Differentiation and integration of implicit functions

[chwala]

1. Given the function $xy+cos y+6xy^2=0$ , it follows that $dy/dx=-y/x-siny+12xy$2. My problem is how do we integrate this derivative $dy/dx=-y/x-siny+12xy$ to get back the original function3.$∫dy/dx dx=y$

 

[RUber]

For your dy/dx, you might have missed one application of the product rule.
I get:
$y + x\frac{dy}{dx}-\sin (y) \frac{dy}{dx}+6y^2 +12xy \frac{dy}{dx}= 0 \\ \frac{dy}{dx}(x-\sin(y)+12xy)=-(6y^2+y)\\ \frac{dy}{dx} = - \frac{6y^2+y}{x-\sin(y)+12xy}$
That might be why you are having trouble recovering the original function.

In general to take the integral, your goal would be to separate the dy and dx and integrate separately. Then you can retrieve the original function with the addition of some extra parts that would have been driven to zero by the derivative.
In applications, exact recovery would only be feasible if you had some additional information about the function (e.g. boundary conditions)

 

[Buzz Bloom]

Hi chwala:

The equation you got for the derivative dy/dx is not right. My guess is that when you differentiated

xy+cosy+6xy2=0​

with respect to x, you forgot that "x" is also a function of x, and/or you forgot about differentiating a product of two functions.

Also, what you said in (2.) about what "My problem is" is unclear. I am guessing that what you want is to solve the original equation for y as a function of x, and you are using a method to find an expression for dy/dx which you can integrate to get this function.

Hope this helps.

Regards,
Buzz

 

[chwala]

Agreed, I have seen my mistake, thanks to Ruber and Buzz. Generally are we saying that the general rule for these kind of derivatives,is to use separation of variables to get original function?

 

[RUber]

That is always the first thing to try.

 

[chwala]

Thanks let me try and see...

 

[Mark44]

1. Given the function $xy+cos y+6xy^2=0$ , it follows that $dy/dx=-y/x-siny+12xy$2. My problem is how do we integrate this derivative $dy/dx=-y/x-siny+12xy$ to get back the original function3.$∫dy/dx dx=y$

@chwala, when you post a question, please do not delete the homework template. Its use is required.

 

[chwala]

Ok ok point noted.

 

[chwala]

Ok ok point noted.

I came up with this other implicit problem $x^2y+x cos y = 6x$ it follows that the derivative $dy/dx= 6-2xy-cosy/x^2-xsin y$ now in getting the original function $y$ , i separated the terms as follows $2xy dx + x^2dy+cosy dx - x siny dy= 6dx$ which is an exact equation of the form $M(x,y) dx + N(x,y) dy = 0$ for lhs integrating this will give you the original function $y$. I hope i have posted in the right place advice if otherwise.

 

[chwala]

I came up with this other implicit problem $x^2y+x cos y = 6x$ it follows that the derivative $dy/dx= 6-2xy-cosy/x^2-xsin y$ now in getting the original function $y$ , i separated the terms as follows $2xy dx + x^2dy+cosy dx - x siny dy= 6dx$ which is an exact equation of the form $M(x,y) dx + N(x,y) dy = 0$ for lhs integrating this will give you the original function $y$. I hope i have posted in the right place advice if otherwise.

Now for my initial problem:
$xy+cos y+6xy^2=0$ we have $dy/dx= -(6y^2+y)/x-siny+12xy$ to get original function i now re-arrange the terms to have $xdy-siny dy+12xy dy=-(6y^2dx+ydx)$ integrating the lhs yields the required solution.