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Question Number 112740 by Mr.D.N. last updated on 09/Sep/20

Solve differential equation the method of variation of parameters: (d^2 y/dx^2 ) +4y =4 cosec 2x

$$\:\mathrm{Solve}\:\mathrm{differential}\:\mathrm{equation}\:\mathrm{the}\:\mathrm{method}\:\mathrm{of} \\ $$$$\:\mathrm{variation}\:\mathrm{of}\:\mathrm{parameters}: \\ $$$$\:\:\frac{\boldsymbol{\mathrm{d}}^{\mathrm{2}} \boldsymbol{\mathrm{y}}}{\boldsymbol{\mathrm{dx}}^{\mathrm{2}} }\:+\mathrm{4}\boldsymbol{\mathrm{y}}\:=\mathrm{4}\:\boldsymbol{\mathrm{cosec}}\:\mathrm{2}\boldsymbol{\mathrm{x}}\: \\ $$

Answered by mathmax by abdo last updated on 09/Sep/20

y^(′′) +4y =(4/(sin(2x))) h→r^2 +4=0 ⇒r =+^− 2i ⇒y_h =ae^(2ix) +be^(−2ix) =α cos(2x) +βsin(2x) =αu_1 +βu_2 W(u_1 ,u_2 )= determinant (((cos(2x) sin(2x))),((−2sin2x 2cos(2x) )))=2(cos^2 2x +sin^2 (2x))=2≠0 W_1 = determinant (((0 sin(2x))),(((4/(sin(2x))) 2cos(2x))))=−4 W_2 = determinant (((cos(2x) 0)),((−2sin(2x) (4/(sin(2x))))))=4cotan(2x) V_1 =∫ (W_1 /W) dx =∫ ((−4)/2) dx =−2x V_2 =∫ (W_2 /W) dx =∫ ((4cotan(2x))/2) dx =2 ∫ ((cos(2x))/(sin(2x))) dx =ln(∣sin(2x)∣) ⇒y_p =u_1 v_1 +u_2 v_2 =cos(2x)(−2x)+sin(2x) ln∣sin(2x)∣ the general solution is y =y_h +y_p y =αcos(2x) +β sin(2x)−2x cos(2x)+sin(2x)ln∣sin(2x)∣

$$\mathrm{y}^{''} \:+\mathrm{4y}\:=\frac{\mathrm{4}}{\mathrm{sin}\left(\mathrm{2x}\right)} \\ $$$$\mathrm{h}\rightarrow\mathrm{r}^{\mathrm{2}} +\mathrm{4}=\mathrm{0}\:\Rightarrow\mathrm{r}\:=\overset{−} {+}\mathrm{2i}\:\:\Rightarrow\mathrm{y}_{\mathrm{h}} =\mathrm{ae}^{\mathrm{2ix}} \:+\mathrm{be}^{−\mathrm{2ix}} \:=\alpha\:\mathrm{cos}\left(\mathrm{2x}\right)\:+\beta\mathrm{sin}\left(\mathrm{2x}\right) \\ $$$$=\alpha\mathrm{u}_{\mathrm{1}} \:+\beta\mathrm{u}_{\mathrm{2}} \\ $$$$\mathrm{W}\left(\mathrm{u}_{\mathrm{1}} ,\mathrm{u}_{\mathrm{2}} \right)=\begin{vmatrix}{\mathrm{cos}\left(\mathrm{2x}\right)\:\:\:\:\:\:\:\:\:\mathrm{sin}\left(\mathrm{2x}\right)}\\{−\mathrm{2sin2x}\:\:\:\:\:\:\:\mathrm{2cos}\left(\mathrm{2x}\right)\:}\end{vmatrix}=\mathrm{2}\left(\mathrm{cos}^{\mathrm{2}} \mathrm{2x}\:+\mathrm{sin}^{\mathrm{2}} \left(\mathrm{2x}\right)\right)=\mathrm{2}\neq\mathrm{0} \\ $$$$\mathrm{W}_{\mathrm{1}} =\begin{vmatrix}{\mathrm{0}\:\:\:\:\:\:\:\:\:\:\mathrm{sin}\left(\mathrm{2x}\right)}\\{\frac{\mathrm{4}}{\mathrm{sin}\left(\mathrm{2x}\right)}\:\:\:\:\:\:\mathrm{2cos}\left(\mathrm{2x}\right)}\end{vmatrix}=−\mathrm{4} \\ $$$$\mathrm{W}_{\mathrm{2}} =\begin{vmatrix}{\mathrm{cos}\left(\mathrm{2x}\right)\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}}\\{−\mathrm{2sin}\left(\mathrm{2x}\right)\:\:\:\:\:\frac{\mathrm{4}}{\mathrm{sin}\left(\mathrm{2x}\right)}}\end{vmatrix}=\mathrm{4cotan}\left(\mathrm{2x}\right) \\ $$$$\mathrm{V}_{\mathrm{1}} =\int\:\:\frac{\mathrm{W}_{\mathrm{1}} }{\mathrm{W}}\:\mathrm{dx}\:=\int\:\:\frac{−\mathrm{4}}{\mathrm{2}}\:\mathrm{dx}\:=−\mathrm{2x} \\ $$$$\mathrm{V}_{\mathrm{2}} =\int\:\:\frac{\mathrm{W}_{\mathrm{2}} }{\mathrm{W}}\:\mathrm{dx}\:=\int\:\:\frac{\mathrm{4cotan}\left(\mathrm{2x}\right)}{\mathrm{2}}\:\mathrm{dx}\:=\mathrm{2}\:\int\:\:\frac{\mathrm{cos}\left(\mathrm{2x}\right)}{\mathrm{sin}\left(\mathrm{2x}\right)}\:\mathrm{dx}\:=\mathrm{ln}\left(\mid\mathrm{sin}\left(\mathrm{2x}\right)\mid\right) \\ $$$$\Rightarrow\mathrm{y}_{\mathrm{p}} =\mathrm{u}_{\mathrm{1}} \mathrm{v}_{\mathrm{1}} +\mathrm{u}_{\mathrm{2}} \mathrm{v}_{\mathrm{2}} =\mathrm{cos}\left(\mathrm{2x}\right)\left(−\mathrm{2x}\right)+\mathrm{sin}\left(\mathrm{2x}\right)\:\mathrm{ln}\mid\mathrm{sin}\left(\mathrm{2x}\right)\mid \\ $$$$\mathrm{the}\:\mathrm{general}\:\mathrm{solution}\:\mathrm{is}\:\mathrm{y}\:=\mathrm{y}_{\mathrm{h}} +\mathrm{y}_{\mathrm{p}} \\ $$$$\mathrm{y}\:=\alpha\mathrm{cos}\left(\mathrm{2x}\right)\:+\beta\:\mathrm{sin}\left(\mathrm{2x}\right)−\mathrm{2x}\:\mathrm{cos}\left(\mathrm{2x}\right)+\mathrm{sin}\left(\mathrm{2x}\right)\mathrm{ln}\mid\mathrm{sin}\left(\mathrm{2x}\right)\mid \\ $$

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