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 April 30th, 2017, 02:52 PM #1 Senior Member   Joined: Dec 2015 From: Earth Posts: 239 Thanks: 27 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.
 April 30th, 2017, 03:05 PM #2 Senior Member   Joined: Dec 2015 From: Earth Posts: 239 Thanks: 27 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 $\displaystyle \mu=\int \phi dx$ Last edited by idontknow; April 30th, 2017 at 03:09 PM.
 April 30th, 2017, 03:12 PM #3 Senior Member   Joined: Dec 2015 From: Earth Posts: 239 Thanks: 27
 April 30th, 2017, 03:32 PM #4 Senior Member   Joined: Feb 2016 From: Australia Posts: 1,638 Thanks: 570 Math Focus: Yet to find out. ??
April 30th, 2017, 05:58 PM   #5
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April 30th, 2017, 07:07 PM   #6
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