# Losing a solution of a 1st order ODE?

1. Mar 10, 2005

### misogynisticfeminist

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

$$y'=y^2-4, y(0)=-2$$where 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?

Last edited: Mar 10, 2005
2. Mar 10, 2005

### 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.

3. Mar 10, 2005

### 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?

Last edited: Mar 10, 2005
4. Mar 10, 2005

### 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 $$y_{1}(t)=2,y_{2}(t)=-2$$

Last edited: Mar 10, 2005
5. Mar 10, 2005

### 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.