Integration of two variable function

Note: This concept idea hasn't been proven yet or maybe mathematically incorrect. Any contribution would be greatly appreciated.

Basic Knowledge of Integration and Functions

In mathematics, usually we take `x` as independent variable and `y` as dependent variable. In General, a function is written as

`y=ax^2+bx+c`

If we differentiate the above function, we get:

`\frac{dy}{dx}=2ax+b `

i.e  `dy=(2ax+b)dx `

Generally, a Integral function is written as:

`f(x)=\int (2ax+b) dx `

Note: There is a wrong perception that, Integration give area under function, but actually, Definite Integration give area under a given function `f(x)` and Indefinite Integration gives an entire family of functions with an infinite number of members.

Therefore, it is understood that: 

`f(x)= \int dy=\int (2ax+b)dx`

So, After clearing the basic knowledge, we can see how to Integrate something like this:

`\int f(x,y) dx`   (where both `x` and `y` are variable) 

 

Integration of two variable function `F(x,y)`

Lets take an example function:

`f'(x,y) = 3x^2+y `

 So, how to find: `\int (3x^2+y)dx`

As mentioned above, it is understood that it means:

`\frac{dy}{dx}=3x^2+y`   (and we have to find its integral)

`\implies \frac{dy}{dx}-y=3x^2`   (Rearranging the Equation)

 

Now we have convert the Integral Question, into a Differential Equation.

So, we can solve this using, P,Q Factor method (What is Factor Method ?)

`\frac{dy}{dx}-y=3x^2`

 Let `P=(-1)` and `Q=3x^2`

Therefore:

Integral Factor `(I.F)`=`e^{\intP dx}`

`\implies (I.F)` = `e^{\int-1 dx}`

`\implies (I.F)` = `e^{-x}`

Hence our result is:

`y(I.F)=\int Q (I.F) dx` 

`\implies ye^{-x}=\int 3x^2e^{-x} dx`

`\implies ye^{-x} =-3(x^2e^-x +2xe^-x +2e^-x) +C`   (using By Parts)

`\implies ye^(-x)+3x^2e^{-x}+6xe^{-x}+6e^{-x}=C`   (rearranging the equation)

`\implies \ln(e^{-x}(y+3x^2+6x+6))=\ln C`    (taking log both the sides)

`\implies -x+\ln(y+3x^2+6x+6)=A`

 

Hence, `f(x,y)` =  `-x+\ln(y+3x^2+6x+6)=A`

Fun Assignment: Differentiate the Above function, and you will get the original `f'(x,y)`