Solve-this-differential-equation-q-t-1-LC-q-t-V-L-with-L-C-V-R-3-

Question Number 125685 by Hassen_Timol last updated on 12/Dec/20

Solve this differential equation

q_{(t)}^{″} + \frac{1}{LC} q_{(t)} = \frac{V}{L}

with (L, C, V) \in R^{3}

Answered by mr W last updated on 13/Dec/20

q (t) = A \cos ω t + B \sin ω t + V C

w i t h ω = \sqrt{\frac{1}{L C}}

Commented by Hassen_Timol last updated on 13/Dec/20

Thank you a lot Sir

Commented by Hassen_Timol last updated on 13/Dec/20

Isn't it ...+VC ? not VC² ? because V*L*C/L=VC

Commented by mr W last updated on 13/Dec/20

y e s, V C .

Answered by mathmax by abdo last updated on 14/Dec/20

h \to r^{2} + \frac{1}{LC} = 0 \Rightarrow r = \bar{+} i \sqrt{\frac{1}{LC}} \Rightarrow q_{h} (t) = {ae}^{i \sqrt{\frac{1}{LC}}} + {be}^{- i \sqrt{\frac{1}{LC}}}

= α \cos (\frac{t}{\sqrt{LC}}) + β \sin (\frac{t}{\sqrt{LC}}) = α u_{1} + β u_{2} let w = \frac{1}{\sqrt{LC}}

W (u_{1}, u_{2}) = | \begin{matrix} \cos (wt) \sin (wt) \\ - wsin (wt) wcos (wt) \end{matrix} | = w \neq 0

W_{1} = | \begin{matrix} o \sin (wt) \\ \frac{V}{L} wcos (wt) \end{matrix} | = - \frac{V}{L} \sin (wt)

W_{2} = | \begin{matrix} \cos (wt) 0 \\ - wsin (wt) \frac{V}{L} \end{matrix} | = \frac{V}{L} \cos (wt)

v_{1} = \int \frac{w_{1}}{w} dt = - \frac{V}{L} \int \frac{\sin (wt)}{w} dt = - \frac{V}{L} \times \frac{1}{w^{2}} \cos (wt)

= \frac{V}{L} \times LC \cos (wt) = VC \cos (wt)

v_{2} = \int \frac{w_{2}}{w} dt = \frac{V}{L} \int \frac{\cos (wt)}{w} dt = \frac{V}{L} \times \frac{1}{w^{2}} \sin (wt) = VC \sin (wt) \Rightarrow

q_{p} = u_{1} v_{1} + u_{2} v_{2} = \cos (wt) VC \cos (wt) + \sin (wt) VC \sin (wt)

= VC (\cos^{2} (wt) + \sin^{2} (wt)) = VC \Rightarrow

q (t) = q_{h} (t) + q_{p} (t) = α \cos (wt) + β \sin (wt) + VC

and w = \frac{1}{\sqrt{LC}}