Find f(x,y) given partial derivative and initial condition

  • #1
songoku
2,319
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Homework Statement
Please see below
Relevant Equations
Partial derivative

Integration (maybe)
1697703842950.png


My attempt:
$$\frac{\partial f}{\partial x}=-\sin y + \frac{1}{1-xy}$$
$$\int \partial f=\int (-\sin y+\frac{1}{1-xy})\partial x$$
$$f=-x~\sin y-\frac{1}{y} \ln |1-xy|+c$$

Using ##f(0, y)=2 \sin y + y^3##:
$$c=2 \sin y + y^3$$

So:
$$f(x,y)=-x~\sin y-\frac{1}{y} \ln |1-xy|+2 \sin y + y^3$$

Is my answer correct? In the lesson itself, there is no integration when learning partial derivative but I can't think of any other way to solve the question without integration.

Thanks
 
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  • #3
Thank you very much Demystifier
 
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  • #4
songoku said:
My attempt:
##\frac{\partial f}{\partial x}=-\sin y + \frac{1}{1-xy}##
##\int \partial f=\int (-\sin y+\frac{1}{1-xy})\partial x##
The second line should be written like this:
##\int \frac{\partial f}{\partial x}dx = \int (-\sin y+\frac{1}{1-xy})dx##
 
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  • #5
Mark44 said:
The second line should be written like this:
##\int \frac{\partial f}{\partial x}dx = \int (-\sin y+\frac{1}{1-xy})dx##
Oh ok. Thank you very much Mark44
 
  • #6
I know it’s obvious and shouldn’t need saying, but just in case...

The final answer is ##f(x,y)=-x~\sin y-\frac{1}{y} \ln |1-xy|+2 \sin y + y^3##.

To check, simply evaluate ##f(0,y)## and ##\frac{\partial f}{\partial x}## to confirm that they match the given expressions.

(I’d guess - not being a mathematician - that there is some theorem which guarantees that the answer for ##f(x,y)## must be unique.)
 
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  • #7
Steve4Physics said:
To check, simply evaluate ##f(0,y)## and ##\frac{\partial f}{\partial x}## to confirm that they match the given expressions.
Good point. This is something that should always be done when solving differential equations.

Steve4Physics said:
(I’d guess - not being a mathematician - that there is some theorem which guarantees that the answer for f(x,y) must be unique.)
If there's a theorem about this, it escapes me at the moment. Without the initial condition, you get a whole family of solutions. The initial condition for f(0, y) nails the family down to a single function.
 
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