Exact Equation
An equation of form
$$M(x,y)dx,\;N(x,y)dy=0$$
has some unique function I(x,y) whose partial derivatives can be written in place of M and N
$$\frac{\partial\;I}{\partial\;x}dx+\frac{\partial\;I}{\partial\;y}dy=0$$
Fact: Our job is to find out I(x,y) if exists
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Step#1: First, we check given equation is exact or not
$$\frac{\partial\;N}{\partial\;x}=\frac{\partial\;M}{\partial\;y}$$
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Step#2: Then, to find I(x,y) we further do any of these
- $$I(x,y)=\int\;M(x,y)dx$$ where x as an independent variable
- $$I(x,y)=\int\;N(x,y)dy$$ where y as an independent variable
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Step#3: then further we do some important which help us to find general solution
$$I(x,y)=C$$
Example:
$$(3x^2-2xy+2)dx+(6y^2-x^2+3)dy=0$$
Solution:
$$M=3x^2-2xy+2$$
$$N=6y^2-x^2+3$$
$$\frac{\partial\;N}{\partial\;x}=\frac{\partial\;M}{\partial\;y}=-2x$$
So, given equation is exact
Now, we try to find I(x,y)
Also,
$$I(x,y)=\int\;N(x,y)dy$$
$$I(x,y)=\int\;(6y^2-x^2+3)dy$$
$$I(x,y)=2y^3-x^2y+3y+f(x)....(a)$$
Now,differentiate I(x,y) and write as equal to M
$$\frac{\partial\;I}{\partial\;x}=M(x,y)$$
$$2xy+f'x=3x^2-2xy+2$$
$$f'x=3x^2+2$$
Then, perform integration
$$fx=x^3+2x+c....(b)$$
Now we replace gx in equation a in equation (b)
$$I(x,y)=2y^3-x^2y+3y+x^3+2x+C$$
$$I(x,y)=C$$
Which is general solution form
Fact: Previous C’s are different constant here then can transform as
$$C=c_1+c_2$$
$$C=2y^3-x^2y+3y+x^3+2x$$