Menu Close

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

Tinku Tara 0 Comments
Pin




Question Number 125685 by Hassen_Timol last updated on 12/Dec/20
Solve this differential equation q_((t)) ′′ + (1/(LC)) q_((t)) = (V/L) with (L,C,V) ∈ R^3
$$ \\ $$$$\mathrm{Solve}\:\mathrm{this}\:\mathrm{differential}\:\mathrm{equation} \\ $$$$ \\ $$$$\:\:\:\:\:{q}_{\left({t}\right)} ''\:\:+\:\:\frac{\mathrm{1}}{\mathrm{LC}}\:{q}_{\left({t}\right)} \:\:=\:\:\frac{{V}}{{L}} \\ $$$$ \\ $$$$\mathrm{with}\:\left(\mathrm{L},\mathrm{C},\mathrm{V}\right)\:\in\:\mathbb{R}^{\mathrm{3}} \\ $$$$ \\ $$
Answered by mr W last updated on 13/Dec/20
q(t)=A cos ωt+B sin ωt+VC with ω=(√(1/(LC)))
$${q}\left({t}\right)={A}\:\mathrm{cos}\:\omega{t}+{B}\:\mathrm{sin}\:\omega{t}+{VC} \\ $$$${with}\:\omega=\sqrt{\frac{\mathrm{1}}{{LC}}} \\ $$
Commented by Hassen_Timol last updated on 13/Dec/20
Thank you a lot Sir
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{a}\:\mathrm{lot}\:\mathrm{Sir} \\ $$
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
yes, VC.
$${yes},\:{VC}. \\ $$
Answered by mathmax by abdo last updated on 14/Dec/20
h→r^2 +(1/(LC))=0 ⇒r=+^− i(√(1/(LC))) ⇒q_h (t)=ae^(i(√(1/(LC)))) +be^(−i(√(1/(LC)))) =αcos((t/( (√(LC)))))+β sin((t/( (√(LC))))) =αu_1 +βu_2 let w=(1/( (√(LC)))) W(u_1 ,u_2 )= determinant (((cos(wt) sin(wt))),((−wsin(wt) wcos(wt))))=w≠0 W_1 = determinant (((o sin(wt))),(((V/L) wcos(wt))))=−(V/L)sin(wt) W_2 = determinant (((cos(wt) 0)),((−wsin(wt) (V/L))))=(V/L)cos(wt) v_1 =∫ (w_1 /w)dt =−(V/L)∫ ((sin(wt))/w) dt =−(V/L)×(1/w^2 )cos(wt) =(V/L)×LC cos(wt) =VC cos(wt) v_2 =∫ (w_2 /w)dt =(V/L)∫((cos(wt))/w)dt =(V/(L ))×(1/w^2 )sin(wt)=VC sin(wt) ⇒ 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 ⇒ q(t) =q_h (t)+q_p (t)=α cos(wt)+βsin(wt)+VC and w =(1/( (√(LC))))
$$\mathrm{h}\rightarrow\mathrm{r}^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{LC}}=\mathrm{0}\:\Rightarrow\mathrm{r}=\overset{−} {+}\mathrm{i}\sqrt{\frac{\mathrm{1}}{\mathrm{LC}}}\:\Rightarrow\mathrm{q}_{\mathrm{h}} \left(\mathrm{t}\right)=\mathrm{ae}^{\mathrm{i}\sqrt{\frac{\mathrm{1}}{\mathrm{LC}}}} \:+\mathrm{be}^{−\mathrm{i}\sqrt{\frac{\mathrm{1}}{\mathrm{LC}}}} \\ $$$$=\alpha\mathrm{cos}\left(\frac{\mathrm{t}}{\:\sqrt{\mathrm{LC}}}\right)+\beta\:\mathrm{sin}\left(\frac{\mathrm{t}}{\:\sqrt{\mathrm{LC}}}\right)\:=\alpha\mathrm{u}_{\mathrm{1}} \:+\beta\mathrm{u}_{\mathrm{2}} \:\:\mathrm{let}\:\mathrm{w}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{LC}}} \\ $$$$\mathrm{W}\left(\mathrm{u}_{\mathrm{1}} ,\mathrm{u}_{\mathrm{2}} \right)=\begin{vmatrix}{\mathrm{cos}\left(\mathrm{wt}\right)\:\:\:\:\:\:\:\:\:\:\mathrm{sin}\left(\mathrm{wt}\right)}\\{−\mathrm{wsin}\left(\mathrm{wt}\right)\:\:\:\:\:\:\:\:\mathrm{wcos}\left(\mathrm{wt}\right)}\end{vmatrix}=\mathrm{w}\neq\mathrm{0} \\ $$$$\mathrm{W}_{\mathrm{1}} =\begin{vmatrix}{\mathrm{o}\:\:\:\:\:\:\:\:\:\mathrm{sin}\left(\mathrm{wt}\right)}\\{\frac{\mathrm{V}}{\mathrm{L}}\:\:\:\:\:\:\:\:\mathrm{wcos}\left(\mathrm{wt}\right)}\end{vmatrix}=−\frac{\mathrm{V}}{\mathrm{L}}\mathrm{sin}\left(\mathrm{wt}\right) \\ $$$$\mathrm{W}_{\mathrm{2}} =\begin{vmatrix}{\mathrm{cos}\left(\mathrm{wt}\right)\:\:\:\:\:\:\:\:\mathrm{0}}\\{−\mathrm{wsin}\left(\mathrm{wt}\right)\:\:\frac{\mathrm{V}}{\mathrm{L}}}\end{vmatrix}=\frac{\mathrm{V}}{\mathrm{L}}\mathrm{cos}\left(\mathrm{wt}\right) \\ $$$$\mathrm{v}_{\mathrm{1}} =\int\:\frac{\mathrm{w}_{\mathrm{1}} }{\mathrm{w}}\mathrm{dt}\:=−\frac{\mathrm{V}}{\mathrm{L}}\int\:\:\frac{\mathrm{sin}\left(\mathrm{wt}\right)}{\mathrm{w}}\:\mathrm{dt}\:=−\frac{\mathrm{V}}{\mathrm{L}}×\frac{\mathrm{1}}{\mathrm{w}^{\mathrm{2}} }\mathrm{cos}\left(\mathrm{wt}\right) \\ $$$$=\frac{\mathrm{V}}{\mathrm{L}}×\mathrm{LC}\:\mathrm{cos}\left(\mathrm{wt}\right)\:=\mathrm{VC}\:\mathrm{cos}\left(\mathrm{wt}\right) \\ $$$$\mathrm{v}_{\mathrm{2}} =\int\:\:\frac{\mathrm{w}_{\mathrm{2}} }{\mathrm{w}}\mathrm{dt}\:=\frac{\mathrm{V}}{\mathrm{L}}\int\frac{\mathrm{cos}\left(\mathrm{wt}\right)}{\mathrm{w}}\mathrm{dt}\:=\frac{\mathrm{V}}{\mathrm{L}\:}×\frac{\mathrm{1}}{\mathrm{w}^{\mathrm{2}} }\mathrm{sin}\left(\mathrm{wt}\right)=\mathrm{VC}\:\mathrm{sin}\left(\mathrm{wt}\right)\:\Rightarrow \\ $$$$\mathrm{q}_{\mathrm{p}} =\mathrm{u}_{\mathrm{1}} \mathrm{v}_{\mathrm{1}} \:+\mathrm{u}_{\mathrm{2}} \mathrm{v}_{\mathrm{2}} =\mathrm{cos}\left(\mathrm{wt}\right)\mathrm{VC}\:\mathrm{cos}\left(\mathrm{wt}\right)+\mathrm{sin}\left(\mathrm{wt}\right)\mathrm{VC}\:\mathrm{sin}\left(\mathrm{wt}\right) \\ $$$$=\mathrm{VC}\left(\mathrm{cos}^{\mathrm{2}} \left(\mathrm{wt}\right)+\mathrm{sin}^{\mathrm{2}} \left(\mathrm{wt}\right)\right)=\mathrm{VC}\:\Rightarrow \\ $$$$\mathrm{q}\left(\mathrm{t}\right)\:=\mathrm{q}_{\mathrm{h}} \left(\mathrm{t}\right)+\mathrm{q}_{\mathrm{p}} \left(\mathrm{t}\right)=\alpha\:\mathrm{cos}\left(\mathrm{wt}\right)+\beta\mathrm{sin}\left(\mathrm{wt}\right)+\mathrm{VC} \\ $$$$\mathrm{and}\:\mathrm{w}\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{LC}}} \\ $$

Pin

Leave a Reply

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