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A differential equationHello, Solve the differential equation $\displaystyle |f(x)-f'(x)|+(f(x))^2=0$. All the best, Integrator |

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

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

$\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? |

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However, some say that applying the module definition also leads to two other function families. All the best, Integrator |

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An idea: It is observed that $\displaystyle f (x) = g (e^{x+c})$... All the best, Integrator |

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 |

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. |

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! |

JOmg yes!!! You mean $\displaystyle let: f(x) = u^{\frac{1}{1-n}} $ Where n here is 2. Forgot that. |

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