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Question Number 182788 by CrispyXYZ last updated on 14/Dec/22

y′′+ay=b

$${y}''+{ay}={b} \\ $$

Answered by qaz last updated on 14/Dec/22

y_p =(1/(D^2 +a))b=(b/(2i(√a)))((1/(D−i(√a)))−(1/(D+i(√a))))  =(b/(2i(√a)))(e^(i(√a)x) (1/D)e^(−i(√a)x) −e^(−i(√a)x) (1/D)e^(i(√a)x) )  =(b/(2i(√a)))(e^(i(√a)x) ∫e^(−i(√a)x) dx−e^(−i(√a)x) ∫e^(i(√a)x) dx)  =(b/(2i(√a)))(−(1/(i(√a)))−(1/(i(√a))))=(b/a)  ⇒y=C_1 cos (√a)x+C_2 sin (√a)x+(b/a)

$${y}_{{p}} =\frac{\mathrm{1}}{{D}^{\mathrm{2}} +{a}}{b}=\frac{{b}}{\mathrm{2}{i}\sqrt{{a}}}\left(\frac{\mathrm{1}}{{D}−{i}\sqrt{{a}}}−\frac{\mathrm{1}}{{D}+{i}\sqrt{{a}}}\right) \\ $$$$=\frac{{b}}{\mathrm{2}{i}\sqrt{{a}}}\left({e}^{{i}\sqrt{{a}}{x}} \frac{\mathrm{1}}{{D}}{e}^{−{i}\sqrt{{a}}{x}} −{e}^{−{i}\sqrt{{a}}{x}} \frac{\mathrm{1}}{{D}}{e}^{{i}\sqrt{{a}}{x}} \right) \\ $$$$=\frac{{b}}{\mathrm{2}{i}\sqrt{{a}}}\left({e}^{{i}\sqrt{{a}}{x}} \int{e}^{−{i}\sqrt{{a}}{x}} {dx}−{e}^{−{i}\sqrt{{a}}{x}} \int{e}^{{i}\sqrt{{a}}{x}} {dx}\right) \\ $$$$=\frac{{b}}{\mathrm{2}{i}\sqrt{{a}}}\left(−\frac{\mathrm{1}}{{i}\sqrt{{a}}}−\frac{\mathrm{1}}{{i}\sqrt{{a}}}\right)=\frac{{b}}{{a}} \\ $$$$\Rightarrow{y}={C}_{\mathrm{1}} \mathrm{cos}\:\sqrt{{a}}{x}+{C}_{\mathrm{2}} \mathrm{sin}\:\sqrt{{a}}{x}+\frac{{b}}{{a}} \\ $$

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