y^{''}-8y^'+16y=((10.5e^{4t}))/((t^2+1))

解

y^{^{''}} - 8 y^{^{'}} + 16 y = \frac{( 10.5 e ^{4 t} )}{( t ^{2} + 1 )}

y = c_{1} e^{4 t} + c_{2} t e^{4 t} - 5.25 e^{4 t} ln (t^{2} + 1) + 10.5 e^{4 t} t arctan (t)

解答ステップ

y^{''} (t) - 8 y^{'} (t) + 16 y = \frac{( 10.5 e ^{4 t} )}{( t ^{2} + 1 )}

線形 ODE を解く:

y = c_{1} e^{4 t} + c_{2} t e^{4 t} - 5.25 e^{4 t} ln (t^{2} + 1) + 10.5 e^{4 t} t arctan (t)

y = c_{1} e^{4 t} + c_{2} t e^{4 t} - 5.25 e^{4 t} ln (t^{2} + 1) + 10.5 e^{4 t} t arctan (t)