\(\frac{dy}{dx}-ay=\sin x\)

１階線形微分方程式の公式（＜１１９８＞参照）より、

\(\int Pdx=\int (-a)dx=-ax+C\)
NAMO_EQN__ 160 1 \int Pdx=\int (-a)dx=-ax+C
\(e^{^{\int Pdx}}=e^{^{-ax+C}}=Ce^{^{-ax}}\)
NAMO_EQN__ 160 1 e^{^{\int Pdx}}=e^{^{-ax+C}}=Ce^{^{-ax}}
\(\int Qe^{^{\int Pdx}}dx=\int \sin x\cdot Ce^{^{-ax}}dx=C\int \sin x\cdot e^{^{-ax}}dx\)
NAMO_EQN__ 160 1 \int Qe^{^{\int Pdx}}dx=\int \sin x\cdot Ce^{^{-ax}}dx=C\int \sin x\cdot e^{^{-ax}}dx
\(I=\int \sin x\cdot e^{^{-ax}}dx=-\cos x\cdot e^{^{-ax}}-\int (-\cos x)(-a)e^{^{-ax}}dx\)
NAMO_EQN__ 160 1 I=\int \sin x\cdot e^{^{-ax}}dx=-\cos x\cdot e^{^{-ax}}-\int (-\cos x)(-a)e^{^{-ax}}dx
\(=-\cos x\cdot e^{^{-ax}}-a\int \cos x\cdot e^{^{-ax}}dx=-\cos x\cdot e^{^{-ax}}-a\{ \sin x\cdot e^{^{-ax}}-\int \sin x\cdot (-a)e^{^{-ax}}dx\}\)
NAMO_EQN__ 160 1 =-\cos x\cdot e^{^{-ax}}-a\int \cos x\cdot e^{^{-ax}}dx=-\cos x\cdot e^{^{-ax}}-a\{ \sin x\cdot e^{^{-ax}}-\int \sin x\cdot (-a)e^{^{-ax}}dx\}
\(=-e^{^{-ax}}(\cos x+a\sin x)-a^{^{2}}\int \sin x\cdot e^{^{-ax}}dx=-e^{^{-ax}}(\cos x+a\sin x)-a^{^{2}}I\)
NAMO_EQN__ 160 1 =-e^{^{-ax}}(\cos x+a\sin x)-a^{^{2}}\int \sin x\cdot e^{^{-ax}}dx=-e^{^{-ax}}(\cos x+a\sin x)-a^{^{2}}I
\((1+a^{^{2}})I=-e^{^{-ax}}(\cos x+a\sin x)\)
NAMO_EQN__ 160 1 (1+a^{^{2}})I=-e^{^{-ax}}(\cos x+a\sin x)
\(I=\frac{-e^{^{-ax}}(\cos x+a\sin x)}{1+a^{^{2}}}\)
NAMO_EQN__ 160 1 I=\frac{-e^{^{-ax}}(\cos x+a\sin x)}{1+a^{^{2}}}
\(y=Ce^{^{ax}}(C\cdot \frac{-e^{^{-ax}}(\cos x+a\sin x)}{1+a^{^{2}}}+C)\)
NAMO_EQN__ 160 1 y=Ce^{^{ax}}(C\cdot \frac{-e^{^{-ax}}(\cos x+a\sin x)}{1+a^{^{2}}}+C)
\(=C_{_{1}}\frac{\cos x+a\sin x}{1+a^{^{2}}}+C_{_{2}}e^{^{ax}}\)
NAMO_EQN__ 160 1 =C_{_{1}}\frac{\cos x+a\sin x}{1+a^{^{2}}}+C_{_{2}}e^{^{ax}}
………（答）