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Integrator September 2nd, 2018 07:08 AM

A differential equation
 
Hello,

Solve the differential equation $\displaystyle |f(x)-f'(x)|+(f(x))^2=0$.

All the best,

Integrator

SDK September 2nd, 2018 07:12 AM

You need initial data to determine signs for $f,f'$.

skipjack September 2nd, 2018 07:28 AM

If f(x) is always real, f(x) = 0.

babaliaris September 2nd, 2018 08:04 AM

$\displaystyle
|f(x) - \frac{df(x)}{dx}| + f^{2}(x) = 0
$

If:
$\displaystyle
f(x) = \frac{df(x)}{dx}
$

Then :

$\displaystyle
0 + f^{2}(x) = 0 \Leftrightarrow f(x) = 0
$

Else if:
$\displaystyle
f(x) > \frac{df(x)}{dx}
$

Then:
$\displaystyle
f(x) - \frac{df(x)}{dx} + f^{2}(x) = 0 \Leftrightarrow
- \frac{df(x)}{dx} + f(x) = -f^{2}(x) \Leftrightarrow
\frac{df(x)}{dx} - f(x) = f^{2}(x)
$

$\displaystyle
μ_{0} = e^{\int -1dx} = e^{-1 \cdot \int dx} \Leftrightarrow
μ_{0} = e^{-x + c}
$

$\displaystyle
\frac{df(x)}{dx} \cdot e^{-x} - f(x) \cdot e^{-x} = f^{2}(x) \cdot e^{-x}
\Leftrightarrow
\frac{d[f(x) \cdot e^{-x}]}{dx} = f^{2}(x) \cdot e^{-x}
\Leftrightarrow
\int \frac{d[f(x) \cdot e^{-x}]}{dx} \cdot dx = \int f^{2}(x) \cdot e^{-x} \cdot dx
\Leftrightarrow
f(x) \cdot e^{-x} + c_{1} = \int f^{2}(x) \cdot e^{-x}
$



Well of course here I'm stuck. Any Ideas?

Integrator September 2nd, 2018 08:20 AM

Quote:

Originally Posted by skipjack (Post 598322)
If f(x) is always real, f(x) = 0.

$\displaystyle f (x) = 0$ is the common solution.
However, some say that applying the module definition also leads to two other function families.

All the best,

Integrator

Integrator September 2nd, 2018 09:33 AM

Quote:

Originally Posted by babaliaris (Post 598329)
$\displaystyle
|f(x) - \frac{df(x)}{dx}| + f^{2}(x) = 0
$

If:
$\displaystyle
f(x) = \frac{df(x)}{dx}
$

Then :

$\displaystyle
0 + f^{2}(x) = 0 \Leftrightarrow f(x) = 0
$

Else if:
$\displaystyle
f(x) > \frac{df(x)}{dx}
$

Then:
$\displaystyle
f(x) - \frac{df(x)}{dx} + f^{2}(x) = 0 \Leftrightarrow
- \frac{df(x)}{dx} + f(x) = -f^{2}(x) \Leftrightarrow
\frac{df(x)}{dx} - f(x) = f^{2}(x)
$

$\displaystyle
μ_{0} = e^{\int -1dx} = e^{-1 \cdot \int dx} \Leftrightarrow
μ_{0} = e^{-x + c}
$

$\displaystyle
\frac{df(x)}{dx} \cdot e^{-x} - f(x) \cdot e^{-x} = f^{2}(x) \cdot e^{-x}
\Leftrightarrow
\frac{d[f(x) \cdot e^{-x}]}{dx} = f^{2}(x) \cdot e^{-x}
\Leftrightarrow
\int \frac{d[f(x) \cdot e^{-x}]}{dx} \cdot dx = \int f^{2}(x) \cdot e^{-x} \cdot dx
\Leftrightarrow
f(x) \cdot e^{-x} + c_{1} = \int f^{2}(x) \cdot e^{-x}
$



Well of course here I'm stuck. Any Ideas?

I think that $\displaystyle μ_{0} = e^{x + c}$.
An idea:
It is observed that $\displaystyle f (x) = g (e^{x+c})$...

All the best,

Integrator

Integrator September 2nd, 2018 09:36 AM

babaliaris,

I think that $\displaystyle μ_{0} = e^{x + c}$.
An idea:
It is observed that $\displaystyle f (x) = g (e^{x+c})=\frac{a_1e^{x+c}+a_2}{a_3e^{x+c}+a_4}$ where $\displaystyle a_i$ are constants to be found...

All the best,

Integrator

skipjack September 2nd, 2018 09:43 AM

You're "overthinking" it. If f(x) is real, both terms are non-negative, so they must both be zero if their sum is zero.

thegrade September 2nd, 2018 11:34 AM

Have you discussed Bernoulli's equation in your Differential Equations course? After separating the absolute value, you get two equations that involve a nonlinear term.

You can use a "change of variables" approach to reduce the order to a linear equation, and then solve from there. This particular equation will have a power of 2, which will yield the "logistic equation" as the solution.

If any of this sounds familiar from your coursework, please feel free to let me know. If you are still unable to solve it, then I don't mind typing it up for you.

Note: If/when you do solve this problem, it will be necessary to check your answer, as an extraneous function may or may not be found.

Hope that helps!

babaliaris September 2nd, 2018 12:07 PM

JOmg yes!!! You mean
$\displaystyle
let:
f(x) = u^{\frac{1}{1-n}}
$

Where n here is 2.

Forgot that.


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