# The loss of y = 0 as a solution

1. Feb 1, 2009

### djeitnstine

When solving a Diff. Eq. how do we know that y=0 is another solution lost when we solved it?

2. Feb 1, 2009

### NoMoreExams

You don't know how to check if y = 0 is a solution of a DE?

3. Feb 2, 2009

### Office_Shredder

Staff Emeritus
Whatever you divided by is potentially lost as a (when set to zero)solution of your differential equation. So for example, y2 = y*y' you divide by y to get y=y' and you know the solution of that, but since you divided both sides by y, you need to check the case when y=0, since you can't divide by y when y=0

4. Feb 2, 2009

### djeitnstine

Thank you for the replies. For NoMoreExams an example here: I did an ODE $$\frac{dy}{dx}-y=e^{2x}y^{3}$$

and solved it using bournulli's method to get

$$y=\sqrt{\frac{2}{-e^{2x} + Ce^{-2x}}}$$

My professor said y = 0 was lost in that solution.

So I place y = 0 in the original diff eq. to get

$$\frac{dy}{dx}-0=e^{2x}0^{3}$$

which is

$$\frac{dy}{dx}=0$$

which seems wierd

5. Feb 2, 2009

### NoMoreExams

Remember how you check solutions, LHS = RHS

Since

$$y = 0 \Rightarrow \frac{dy}{dx} = 0$$

So let's look at LHS:

$$\frac{dy}{dx} - y = 0 - 0 = 0$$

Now let's look at RHS:

$$e^{2x}y^{3} = e^{2x}0^{3} = 0$$

Since $$0 = 0$$, we have shown that $$y = 0$$ is a solution to our DE.

6. Feb 2, 2009

### djeitnstine

Wow, I can't believe I completely missed that...thanks.

7. Feb 2, 2009

### Cvan

Never ignore the trivial equilibrium soln.
They're always the hardest, since you're never really "looking hard" for them.