# Homework Help: Two questions on ODEs

1. Dec 27, 2005

### twoflower

Hi all,
I'd be happy if someone could clarify these two things to me:

1. While solving linear first-order ODE, I first solve homogenous equation (with the right side equal to 0) and eventually I get to the point (just an example):

$$\log |y| = \log C(e^{x} - 1)$$

Now, is it ok to compute $C$ for given non-homogenous equation and then write there two solutions

$$y = C(e^{x} - 1)$$

$$y = -C(e^{x} - 1)\mbox{ ?}$$

Because, you know, both satisfy

$$\log |y| = \log C(e^{x} - 1)$$

Anyway, sometimes it gave me correct results (both $C(e^{x} - 1)$ and $-C(e^{x} - 1)$ were solutions) while other times only plus-signed solution was ok.
2. Solving Bernoulli's ODE, let's say

$$y' = 2y + 2x\sqrt{y}$$

and substituting

$$z = \sqrt{y}$$

$$y = z^2$$

$$y' = 2zz'$$

We get

$$2zz' = 2z^2 + 2xz$$

and first thing I do is dividing with $2z$ and so getting the condition $z \neq 0$
After some computing, I get the result

$$z = Ce^{x} - x - 1$$

and thus

$$y = (Ce^{x} - x - 1)^2$$

a) first question
I'd suppose this expression must not get zeroed because at the beginning we divided the equation with $2z$. Anyway, even if $C=1$ and $x=0$ (and thus $z = 0 = y$), the original equation holds true. How it comes?

b) second question
Having the substitution above in mind, I know that $z$ itself must be $\geq 0$. Anyway, if we take $C \in (0, 1)$, then the equation

$$Ce^{x} -x - 1 = 0$$

has two roots, $x_0$ and $x_1$. Then we know that
for $x \in (x_0, x_1)$

$$Ce^{x} -x - 1 < 0$$

Ok, still remembering that this is $z$ and it must be $\geq 0$, we shouldn't accept this interval for $x$. However, what's the problem with this solution on this interval? When you put it in the original equation, it's ok I think.

I can't understand why the interval $(x_0,x_1)$ is excluded from the solution (according to our professor).

Thank you for any suggestions.

Last edited: Dec 27, 2005
2. Dec 27, 2005

### saltydog

Well, I would've lost money on that one Twoflower. Apparently when you assign:

$$z=y^{1/2}$$

You implicitly define z$\geq 0$.

Note that the solution:

$$y(t)=(ce^x-(1+x))^2$$

does NOT satisfiy the ODE in the interval where z is less than zero. For example, take the plot below for c=0.5 and:

$$y(t)=(ce^x-(1+x))^2$$

For (approx):

$$x\in (-0.77,1.68)$$

the curve does not satisfy the ODE. New for me . Thanks!

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3. Dec 28, 2005

### saltydog

I wish to clear up something with this problem from the perspective of existence and uniqueness:

Consider:

$$y^{'}=f(x,y)=2y+2x\sqrt{y},\quad y(0)=0.25$$

Now,since f(x,y) and the partial of f with respect to y exists in a bounded region about the point (0,0.25), we can expect a unique solution passing through this point.

The solution above was determined to be:

$$y(x)=(ce^x-1-x)^2\;\text{with}\;ce^x-1-x\geq 0$$

Solving for c, I obtain:

$$c=0.5\;\text{and}\;c=1.5$$

Now, initially this would suggest two solutions satisfy the equation. However, the value of c=0.5 would not satisify the restiction $ce^x-1-x\geq 0[/tex] and so must be discarded. I'm left then with the "unique" solution: $$y(x)=(1.5e^x-1-x)^2$$ 4. Dec 29, 2005 ### twoflower Thank you Saltydog, now I'm clear about that, it helped me a lot. What does this follow from? Is there some theorem saying this? Could you clarify those two other questions to me too? Thank you very much. Last edited: Dec 29, 2005 5. Dec 29, 2005 ### saltydog This is based on Existence and Uniqueness Theorems for ordinary differential equations: Consider: $$y^{'}=f(x,y)$$ Let T denote the rectangular region defined by: $$|x-x_0|\leq a\quad |y-y_0|\leq b$$ with the point [itex](x_0,y_0)$ at its center. Let f and $\frac{\partial f}{\partial y}$ be continuous at each point in T, then there exists an interval$|x-x_0|\leq h$ and a function y(x) such that:

1. y(x) is a solution to the ODE on the interval $|x-x_0|\leq h$.

2. y(x) is unique in the sense that if another equation h(x) satisfies the ODE, then y(x)=h(x).

Note that the interval $|x-x_0|\leq h$ can be very small and nothing is said about uniqueness and existence if f and the partial do not exist at the point x0.

Thus for a particular IVP, I know if the above is satisfied, I can be guaranteed a unique solution. However, I'm not sure about the case of y=0. Note that for the IVP:

$$y^{'}=2y+2x\sqrt{y};\quad y(0)=0$$

The partial is indeterminant at the initial condition so uniqueness and existence cannot be guaranteed.

Not sure I can do better than the above. Perhaps someone in here more knowledgable than I could elaborate on the matter.

Last edited: Dec 29, 2005