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Question Number 95600 by Tony Lin last updated on 26/May/20

s o l v e t h e d i f f e r e n t i a l e q u a t i o n

L \frac{d^{2} q}{{d t}^{2}} + R \frac{d q}{d t} + \frac{1}{C} q = ε c o s ω t

w h i c h i s i n R . L . C c i r c u i t w i t h f o r c e d o s c i l l a t i o n

w h e r e L i s i n d u c t a n c e

R i s r e s i s t a n c e

C i s c a p a c i t a n a c e

q i s c h a r g e

ε i s m o t i o n e m f

t i s t i m e

Answered by mathmax by abdo last updated on 26/May/20

we take q = x and L = δ so (e) \Rightarrow δ x^{^{″}} + r x^{^{'}} + cx = ξ \cos (wt) (c = \frac{1}{C})

let solve by laplace transform \Rightarrow

δ L (x^{^{″}}) + rL (x^{^{'}}) + cL (x) = ξ L (\cos (wt)) \Rightarrow

δ (t^{2} L (x) - tx (0) - x^{^{'}} (0)) + r (tL (x) - x (0)) + cL (x) = ξ L (\cos (wt)) \Rightarrow

(δ t^{2} + rt + c) L (x) - δ t x (0) - δ x^{^{'}} (0) - rx (0) = ξ L (\cos (wt)

L (\cos (wt) = \int_{0}^{\infty} \cos (wu) e^{- tu} du

= Re (\int_{0}^{\infty} e^{iwu - tu} du) but \int_{0}^{\infty} e^{(iw - t) u} du = {[\frac{1}{iw - t} e^{(- t + iw) u}]}_{0}^{\infty}

= - \frac{1}{iw - t} = \frac{1}{t - iw} = \frac{t + iw}{t^{2} + w^{2}} \Rightarrow L (\cos (wt)) = \frac{t}{t^{2} + w^{2}} \Rightarrow

(δ t^{2} + rt + c) L (x) = \frac{ξ t}{t^{2} + w^{2}} + (δ t + r) x (0) + δ x^{^{'}} (0) \Rightarrow

L (x) = \frac{ξ t}{(t^{2} + w^{2}) (δ t^{2} + rt + c)} + \frac{δ t + r}{δ t^{2} + rt + c} x (0) + \frac{δ x^{^{'}} (0)}{δ t^{2} + rt + c} \Rightarrow

x (t) = ξ L^{- 1} (\frac{1}{(t^{2} + w^{2}) (δ t^{2} + rt + c)}) + L^{- 1} (\frac{δ t + r}{δ t^{2} + rt + c} x (0))

+ L^{- 1} (\frac{δ x^{^{'}} (0)}{δ t^{2} + rt + c})

δ t^{2} + rt + c = 0 \to Δ = r^{2} - 4 δ c

if Δ > 0 \Rightarrow t_{1} = \frac{- r + \sqrt{r^{2} - 4 δ c}}{2 δ} and t_{2} = \frac{- r - \sqrt{r^{2} - 4 δ c}}{2 δ} \Rightarrow

\frac{δ x^{^{'}} (0)}{δ t^{2} + rt + c} = \frac{x^{^{'}} (0)}{(t - t_{1}) (t - t_{2})} = δ \frac{x^{^{'}} (0)}{\sqrt{r^{2} - 4 δ c}} (\frac{1}{t - t_{1}} - \frac{1}{t - t_{2}}) \Rightarrow

L^{- 1} (\dots .) = \frac{δ x^{^{'}} (0)}{\sqrt{r^{2} - 4 δ c}} {L^{- 1} (\frac{1}{t - t_{1}}) - L^{- 1} (\frac{1}{t - t_{2}})}

= \frac{δ x^{^{'}} (0)}{\sqrt{r^{2} - 4 δ c}} {e^{t_{1} t} - e^{t_{2} t}} \dots be continued \dots .

Commented by mathmax by abdo last updated on 26/May/20

wroskien method (e) \Rightarrow x^{^{″}} + \frac{r}{δ} x^{^{'}} + \frac{c}{δ} x = \frac{ξ}{δ} \cos (wt)

(he) \to r^{2} + \frac{r}{δ} r + \frac{c}{δ} = 0 \to Δ = \frac{r^{2}}{δ^{2}} - \frac{4 c}{δ}

if Δ > 0 \Rightarrow r_{1} = \frac{- \frac{r}{δ} + \sqrt{Δ}}{2} and r_{2} = \frac{- \frac{r}{δ} - \sqrt{Δ}}{2}

y_{h} = α e^{r_{1} t} + β e^{r_{2} t} = α u_{1} + β u_{2}

W (u_{1}, u_{2}) = | \begin{matrix} u_{1} u_{2} \\ u_{1}^{^{'}} u_{2}^{^{'}} \end{matrix} | = | \begin{matrix} e^{r_{1} t} e^{r_{2} t} \\ r_{1} e^{r_{1} t} r_{2} e^{r_{2} t} \end{matrix} |

= (r_{2} - r_{1}) e^{(r_{1} + r_{2}) t} = \sqrt{Δ} e^{- \frac{r}{δ} t}

W_{1} = | \begin{matrix} 0 e^{r_{2} t} \\ \frac{ξ}{δ} \cos (wt) r_{2} e^{r_{2} t} \end{matrix} | = - \frac{ξ}{δ} \cos (wt) e^{r_{2} t}

W_{2} = | \begin{matrix} e^{r_{1} t} 0 \\ r_{1} e^{r_{1} t} \frac{ξ}{δ} \cos (wt) \end{matrix} | = \frac{ξ}{δ} \cos (wt) e^{r_{1} t}

v_{1} = \int \frac{w_{1}}{W} dt = - \frac{ξ}{δ} \int \frac{\cos (wt) e^{r_{2} t}}{\sqrt{Δ} e^{- \frac{r}{δ} t}} = - \frac{ξ}{δ \sqrt{Δ}} \int \cos (wt) e^{(r_{2} + \frac{r}{δ}) t} dt

= - \frac{ξ}{δ \sqrt{Δ}} Re (\int e^{iwt + (r_{2} + \frac{r}{δ}) t} dt) = \dots .

v_{2} = \int \frac{w_{2}}{W} dt = \frac{ξ}{δ \sqrt{Δ}} \int \frac{\cos (wt) e^{r_{1} t}}{e^{- \frac{r}{δ}}} dt = \frac{ξ}{δ \sqrt{Δ}} \int \cos (wt) e^{(r_{1} + \frac{r}{δ}) t} dt

\dots \Rightarrow y_{p} = u_{1} v_{1} + u_{2} v_{2} and y = y_{h} + y_{p}

Commented by Tony Lin last updated on 26/May/20

t h a n k s s i r

Commented by mathmax by abdo last updated on 27/May/20

you are welcome sir .