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

$$y^{″}+ay=b$$

Answered by qaz last updated on 14/Dec/22

$$y_p=\frac{1}{D^2+a}b=\frac{b}{2i\sqrt{a}}(\frac{1}{D-i\sqrt{a}}-\frac{1}{D+i\sqrt{a}})$$$$=\frac{b}{2i\sqrt{a}}(e^{i\sqrt{a}x}\frac{1}{D}{e}^{-i\sqrt{a}x}-e^{-i\sqrt{a}x}\frac{1}{D}{e}^{i\sqrt{a}x})$$$$=\frac{b}{2i\sqrt{a}}(e^{i\sqrt{a}x}\int e^{-i\sqrt{a}x}dx-e^{-i\sqrt{a}x}\int e^{i\sqrt{a}x}dx)$$$$=\frac{b}{2i\sqrt{a}}(-\frac{1}{i\sqrt{a}}-\frac{1}{i\sqrt{a}})=\frac{b}{a}$$$$\Rightarrow y=C_1\cos \sqrt{a}x+C_2\sin \sqrt{a}x+\frac{b}{a}$$

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