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Question Number 108353 by bemath last updated on 16/Aug/20

((△BeMath△)/…) General solution of (d^2 y/dx^2 ) + (dy/dx)−2y = sin x

$$\:\:\:\:\:\:\frac{\bigtriangleup\mathcal{B}{e}\mathcal{M}{ath}\bigtriangleup}{\ldots} \\ $$$$\:{General}\:{solution}\:{of}\: \\ $$$$\:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:+\:\frac{{dy}}{{dx}}−\mathrm{2}{y}\:=\:\mathrm{sin}\:{x}\: \\ $$

Commented by bemath last updated on 16/Aug/20

Answered by Aziztisffola last updated on 16/Aug/20

r^2 +r−2=0 ⇒r=−2 or r=1 ⇒y_h =αe^(−2x) +βe^x y_p =u_1 e^(−2x) +u_2 e^x w(u_1 ;u_2 )= determinant ((e^(−2x) ,e^x ),((−2e^(−2x) ),e^x ))=e^(−x) +2e^(−x) =3e^(−x) w_1 = determinant ((0,e^x ),((sinx),e^x ))=−e^x sin(x) w_2 = determinant ((e^(−2x) ,0),((−2e^(−2x) ),(sin(x))))=e^(−2x) sin(x) u_1 =∫(w_1 /w) dx=∫((−e^x sin(x))/(3e^(−x) )) dx=((−1)/3)∫e^(2x) sin(x)dx =((−1)/3)(−e^(2x) cos(x)+2∫e^(2x) cos(x)dx) =−(1/(15))e^(2x) (cos(x)+2sin(x)) u_2 =∫(w_2 /w) dx=∫((e^(−2x) sin(x))/(3e^(−x) )) dx=(1/3)∫e^(−x) sin(x)dx =(1/3)(−e^(−x) cos(x)+∫e^(−x) cos(x))=−(1/3)e^(−x) cos(x) y=y_h +y_p

$$\mathrm{r}^{\mathrm{2}} +\mathrm{r}−\mathrm{2}=\mathrm{0}\:\Rightarrow\mathrm{r}=−\mathrm{2}\:\mathrm{or}\:\mathrm{r}=\mathrm{1} \\ $$$$\Rightarrow\mathrm{y}_{\mathrm{h}} =\alpha\mathrm{e}^{−\mathrm{2x}} +\beta\mathrm{e}^{\mathrm{x}} \\ $$$$\:\mathrm{y}_{\mathrm{p}} =\mathrm{u}_{\mathrm{1}} \mathrm{e}^{−\mathrm{2x}} +\mathrm{u}_{\mathrm{2}} \mathrm{e}^{\mathrm{x}} \\ $$$$\:\mathrm{w}\left(\mathrm{u}_{\mathrm{1}} ;\mathrm{u}_{\mathrm{2}} \right)=\begin{vmatrix}{\mathrm{e}^{−\mathrm{2x}} }&{\mathrm{e}^{\mathrm{x}} }\\{−\mathrm{2e}^{−\mathrm{2x}} }&{\mathrm{e}^{\mathrm{x}} }\end{vmatrix}=\mathrm{e}^{−\mathrm{x}} +\mathrm{2e}^{−\mathrm{x}} =\mathrm{3e}^{−\mathrm{x}} \\ $$$$\:\mathrm{w}_{\mathrm{1}} =\begin{vmatrix}{\mathrm{0}}&{\mathrm{e}^{\mathrm{x}} }\\{\mathrm{sin}{x}}&{\mathrm{e}^{\mathrm{x}} }\end{vmatrix}=−\mathrm{e}^{\mathrm{x}} \mathrm{sin}\left(\mathrm{x}\right) \\ $$$$\:\mathrm{w}_{\mathrm{2}} =\begin{vmatrix}{\mathrm{e}^{−\mathrm{2x}} }&{\mathrm{0}}\\{−\mathrm{2e}^{−\mathrm{2x}} }&{\mathrm{sin}\left(\mathrm{x}\right)}\end{vmatrix}=\mathrm{e}^{−\mathrm{2x}} \mathrm{sin}\left(\mathrm{x}\right) \\ $$$$\:\mathrm{u}_{\mathrm{1}} =\int\frac{\mathrm{w}_{\mathrm{1}} }{\mathrm{w}}\:\mathrm{dx}=\int\frac{−\mathrm{e}^{\mathrm{x}} \mathrm{sin}\left(\mathrm{x}\right)}{\mathrm{3e}^{−\mathrm{x}} }\:\mathrm{dx}=\frac{−\mathrm{1}}{\mathrm{3}}\int\mathrm{e}^{\mathrm{2x}} \mathrm{sin}\left(\mathrm{x}\right)\mathrm{dx} \\ $$$$\:=\frac{−\mathrm{1}}{\mathrm{3}}\left(−\mathrm{e}^{\mathrm{2x}} \mathrm{cos}\left(\mathrm{x}\right)+\mathrm{2}\int\mathrm{e}^{\mathrm{2x}} \mathrm{cos}\left(\mathrm{x}\right)\mathrm{dx}\right) \\ $$$$=−\frac{\mathrm{1}}{\mathrm{15}}\mathrm{e}^{\mathrm{2x}} \left(\mathrm{cos}\left(\mathrm{x}\right)+\mathrm{2sin}\left(\mathrm{x}\right)\right) \\ $$$$\:\mathrm{u}_{\mathrm{2}} =\int\frac{\mathrm{w}_{\mathrm{2}} }{\mathrm{w}}\:\mathrm{dx}=\int\frac{\mathrm{e}^{−\mathrm{2x}} \mathrm{sin}\left(\mathrm{x}\right)}{\mathrm{3e}^{−\mathrm{x}} }\:\mathrm{dx}=\frac{\mathrm{1}}{\mathrm{3}}\int\mathrm{e}^{−\mathrm{x}} \mathrm{sin}\left(\mathrm{x}\right)\mathrm{dx} \\ $$$$\:=\frac{\mathrm{1}}{\mathrm{3}}\left(−\mathrm{e}^{−\mathrm{x}} \mathrm{cos}\left(\mathrm{x}\right)+\int\mathrm{e}^{−\mathrm{x}} \mathrm{cos}\left(\mathrm{x}\right)\right)=−\frac{\mathrm{1}}{\mathrm{3}}\mathrm{e}^{−\mathrm{x}} \mathrm{cos}\left(\mathrm{x}\right) \\ $$$$\:\mathrm{y}=\mathrm{y}_{\mathrm{h}} +\mathrm{y}_{\mathrm{p}} \\ $$

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