10. (Original post by Zacken)
Why don't you think about writing e^{i...} in cosines and sines and extracting the real part from that?
not sure that would help. f(y) is a complex valued function

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