- #1

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- TL;DR Summary
- I am going through these notes; i want to check that i am gettting it right...

Solve the pde;

##U_{xy}=0##

This is part of the notes;

My own way of thought;

Given;

##U_{xy}=0##

then considering ##U_x## as on ode in the ##y## variable; we integrate both sides with respect to ##y## i.e

##\dfrac{du}{dx} \int \dfrac{1}{dy} dy=\int 0 dy##

this is the part i need insight...the original problem involves partial derivatives but in this case when we integrate with respect to ##y## we are integrating with respect to ##dy## and not ##∂y## ...correct?

##U_x= 0 + k##, where ##k## is a constant in terms of ##x## therefore,

##U_x= f(x)##, an arbitrary function of ##x##... which is an ode for ##u## in the ##x## variable. On integrating again with respect to ##x## we get,

##U(x,y) = F(x) + k##, where ##k## is a constant in terms of ##y## therefore,

##U(x,y) = F(x) + H(y)##

cheers!

My own way of thought;

Given;

##U_{xy}=0##

then considering ##U_x## as on ode in the ##y## variable; we integrate both sides with respect to ##y## i.e

##\dfrac{du}{dx} \int \dfrac{1}{dy} dy=\int 0 dy##

this is the part i need insight...the original problem involves partial derivatives but in this case when we integrate with respect to ##y## we are integrating with respect to ##dy## and not ##∂y## ...correct?

##U_x= 0 + k##, where ##k## is a constant in terms of ##x## therefore,

##U_x= f(x)##, an arbitrary function of ##x##... which is an ode for ##u## in the ##x## variable. On integrating again with respect to ##x## we get,

##U(x,y) = F(x) + k##, where ##k## is a constant in terms of ##y## therefore,

##U(x,y) = F(x) + H(y)##

cheers!

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