What is the solution for y^{''}+3y^'+2y=10u,y(0)=0,y^'(0)=0 ?

The solution for y^{''}+3y^'+2y=10u,y(0)=0,y^'(0)=0 is y=-10ue^{-t}+5ue^{-2t}+5u

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y^{''}+3y^'+2y=10u,y(0)=0,y^'(0)=0

y^{^{''}} + 3 y^{^{'}} + 2 y = 10 u, y (0) = 0, y^{^{'}} (0) = 0

y = - 10 u e^{- t} + 5 u e^{- 2 t} + 5 u

Solution steps

y^{''} (t) + 3 y^{'} (t) + 2 y = 10 u

Solve linear ODE:

y = - 10 u e^{- t} + 5 u e^{- 2 t} + 5 u

y = - 10 u e^{- t} + 5 u e^{- 2 t} + 5 u

y^{^{''}} - 2 y^{^{'}} - 2 = 0

\frac{d ^{2} y}{d x ^{2}} - y - 6 = 1

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

\frac{d ^{2} y}{d x ^{2}} - 5 \frac{d y}{d x} + 9 y = x e^{x}

y^{^{''}} + y^{^{'}} + 2 y = x + 3, y (0) = 0, y^{^{'}} (0) = 1

What is the solution for y^{''}+3y^'+2y=10u,y(0)=0,y^'(0)=0 ?
The solution for y^{''}+3y^'+2y=10u,y(0)=0,y^'(0)=0 is y=-10ue^{-t}+5ue^{-2t}+5u