Question and Answers Forum

All Questions      Topic List

Differential Equation Questions

Previous in All Question      Next in All Question      

Previous in Differential Equation      Next in Differential Equation      

Question Number 145578 by rexford last updated on 06/Jul/21

y′′_y=xsin2x solve the differential eqn..

$${y}''\_{y}={xsin}\mathrm{2}{x} \\ $$$${solve}\:{the}\:{differential}\:{eqn}.. \\ $$

Answered by mathmax by abdo last updated on 06/Jul/21

y^(′′) −y =xsin(2x) e→r^2 −1=0 ⇒r=+^− 1 ⇒y=ae^x +be^(−x ) =au_1 +bu_2 w(u_1 ,u_2 )= determinant (((e^x e^(−x) )),((e^(x ) −e^(−x) )))=−2≠0 w_1 = determinant (((o e^(−x) )),((xsin(2x) −e^(−x) )))=−xe^(−x) sin(2x) w_2 = determinant (((e^x 0)),((e^x xsin(2x))))=xe^x sin(2x) v_1 =∫ (w_1 /w)dx=(1/2)∫xe^(−x) sin(2x)dx =(1/2)Im(∫ xe^(−x+2ix) dx) ∫ xe^((−1+2i)x) dx =(x/(−1+2i))e^((−1+2i)x) −∫ (1/(−1+2i))e^((−1+2i)x) dx =((−x)/(1−2i))e^((−1+2i)x) +(1/(1−2i))×(1/(−1+2i))e^((−1+2i)x) =((−x(1+2i))/5)e^(−x) (cos(2x)+isin(2x))−(1/((1−2i)^2 ))e^(−x) (cos(2x)+isin(2x)) =((−xe^(−x) )/5)(cos(2x)+isin(2x)+2icos(2x)−2sin(2x)) −(((1+2i)^2 )/(25))e^(−x) (cos(2x)+isin(2x))=.... v_2 =∫ (w_2 /w)dx=−(1/2)∫ xe^x sin(2x)dx =−(1/2)Im(∫ xe^(x+2ix) dx) ∫ x e^((1+2i)x) dx =(x/(1+2i))e^((1+2i)x) −∫(1/(1+2i))e^((1+2i)x) dx=.... ⇒y_p =u_1 v_1 +u_2 v_2 and general solution is y=ae^x +be^(−x) +u_1 v_1 +u_2 v_2

$$\mathrm{y}^{''} −\mathrm{y}\:=\mathrm{xsin}\left(\mathrm{2x}\right) \\ $$$$\mathrm{e}\rightarrow\mathrm{r}^{\mathrm{2}} −\mathrm{1}=\mathrm{0}\:\Rightarrow\mathrm{r}=\overset{−} {+}\mathrm{1}\:\Rightarrow\mathrm{y}=\mathrm{ae}^{\mathrm{x}} \:+\mathrm{be}^{−\mathrm{x}\:} =\mathrm{au}_{\mathrm{1}} +\mathrm{bu}_{\mathrm{2}} \\ $$$$\mathrm{w}\left(\mathrm{u}_{\mathrm{1}} ,\mathrm{u}_{\mathrm{2}} \right)=\begin{vmatrix}{\mathrm{e}^{\mathrm{x}} \:\:\:\:\:\:\:\:\:\mathrm{e}^{−\mathrm{x}} }\\{\mathrm{e}^{\mathrm{x}\:\:\:\:} \:\:\:\:\:\:−\mathrm{e}^{−\mathrm{x}} }\end{vmatrix}=−\mathrm{2}\neq\mathrm{0} \\ $$$$\mathrm{w}_{\mathrm{1}} =\begin{vmatrix}{\mathrm{o}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{e}^{−\mathrm{x}} }\\{\mathrm{xsin}\left(\mathrm{2x}\right)\:\:\:\:\:\:\:−\mathrm{e}^{−\mathrm{x}} }\end{vmatrix}=−\mathrm{xe}^{−\mathrm{x}} \mathrm{sin}\left(\mathrm{2x}\right) \\ $$$$\mathrm{w}_{\mathrm{2}} =\begin{vmatrix}{\mathrm{e}^{\mathrm{x}} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}}\\{\mathrm{e}^{\mathrm{x}} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{xsin}\left(\mathrm{2x}\right)}\end{vmatrix}=\mathrm{xe}^{\mathrm{x}} \:\mathrm{sin}\left(\mathrm{2x}\right) \\ $$$$\mathrm{v}_{\mathrm{1}} =\int\:\frac{\mathrm{w}_{\mathrm{1}} }{\mathrm{w}}\mathrm{dx}=\frac{\mathrm{1}}{\mathrm{2}}\int\mathrm{xe}^{−\mathrm{x}} \mathrm{sin}\left(\mathrm{2x}\right)\mathrm{dx}\:=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{Im}\left(\int\:\mathrm{xe}^{−\mathrm{x}+\mathrm{2ix}} \mathrm{dx}\right) \\ $$$$\int\:\mathrm{xe}^{\left(−\mathrm{1}+\mathrm{2i}\right)\mathrm{x}} \mathrm{dx}\:=\frac{\mathrm{x}}{−\mathrm{1}+\mathrm{2i}}\mathrm{e}^{\left(−\mathrm{1}+\mathrm{2i}\right)\mathrm{x}} −\int\:\:\frac{\mathrm{1}}{−\mathrm{1}+\mathrm{2i}}\mathrm{e}^{\left(−\mathrm{1}+\mathrm{2i}\right)\mathrm{x}} \mathrm{dx} \\ $$$$=\frac{−\mathrm{x}}{\mathrm{1}−\mathrm{2i}}\mathrm{e}^{\left(−\mathrm{1}+\mathrm{2i}\right)\mathrm{x}} \:+\frac{\mathrm{1}}{\mathrm{1}−\mathrm{2i}}×\frac{\mathrm{1}}{−\mathrm{1}+\mathrm{2i}}\mathrm{e}^{\left(−\mathrm{1}+\mathrm{2i}\right)\mathrm{x}} \\ $$$$=\frac{−\mathrm{x}\left(\mathrm{1}+\mathrm{2i}\right)}{\mathrm{5}}\mathrm{e}^{−\mathrm{x}} \left(\mathrm{cos}\left(\mathrm{2x}\right)+\mathrm{isin}\left(\mathrm{2x}\right)\right)−\frac{\mathrm{1}}{\left(\mathrm{1}−\mathrm{2i}\right)^{\mathrm{2}} }\mathrm{e}^{−\mathrm{x}} \left(\mathrm{cos}\left(\mathrm{2x}\right)+\mathrm{isin}\left(\mathrm{2x}\right)\right) \\ $$$$ \\ $$$$=\frac{−\mathrm{xe}^{−\mathrm{x}} }{\mathrm{5}}\left(\mathrm{cos}\left(\mathrm{2x}\right)+\mathrm{isin}\left(\mathrm{2x}\right)+\mathrm{2icos}\left(\mathrm{2x}\right)−\mathrm{2sin}\left(\mathrm{2x}\right)\right) \\ $$$$−\frac{\left(\mathrm{1}+\mathrm{2i}\right)^{\mathrm{2}} }{\mathrm{25}}\mathrm{e}^{−\mathrm{x}} \left(\mathrm{cos}\left(\mathrm{2x}\right)+\mathrm{isin}\left(\mathrm{2x}\right)\right)=.... \\ $$$$\mathrm{v}_{\mathrm{2}} =\int\:\frac{\mathrm{w}_{\mathrm{2}} }{\mathrm{w}}\mathrm{dx}=−\frac{\mathrm{1}}{\mathrm{2}}\int\:\:\mathrm{xe}^{\mathrm{x}} \mathrm{sin}\left(\mathrm{2x}\right)\mathrm{dx} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{Im}\left(\int\:\:\mathrm{xe}^{\mathrm{x}+\mathrm{2ix}} \mathrm{dx}\right) \\ $$$$\int\:\mathrm{x}\:\mathrm{e}^{\left(\mathrm{1}+\mathrm{2i}\right)\mathrm{x}} \mathrm{dx}\:=\frac{\mathrm{x}}{\mathrm{1}+\mathrm{2i}}\mathrm{e}^{\left(\mathrm{1}+\mathrm{2i}\right)\mathrm{x}} −\int\frac{\mathrm{1}}{\mathrm{1}+\mathrm{2i}}\mathrm{e}^{\left(\mathrm{1}+\mathrm{2i}\right)\mathrm{x}} \mathrm{dx}=.... \\ $$$$\Rightarrow\mathrm{y}_{\mathrm{p}} =\mathrm{u}_{\mathrm{1}} \mathrm{v}_{\mathrm{1}} +\mathrm{u}_{\mathrm{2}} \mathrm{v}_{\mathrm{2}} \:\:\:\mathrm{and}\:\mathrm{general}\:\mathrm{solution}\:\mathrm{is} \\ $$$$\mathrm{y}=\mathrm{ae}^{\mathrm{x}} \:+\mathrm{be}^{−\mathrm{x}} \:+\mathrm{u}_{\mathrm{1}} \mathrm{v}_{\mathrm{1}} \:+\mathrm{u}_{\mathrm{2}} \mathrm{v}_{\mathrm{2}} \\ $$

Commented by mathmax by abdo last updated on 06/Jul/21

you can use mvc method but contains alos of calculus the best is wronskien...

$$\mathrm{you}\:\mathrm{can}\:\mathrm{use}\:\mathrm{mvc}\:\mathrm{method}\:\mathrm{but}\:\mathrm{contains}\:\mathrm{alos}\:\mathrm{of}\:\mathrm{calculus} \\ $$$$\mathrm{the}\:\mathrm{best}\:\mathrm{is}\:\mathrm{wronskien}... \\ $$

Commented by rexford last updated on 06/Jul/21

Please,is there a different way of solving this as in introducing constants as coefficients for the particular solution

$${Please},{is}\:{there}\:{a}\:{different}\:{way}\:{of}\:{solving}\:{this}\:{as}\:{in}\:{introducing}\:{constants}\:{as}\:{coefficients}\:{for}\:{the}\:{particular}\:{solution} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com