Question

Question #dbf40

Answer 1

Separate variables, perform partial fraction decomposition, integrate.

Explanation:

Rearranging,

`1/(1-y^2)"d"y="d"x`.

Integrating both sides and factorising gives,

`int 1/((1-y)(1+y))"d"y=x`.

Performing a partial fraction decomposition gives,

`1/2int 1/(1-y) + 1/(1+y) = x`
`-1/2ln(1-y)+1/2ln(1+y) = x`
`1/2ln((1+y)/(1-y))=x`

Adding in the constant of integration gives,

`1/2ln(C(1+y)/(1-y))=x`
`ln(C(1+y)/(1-y))=2x`
`e^(2x)=C*((1+y)/(1-y))`

Redefining the constant,

`Ke^(2x) = (1+y)/(1-y)`.

Solving for y,

`y(1+Ke^(2x))=(Ke^(2x)-1)`
`y=(Ke^(2x)-1)/(Ke^(2x)+1)`.

Then choose K=1.

Answer 2

Given: `y=(e^(2x)-1)/(e^(2x)+1)`

Use the quotient rule:

`dy/dx = ((d(e^(2x)-1))/dx(e^(2x)+1)- (e^(2x)-1)(d(e^(2x)+1))/dx)/(e^(2x)+1)^2`

`dy/dx = ((2e^(2x))(e^(2x)+1)- (e^(2x)-1)(2e^(2x)))/(e^(2x)+1)^2`

`dy/dx = (4e^(2x))/(e^(2x)+1)^2" [1]"`

`1 - y^2 = 1 - ((e^(2x)-1)/(e^(2x)+1))^2`

`1 - y^2 = 1 - (e^(2x)-1)^2/(e^(2x)+1)^2`

`1 - y^2 = (e^(2x)+1)^2/(e^(2x)+1)^2 - (e^(2x)-1)^2/(e^(2x)+1)^2`

`1 - y^2 = ((e^(2x)+1)^2 - (e^(2x)-1)^2)/(e^(2x)+1)^2`

`1 - y^2 = ((e^(4x)+ 2e^(2x)+1) - (e^(4x)- 2e^(2x)+1))/(e^(2x)+1)^2`

`1 - y^2 = (e^(4x)+ 2e^(2x)+1 - e^(4x)+ 2e^(2x)-1)/(e^(2x)+1)^2`

`1 - y^2 = (2e^(2x)+ 2e^(2x))/(e^(2x)+1)^2`

`1 - y^2 = (4e^(2x))/(e^(2x)+1)^2" [2]"`

Equation [1] is the same as equation [2]. Q.E.D.