Question on Ordinary Differential Equation (ODE)

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SUMMARY

The discussion focuses on solving ordinary differential equations (ODEs) and their relationship to partial differential equations (PDEs). The first equation, du/dy = -u, yields the solution u = A(x)e^(-y), while the second equation, d^2u/dxdy = -du/dx, results in u = e^(-y)(B(X) + c(Y)). The presence of the c(Y) term in the second solution is explained as an additive term that disappears during partial differentiation, analogous to the constant of integration in ODEs.

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Homework Statement



Find the ODE of the following
(1) du/dy = -u
(2) d^2u/dxdy = -du/dx


Homework Equations



For question 1, the answer is u= A(x)e^(-y)
while for question 2, the answer is u= e^(-y)(B(X) + c(Y))


The Attempt at a Solution



I've already solved the question, but just want to ask, if i change (1) to (2), its just a differentiation with respect to dx, but yet the solution contain a c(Y) term, any idea why is this so?
 
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To get those answers, these are PDEs, surely?
In (2), both sides have been differentiated partially wrt x. So any additive term that depends solely on y will have disappeared and cannot be reconstructed from the PDE. It's analogous to the constant of integration in ODEs.
 
If you differentiate (1) wrt x, then
d/dx (du/dy) = d/dy (du/dy) dy/dx = d2u/dy2 dy/dx, by chain rule.
 

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