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idontknow April 30th, 2017 02:52 PM

New MATHbrained sound

Formal Function $\displaystyle \;$$\displaystyle \frac{d^2 y}{d x ^2}+j \frac{dy}{dx}+aY\models JY''-Y=0\leftrightarrow d\phi +\phi=0$
$\displaystyle d\phi+\phi=0 \leftrightarrow \phi=Ye^\omega=e^{rx}F(x)$
$\displaystyle jy'-iy \models e^{rx}F(x)\Rightarrow y'-y\models \phi '$
$\displaystyle \gamma ' +\gamma=\phi '$ $\displaystyle \; \:$ $\displaystyle (\gamma - \phi)'=y\precapprox e^{rx}\omega$
$\displaystyle \gamma=\phi(\eta) +\int e^{Rx}dx=e^{Tx}+\phi(\eta,\theta)$
$\displaystyle \gamma \precapprox Y=e^{TX} + \bigtriangleup \phi(\eta,theta)=\Delta (e^{TX} + \phi(\eta,theta))$
Examples of Formal Function , Let $\displaystyle \psi(S) $ be formal function of S
1) $\displaystyle \psi(e^2\beta)=e^\beta$ $\displaystyle \;$ , $\displaystyle \psi(a\beta+b)=\xi(a,b)\beta \; \;$, $\displaystyle \psi(\frac{\beta}{c})=C_1 \beta \; \; $ , $\displaystyle \psi(2\beta)=\beta$
So differential equation $\displaystyle \psi(jy''+by'+iy)=\gamma''-\eta \gamma$ has constant exponential solution of form $\displaystyle e^TX$
Other questions for Universal Definable Structure differential equation $\displaystyle \; \;$ $\displaystyle \delta^{\eta_1}=\delta{\eta_ 2}\leftrightarrow max\{\Delta \eta\}=2$ $\displaystyle \;$ Equation $\displaystyle F(j_i y^{(i)},.....,j_0y) \;$ to have solution it must equal to $\displaystyle d\mu=0$$\displaystyle \;$ such that $\displaystyle \mu=\int fdx$
All in one $\displaystyle \sum {_i} j_0 d^i \gamma=d\mu=d\mu_1 +d\mu_2=d\mu_1 +d\mu_2 +d\mu_3 =......=d\mu_1+d\mu_2+...+d\mu_n=\sum _{i} d\mu_i$
To define universal structure of DE-solution , we must insert two equations with two different solutions $\displaystyle \gamma_1 , \gamma_2$
$\displaystyle \begin{cases} \phi_{\delta_1}=s \\ \phi_{\delta_2}=s \end{cases}$ Wronskian tells formal function $\displaystyle \;$ $\displaystyle \begin{cases} j\gamma '' +i \gamma ' +z\gamma=0 \\ j\gamma_0 '' +i\gamma_0 ' + z\gamma_0=0 \end{cases} \Rightarrow$$\displaystyle W=e^{-\lambda x}$ ; $\displaystyle y=\sqrt{e^{\xi x}\varphi}$
$\displaystyle y=\varphi ' e^{px}=\mu(x) e^{px}$ $\displaystyle \psi_y=y''+zy'+gy=0$ $\displaystyle \psi(\mu e^{px})=\psi(t,\mu)=0$
So this is a proof that we can take : $\displaystyle y=\mu e^{px}$
$\displaystyle \psi(y)=0;;; \exists \mu(x)=\chi$ $\displaystyle y=\chi e^{px}=e^{rx}$
So solution can be found in $\displaystyle \exists y=e^{rx}$
But in general $\displaystyle y=f(x)e^{px}$
I'm not having time to explain which are constants and which are functions and to write it step-by-step, but still it can be understood.

idontknow April 30th, 2017 03:05 PM

For 2-SDE $\displaystyle jy''+py'+ky=0 \; $ to have solution ,there must exist $\displaystyle \{ \exists \mu | (jy''+py'=d\mu) , (jy''+ky=d\mu),(py'+ky=d\mu)\} such that $ $\displaystyle \; d\mu_0 +d\mu=0 \;$ or $\displaystyle \mu_0+\mu=C$
So one of these $\displaystyle \phi=\begin{cases} (jy'',ky) \\ (ky,py') \\ (jy'',py')\end{cases} \; $ must equal to a differential function $\displaystyle d\mu \; $ such that $\displaystyle \mu=\int \phi dx$

idontknow April 30th, 2017 03:12 PM

Joppy April 30th, 2017 03:32 PM


Denis April 30th, 2017 05:58 PM


Originally Posted by Joppy (Post 568758)


Joppy April 30th, 2017 07:07 PM


Originally Posted by Denis (Post 568766)

??? + 1

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