The loss of y = 0 as a solution

[djeitnstine]

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

 

[NoMoreExams]

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

 

[Office_Shredder]

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

 

[djeitnstine]

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

and solved it using bournulli's method to get

[tex]y=\sqrt{\frac{2}{-e^{2x} + Ce^{-2x}}}[/tex]

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

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

[tex]\frac{dy}{dx}-0=e^{2x}0^{3}[/tex]

which is

[tex]\frac{dy}{dx}=0[/tex]

which seems wierd

 

[NoMoreExams]

Remember how you check solutions, LHS = RHS

Since

[tex] y = 0 \Rightarrow \frac{dy}{dx} = 0 [/tex]

So let's look at LHS:

[tex] \frac{dy}{dx} - y = 0 - 0 = 0 [/tex]

Now let's look at RHS:

[tex] e^{2x}y^{3} = e^{2x}0^{3} = 0 [/tex]

Since [tex] 0 = 0 [/tex], we have shown that [tex] y = 0 [/tex] is a solution to our DE.

 

[djeitnstine]

[tex] y = 0 \Rightarrow \frac{dy}{dx} = 0 [/tex]

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

 

[Cvan]

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