Losing a solution of a 1st order ODE?

[misogynisticfeminist]

There is an example in a book regarding DEs which I do not understand. Solve the IVP

y'=y^2-4, y(0)=-2where t is the independent variable

We first solve by separation of variables to arrive at the 1-parameter solution.

-\frac{1}{4}ln (y+2)+\frac{1}{4}ln (y-2)=t+c

Simplifying and expressing the solution explicitly, we find that,

y=2\frac{1+ce^{4t}}{1-ce^{4t}}

Taking the initial condition,

-2=2\frac{1+c}{1-c} which simplifies to,

-1=1.

They said that the solution is wrong because:

we can express the DE as, y'=(y+2)(y-2) and that the when y=-2, and y=2 satisfies this equation (what does it mean?). How do we "preclude" y=-2 and y=-2 before solving starting to solve the DE?

 

[nd]

Your initial condition y(0)=2 seems to imply that c = 0, not -1=1?
You can preclude y=-2 simply by noticing that it fails to satisfy the initial condition.

 

[misogynisticfeminist]

sorry, my initial condition is y(0)=-2. Do we ignore y=-2 because firstly, it doesn't tell us the value of c, and it gives us the illogical -1=1?

Also, do we use the other alternative y=2 because it is the other "alternative" in y'=(y+2)(y-2)? If it is so, why?

 

[arildno]

The "separable", non-constant solutions involve a division of both sides of the equation with the expression y^{2}-4
But, since you can't divide by zero, you have implicitly assumed that y\neq\pm2

In addition to the non-constant solutions, you've got the constant solutions y_{1}(t)=2,y_{2}(t)=-2

 

[dextercioby]

y(0)=-2 is not a valid initial condition.When separating variables YOU ASUMED y\neq 2;y\neq -2 Trying to impose the initial condition to

\frac{y-2}{y+2}=Ce^{4t}

fails...

Daniel.