Y=f(x) NOT y=f(x)f(y)?
dy/dx = 2(x^3 + xy + x)/(x^2 + 1)
After integrating I get:
y = ylog(x^2 + 1) + x^2 + C
Which is not expressing y as a function of x, as it is showing a function of x and y (i.e. both x and y are on the right hand side of the equation)?
How do you obtain y = f(x)?
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M$2 Answers
Dividing x^2 + y + 1 by x^2 + 1 yields 1 + y / (x^2 + 1) so:
dy / dx = 2x (1 + y / (x^2 + 1) )
EQUATION A: dy / dx = 2x + 2xy / (x^2 + 1)
This is an inseparable differential equation because you can't isolate the y variable so you will need to use a special factor which we will define as μ (pronounced "mu").
EQUATION 1: μ = e ^( ∫p(x) dx )
Another equation we will need is:
EQUATION 2: dy / dx + p(x) y = q(x)
And last but not least we will need to use:
EQUATION 3: y = ( ∫μ q(x) dx ) / μ
First we will manipulate EQUATION A into the form of EQUATION 2:
dy / dx = 2x + 2xy / (x^2 + 1)
dy / dx + ( -2x / (x^2 + 1) ) y = 2x
Now we know that:
p(x) = -2x / (x^2 + 1)
and
q(x) = 2x
Next we solve for μ using EQUATION 1:
μ = e ^( ∫p(x) dx ) = e ^( ∫ -2x / (x^2 + 1) dx )
Solve the integral using substitution:
∫ -2x / (x^2 + 1) dx = -2( 1/2 ) ∫ da / a = -ln a = - ln (x^2 + 1)
a = x^2 + 1
da = 2x dx
xdx = da / 2
Then continuing with EQUATION 1:
μ = e ^( ∫ -2x / (x^2 + 1) dx ) = e^( - ln (x^2 + 1) ) = e^( ln (x^2 + 1) )^(-1)
μ = 1 / (x^2 + 1)
Plug μ and q(x) into EQUATION 3 to solve for y:
y = ( ∫μ q(x) dx ) / μ = ( ∫ 1 / (x^2 + 1) 2x dx ) / (1 / (x^2 + 1))
Using substitution just as you did before you will obtain:
y = (ln (x^2 + 1) + C) / (1 / (x^2 + 1))
y = (x^2 + 1) * ln (x^2 + 1) + C * (x^2 + 1)
Stewart, J. (2003). Calculus: Early Transcendentals, fifth edition.
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M$y-ylog(x^2+1)=x^2+C
Take y common factor
y(1-log(x^2+1))=x^2+C
Divide
y=(x^2+C)/(1-log(x^2+1))
But there is a problem
I suppose you rewrote like
dy/dx=2x+2xy/(x^2+1)
when you integrated in the right side, you treated y as a constant
You can't do that since y is a function of x
How do you solve that?
You rewrite
dy/dx-y*2x/(x^2+1)=2x
That is a linear differential equation
See at this website http://tutorial.math.lamar.edu/Classes/DE/Linear.aspx
how to solve a linear differential equation using the integrating factor.
my mind
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M$
According to wolfram alpha the solution is:
y(x) = c*(x^2+1)+(x^2+1)*ln(x^2+1)
So it appears that the solution to the integral in the question is not correct as this answer states.
http://www.wolframalpha.com/input/?i=integrate+dy%2Fdx+%3D+2%28x^3+%2B+xy+%2B+x%29%2F%28x^2+%2B+1%29+
:) wolfram alpha rules
Well, you get the result but it doesn't explain from where you obtained the formula 3
Instead of memorizing formulas (you may not remember at exams), it is better to follow the logical steps. This steps are given in the webpage I provided.
From here
dy / dx + p(x) y = q(x)
if you multiply by
e ^( ∫p(x) dx )
You get
dy/dx*e ^( ∫p(x) dx ) +p(x) y *e ^( ∫p(x) dx )=q(x) *e ^( ∫p(x) dx )
which is by the product rule
d/dx(y* e^( ∫p(x) dx ))=q(x) *e ^( ∫p(x) dx )
If you integrate, you get
y* e^( ∫p(x) dx ) = ∫q(x) *e ^( ∫p(x) dx )+C from which by division you obtain formula of equation 3
This method always works
Let's see on your example
By multiplying by e^( - ln (x^2 + 1) ) (let's not simplify now), you get
dy / dx *e^( - ln (x^2 + 1) ) + ( -2x / (x^2 + 1) ) y*e^( - ln (x^2 + 1) ) = 2x *e^( - ln (x^2 + 1) )
Note that ( -2x / (x^2 + 1) ) *e^( - ln (x^2 + 1) )=d/dx(e^( - ln (x^2 + 1) ))
Then you rewrite the left part using the product rule backwards-that was the purpose of multiplying by e^( - ln (x^2 + 1) ):
d/dx(y*e^( - ln (x^2 + 1) ))=2x *e^( - ln (x^2 + 1) )
Now we integrate on both sides and we have
y*e^( - ln (x^2 + 1) )=∫ 2x *e^( - ln (x^2 + 1) )+C
So
y=(∫ 2x *e^( - ln (x^2 + 1) )+C)/e^( - ln (x^2 + 1) )
Also, you should have added another bracket here: (e^( ln (x^2 + 1) ))^(-1)
because if you write x^y^z, you should clarify if it is x^(y^z) or (x^y)^z
They are not the same thing