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

[songoku]

Homework Statement

Please see below

Relevant Equations

Partial derivative

Integration (maybe)

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

 

[Demystifier]

It's correct.

 

[songoku]

Thank you very much Demystifier

 

[Mark44]

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:
$\int \frac{\partial f}{\partial x}dx = \int (-\sin y+\frac{1}{1-xy})dx$

 

[songoku]

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

 

[Steve4Physics]

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.)

 

[Mark44]

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.

(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.