Hi all,(adsbygoogle = window.adsbygoogle || []).push({});

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):

[tex]

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

[/tex]

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

[tex]

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

[/tex]

[tex]

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

[/tex]

Because, you know, both satisfy

[tex]

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

[/tex]

Anyway, sometimes it gave me correct results (both [itex]C(e^{x} - 1)[/itex] and [itex]-C(e^{x} - 1)[/itex] were solutions) while other times only plus-signed solution was ok.

2.Solving Bernoulli's ODE, let's say

[tex]

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

[/tex]

and substituting

[tex]

z = \sqrt{y}

[/tex]

[tex]

y = z^2

[/tex]

[tex]

y' = 2zz'

[/tex]

We get

[tex]

2zz' = 2z^2 + 2xz

[/tex]

and first thing I do is dividing with [itex]2z[/itex] and so getting the condition [itex]z \neq 0[/itex]

After some computing, I get the result

[tex]

z = Ce^{x} - x - 1

[/tex]

and thus

[tex]

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

[/tex]

a) first question

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

b) second question

Having the substitution above in mind, I know that [itex]z[/itex] itself must be [itex]\geq 0[/itex]. Anyway, if we take [itex]C \in (0, 1)[/itex], then the equation

[tex]

Ce^{x} -x - 1 = 0

[/tex]

has two roots, [itex]x_0[/itex] and [itex]x_1[/itex]. Then we know that

for [itex]x \in (x_0, x_1)[/itex]

[tex]

Ce^{x} -x - 1 < 0

[/tex]

Ok, still remembering that this is [itex]z[/itex] and it must be [itex]\geq 0[/itex], we shouldn't accept this interval for [itex]x[/itex]. 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 [itex](x_0,x_1)[/itex] is excluded from the solution (according to our professor).

Thank you for any suggestions.

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# Two questions on ODEs

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