y^{''}+y^'-2y=upi(t)-u^2pi(t)

解答

y^{^{''}} + y^{^{'}} - 2 y = u π (t) - u^{2} π (t)

y = c_{1} e^{t} + c_{2} e^{- 2 t} - \frac{t ( π u - π u ^{2} )}{2} - \frac{π u - π u ^{2}}{4}

求解步骤

y^{''} (t) + y^{'} (t) - 2 y = u π (t) - u^{2} π (t)

求解线性方程:

y = c_{1} e^{t} + c_{2} e^{- 2 t} - \frac{t ( π u - π u ^{2} )}{2} - \frac{π u - π u ^{2}}{4}

y = c_{1} e^{t} + c_{2} e^{- 2 t} - \frac{t ( π u - π u ^{2} )}{2} - \frac{π u - π u ^{2}}{4}

y^{^{''}} + 4 y^{^{'}} + 3 y = - 19 t e^{3 t}

y^{^{''}} + 4 y^{^{'}} = sin (t)

y^{^{''}} - 5 y^{^{'}} + 6 y = - 5 e^{3 x}

y^{^{''}} - 2 y^{^{'}} + y = t^{2} + 1

y^{^{''}} + 8 y^{^{'}} + 16 y = 6 cos (4 t)