y^{''}+a^2y^'=cos(bx)

解

y^{^{''}} + a^{2} y^{^{'}} = cos (b x)

y = c_{1} + c_{2} e^{- a^{2} x} + \frac{a ^{2} sin ( b x )}{b ( b ^{2} + a ^{4} )} - \frac{cos ( b x )}{b ^{2} + a ^{4}}

解答ステップ

y^{''} (x) + a^{2} y^{'} (x) = cos (b x)

線形 ODE を解く:

y = c_{1} + c_{2} e^{- a^{2} x} + \frac{a ^{2} sin ( b x )}{b ( b ^{2} + a ^{4} )} - \frac{cos ( b x )}{b ^{2} + a ^{4}}

y = c_{1} + c_{2} e^{- a^{2} x} + \frac{a ^{2} sin ( b x )}{b ( b ^{2} + a ^{4} )} - \frac{cos ( b x )}{b ^{2} + a ^{4}}