I have a differential equation stating
solve: y'=(y-1)(y+1) when y(0)=3
I separated the dy/dx so I got dy/(y-1)(y+1) = dx took the integral of both sides and solved with PFD
I then got that
x= 1/2 ln(y-1) - 1/2ln(y+1) +c
I solved for C and got that C=.3466
I moved the C to the left side subtracting it from x
x-c = 1/2 ln(y-1) - 1/2ln(y+1)
I raised e to the power of both sides
e^(x-c) = (y-1)^1/2/(y+1)^1/2
Now I do not know how to finish simplifying in order to get the equation down to a form y=x
(y-1)^1/2/(y+1)^1/2 = e^(x-c)
(y-1)/(y+1) = e^(2(x-c))
Using a different c, that is just
(y-1)/(y+1) = ce^(2x) where the new c = e^(- old c)
y-1 = y*ce^(2x) + ce^(2x)
y(1-ce^2x) = 1+ce^2x
y = (1+ce^2x)/(1-ce^2x)
Well, it seems like you're almost there! Let me give you a helping hand to simplify the equation further.
Starting from where you left off: e^(x-c) = (y-1)^(1/2) / (y+1)^(1/2)
Now, let's multiply both sides of the equation by (y+1)^(1/2) to get rid of the denominator on the right side:
e^(x-c) * (y+1)^(1/2) = (y-1)^(1/2)
Next, raise both sides of the equation to the power of 2 to eliminate the square root:
[e^(x-c) * (y+1)^(1/2)]^2 = [(y-1)^(1/2)]^2
Simplifying further:
e^(2(x-c)) * (y+1) = y-1
Expand the left side:
e^(2x-2c) * (y+1) = y-1
Now, let's isolate y by moving terms around:
e^(2x-2c) * y + e^(2x-2c) = y - 1
Subtract y from both sides:
e^(2x-2c) * y - y + e^(2x-2c) = -1
Factor out y:
y * (e^(2x-2c) - 1) + e^(2x-2c) = -1
Finally, we can express this in the form y = x by rearranging the terms:
y = (-1 - e^(2x-2c)) / (1 - e^(2x-2c))
So, this is the simplified form of your differential equation y' = (y-1)(y+1) with the initial condition y(0) = 3.
To continue simplifying the expression, you can apply some algebraic manipulations.
1. Start by simplifying the left side of the equation using the properties of exponents. Recall that e^(x-c) can be written as e^x / e^c.
e^(x-c) = (y-1)^(1/2) / (y+1)^(1/2)
2. Multiply both sides by (y+1)^(1/2) to eliminate the denominator on the right side.
e^(x-c) * (y+1)^(1/2) = (y-1)^(1/2)
3. Square both sides of the equation to get rid of the square root.
[e^(x-c) * (y+1)^(1/2)]^2 = [(y-1)^(1/2)]^2
e^(2(x-c)) * (y+1) = y - 1
4. Expand the left side and isolate y terms.
e^(2x-2c) * (y+1) = y - 1
e^(2x-2c)y + e^(2x-2c) = y - 1
e^(2x-2c)y - y = -1 - e^(2x-2c)
(e^(2x-2c) - 1)y = -1 - e^(2x-2c)
5. Finally, divide both sides by (e^(2x-2c) - 1) to isolate the y term.
y = (-1 - e^(2x-2c)) / (e^(2x-2c) - 1)
This is the final form of the solution to the differential equation y' = (y - 1)(y + 1) given y(0) = 3.
To simplify the equation further and express it as y = f(x), we can start by eliminating the exponential term on the left side of the equation.
e^(x - c) = (y - 1)^(1/2) / (y + 1)^(1/2)
Since c is a constant, we can use the property of exponents: e^(x - c) = e^x / e^c.
e^x / e^c = (y - 1)^(1/2) / (y + 1)^(1/2)
Next, we can square both sides to remove the square roots:
( e^x / e^c )^2 = ( (y - 1) / (y + 1) )^1/2
Simplifying the left side by squaring the fraction:
( e^x )^2 / ( e^c )^2 = ( y - 1 ) / ( y + 1 )
Expanding the squares:
e^(2x) / e^(2c) = ( y - 1 ) / ( y + 1 )
Next, we can cross-multiply to separate y terms:
e^(2x) * ( y + 1 ) = ( y - 1 ) * e^(2c)
Expanding the left side:
e^(2x) * y + e^(2x) = y * e^(2c) - e^(2c)
Moving y terms to one side:
e^(2x) * y - y * e^(2c) = - e^(2c) - e^(2x)
Factoring out y:
y * ( e^(2x) - e^(2c) ) = - e^(2c) - e^(2x)
Finally, divide both sides by (e^(2x) - e^(2c)):
y = (- e^(2c) - e^(2x)) / (e^(2x) - e^(2c))
Now, if you substitute the value of c (0.3466) into the equation, you should have y = f(x).