Menu Close

Pls-i-need-a-help-from-Ordinary-differantial-Equation-t-2-y-2-t-t-y-1-t-t-2-2-y-t-0-we-Already-Know-Solution-y-t-C-1-J-t-C-2-J-t-But-J-t-can-t-Satisfy-as-Solutio

Tinku Tara 0 Comments
Pin




Question Number 212906 by issac last updated on 26/Oct/24
Pls i need a help.. from Ordinary differantial Equation t^2 y^((2)) (t)+t∙y^((1)) (t)+(t^2 −ν^2 )y(t)=0 we Already Know Solution y(t)=C_1 J_ν (t)+C_2 J_(−ν) (t) But J_(−ν) (t) can′t Satisfy as Solution Cus J_ν (t) and J_(−ν) (t) are Not Linear independent. Wronskian W∈mat(m,m) det W=0 thus Solution y(t)=C_1 J_ν (t)+C_2 Y_ν (t) i already undertand above indentity i wrote my question is prove Abel′s identity W(J_ν (t),Y_ν (t))=(2/(πt)) Pls Help :(
$$\mathrm{Pls}\:\mathrm{i}\:\mathrm{need}\:\mathrm{a}\:\mathrm{help}.. \\ $$$$\mathrm{from}\:\mathrm{Ordinary}\:\mathrm{differantial}\:\mathrm{Equation} \\ $$$${t}^{\mathrm{2}} {y}^{\left(\mathrm{2}\right)} \left({t}\right)+{t}\centerdot{y}^{\left(\mathrm{1}\right)} \left({t}\right)+\left({t}^{\mathrm{2}} −\nu^{\mathrm{2}} \right){y}\left({t}\right)=\mathrm{0} \\ $$$$\mathrm{we}\:\mathrm{Already}\:\mathrm{Know} \\ $$$$\mathrm{Solution}\:{y}\left({t}\right)={C}_{\mathrm{1}} {J}_{\nu} \left({t}\right)+{C}_{\mathrm{2}} {J}_{−\nu} \left({t}\right) \\ $$$$\mathrm{But}\:{J}_{−\nu} \left({t}\right)\:\mathrm{can}'\mathrm{t}\:\mathrm{Satisfy}\:\mathrm{as}\:\mathrm{Solution} \\ $$$$\mathrm{Cus}\:{J}_{\nu} \left({t}\right)\:\mathrm{and}\:{J}_{−\nu} \left({t}\right)\:\mathrm{are}\:\mathrm{Not}\:\mathrm{Linear}\:\mathrm{independent}. \\ $$$$\mathrm{Wronskian}\:\mathcal{W}\in\mathrm{mat}\left({m},{m}\right) \\ $$$$\mathrm{det}\:\mathcal{W}=\mathrm{0} \\ $$$$\mathrm{thus}\:\mathrm{Solution}\:{y}\left({t}\right)={C}_{\mathrm{1}} {J}_{\nu} \left({t}\right)+{C}_{\mathrm{2}} {Y}_{\nu} \left({t}\right) \\ $$$$\mathrm{i}\:\mathrm{already}\:\mathrm{undertand}\:\mathrm{above}\:\mathrm{indentity} \\ $$$$\mathrm{i}\:\mathrm{wrote} \\ $$$$\mathrm{my}\:\mathrm{question}\:\mathrm{is}\:\mathrm{prove}\:\mathrm{Abel}'\mathrm{s}\:\mathrm{identity} \\ $$$$\:\mathcal{W}\left({J}_{\nu} \left({t}\right),{Y}_{\nu} \left({t}\right)\right)=\frac{\mathrm{2}}{\pi{t}} \\ $$$$\boldsymbol{\mathrm{Pls}}\:\boldsymbol{\mathrm{Help}}\:\::\left(\right. \\ $$
Commented by issac last updated on 26/Oct/24
Wronskian mat(n,n) W= ((f_1 ,f_2 ,…,f_n ),(f_1 ^((1)) ,f_2 ^((1)) ,…,f_n ^((1)) ),(⋮, ,⋮, ),(f_1 ^((m−1)) ,f_2 ^((m−1)) ,…,f_n ^((m−1)) ) ) det W=0 →linear dependence. det W≠0 →linear independence.
$$\: \\ $$$$\mathrm{Wronskian}\:\mathrm{mat}\left({n},{n}\right)\: \\ $$$$\:\mathcal{W}=\begin{pmatrix}{{f}_{\mathrm{1}} }&{{f}_{\mathrm{2}} }&{\ldots}&{{f}_{{n}} }\\{{f}_{\mathrm{1}} ^{\left(\mathrm{1}\right)} }&{{f}_{\mathrm{2}} ^{\left(\mathrm{1}\right)} }&{\ldots}&{{f}_{{n}} ^{\left(\mathrm{1}\right)} }\\{\vdots}&{\:}&{\vdots}&{\:}\\{{f}_{\mathrm{1}} ^{\left({m}−\mathrm{1}\right)} }&{{f}_{\mathrm{2}} ^{\left({m}−\mathrm{1}\right)} }&{\ldots}&{{f}_{{n}} ^{\left({m}−\mathrm{1}\right)} }\end{pmatrix} \\ $$$$\mathrm{det}\:\mathcal{W}=\mathrm{0}\:\:\rightarrow\mathrm{linear}\:\mathrm{dependence}. \\ $$$$\mathrm{det}\:\mathcal{W}\neq\mathrm{0}\:\:\rightarrow\mathrm{linear}\:\mathrm{independence}. \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *