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Question Number 207991 by efronzo1 last updated on 02/Jun/24
y W
$$\:\:\:\:\:\mathrm{y}\:\underline{\underbrace{\mathcal{W}}} \\ $$
Answered by mr W last updated on 02/Jun/24
particular solution: y=A sin x+B cos x y′=A cos x−B sin x y′′=−A sin x−B cos x −3A−B=1 −A−3B=0 A=(3/(10)) B=−(1/(10)) ⇒y=(1/(10))(3 sin x−cos x) general solution: y′′+y′−2y=0 r^2 +r−2=0 (r+2)(r−1)=0 r=−2, 1 y=C_1 e^(−2x) +C_2 e^x ⇒y=C_1 e^(−2x) +C_2 e^x +(1/(10))(3 sin x−cos x)
$${particular}\:{solution}: \\ $$$${y}={A}\:\mathrm{sin}\:{x}+{B}\:\mathrm{cos}\:{x} \\ $$$${y}'={A}\:\mathrm{cos}\:{x}−{B}\:\mathrm{sin}\:{x} \\ $$$${y}''=−{A}\:\mathrm{sin}\:{x}−{B}\:\mathrm{cos}\:{x} \\ $$$$−\mathrm{3}{A}−{B}=\mathrm{1} \\ $$$$−{A}−\mathrm{3}{B}=\mathrm{0} \\ $$$${A}=\frac{\mathrm{3}}{\mathrm{10}} \\ $$$${B}=−\frac{\mathrm{1}}{\mathrm{10}} \\ $$$$\Rightarrow{y}=\frac{\mathrm{1}}{\mathrm{10}}\left(\mathrm{3}\:\mathrm{sin}\:{x}−\mathrm{cos}\:{x}\right) \\ $$$${general}\:{solution}: \\ $$$${y}''+{y}'−\mathrm{2}{y}=\mathrm{0} \\ $$$${r}^{\mathrm{2}} +{r}−\mathrm{2}=\mathrm{0} \\ $$$$\left({r}+\mathrm{2}\right)\left({r}−\mathrm{1}\right)=\mathrm{0} \\ $$$${r}=−\mathrm{2},\:\mathrm{1} \\ $$$${y}={C}_{\mathrm{1}} {e}^{−\mathrm{2}{x}} +{C}_{\mathrm{2}} {e}^{{x}} \\ $$$$ \\ $$$$\Rightarrow{y}={C}_{\mathrm{1}} {e}^{−\mathrm{2}{x}} +{C}_{\mathrm{2}} {e}^{{x}} +\frac{\mathrm{1}}{\mathrm{10}}\left(\mathrm{3}\:\mathrm{sin}\:{x}−\mathrm{cos}\:{x}\right) \\ $$
Answered by efronzo1 last updated on 02/Jun/24
(D^2 +D−2)y = sin x Let D^2 +D−2 = 0 (D+2)(D−1)=0 D=−2 , D=1 y_h = C_1 e^(−2x ) + C_2 e^x y_p = (1/(D^2 +D−1)) (sin x) y_p = (1/(D−3)) (sin x) y_p = ((D+3)/(D^2 −9)) (sin x)=−((D+3)/(10)) (sin x) y_p =−(1/(10))(cos x+3sin x) y_g = C_1 e^(−2x) +C_2 e^x −(1/(10))(cos x+3sin x)
$$\:\left(\mathrm{D}^{\mathrm{2}} +\mathrm{D}−\mathrm{2}\right)\mathrm{y}\:=\:\mathrm{sin}\:\mathrm{x} \\ $$$$\:\mathrm{Let}\:\mathrm{D}^{\mathrm{2}} +\mathrm{D}−\mathrm{2}\:=\:\mathrm{0} \\ $$$$\:\:\left(\mathrm{D}+\mathrm{2}\right)\left(\mathrm{D}−\mathrm{1}\right)=\mathrm{0} \\ $$$$\:\:\:\mathrm{D}=−\mathrm{2}\:,\:\mathrm{D}=\mathrm{1} \\ $$$$\:\mathrm{y}_{\mathrm{h}} \:=\:\mathrm{C}_{\mathrm{1}} \mathrm{e}^{−\mathrm{2x}\:} +\:\mathrm{C}_{\mathrm{2}} \mathrm{e}^{\mathrm{x}} \\ $$$$\:\: \\ $$$$\:\:\mathrm{y}_{\mathrm{p}} =\:\frac{\mathrm{1}}{\mathrm{D}^{\mathrm{2}} +\mathrm{D}−\mathrm{1}}\:\left(\mathrm{sin}\:\mathrm{x}\right) \\ $$$$\:\:\:\mathrm{y}_{\mathrm{p}} =\:\frac{\mathrm{1}}{\mathrm{D}−\mathrm{3}}\:\left(\mathrm{sin}\:\mathrm{x}\right) \\ $$$$\:\:\:\mathrm{y}_{\mathrm{p}} =\:\frac{\mathrm{D}+\mathrm{3}}{\mathrm{D}^{\mathrm{2}} −\mathrm{9}}\:\left(\mathrm{sin}\:\mathrm{x}\right)=−\frac{\mathrm{D}+\mathrm{3}}{\mathrm{10}}\:\left(\mathrm{sin}\:\mathrm{x}\right) \\ $$$$\:\:\:\mathrm{y}_{\mathrm{p}} =−\frac{\mathrm{1}}{\mathrm{10}}\left(\mathrm{cos}\:\mathrm{x}+\mathrm{3sin}\:\mathrm{x}\right) \\ $$$$\:\mathrm{y}_{\mathrm{g}} =\:\mathrm{C}_{\mathrm{1}} \mathrm{e}^{−\mathrm{2x}} +\mathrm{C}_{\mathrm{2}} \mathrm{e}^{\mathrm{x}} −\frac{\mathrm{1}}{\mathrm{10}}\left(\mathrm{cos}\:\mathrm{x}+\mathrm{3sin}\:\mathrm{x}\right) \\ $$

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