Question

Find the general solution of `(dy/dx) = y^2-4` ?

Differential equations

Answer 1

The general solution is `y=(-2(e^(4(x+C))+1))/(e^(4(x+C))-1)`

Explanation:

This is a first order separable differential equation

`dy/dx=y^2-4`

`dy/(y^2-4)=dx`

`intdy/(y^2-4)=intdx`

Perform partial fraction

`1/(y^2-4)=A/(y+2)+B/(y-2)`

`=(A(y-2)+B(y+2))/((y^2-4))`

Compare the numerators

`1=A(y-2)+B(y+2)`

Let `y=-2`, `=>`, `1=-4A`, `A=-1/4`

Let `y=2`, `=>`, `1=4B`, `B=1/4`

Therefore,

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

So,

`intdy/(y^2-4)=int(-1/4dy)/(y+2)+int(1/4dy)/(y-2)=intdx`

`1/4ln(|y-2|)-1/4ln(|y+2|)=x+C`

`1/4ln((y-2)/(y+2))=x+C`

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

Solving for `y`

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

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

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

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

Dividing equation `(1)` by equation `(2)`

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

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

Answer 2

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

Explanation:

This is separable:

`(dy)/(y^2 - 4) = dx`

And

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

Which leaves you with:

`1/4 int \ 1/( y - 2) - 1/( y + 2) \ dy = int \ dx`

`lnabs( y - 2) - ln abs( y + 2) = 4x+C`

`abs( y - 2)/abs( y + 2) = Ce^(4x)`

For `y gt 2` or `y lt - 2`, that's:

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

Otherwise it's:

`( y - 2)/( y + 2) = -Ce^(4x)`, ie just some constant with a different sign.

So proceeding with `C`:

`( y - 2)/( y + 2) = Ce^(4x) implies y = 2\ (1+ Ce^(4x) )/(1- Ce^(4x))`