What is the solution for 2y^{''}-y^'-y=t+1,y(0)=1,y^'(0)=0 ?

The solution for 2y^{''}-y^'-y=t+1,y(0)=1,y^'(0)=0 is y=-t+e^t

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2y^{''}-y^'-y=t+1,y(0)=1,y^'(0)=0

2 y^{^{''}} - y^{^{'}} - y = t + 1, y (0) = 1, y^{^{'}} (0) = 0

y = - t + e^{t}

Solution steps

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

Solve linear ODE:

y = - t + e^{t}

y = - t + e^{t}

y^{^{''}} + 4 y = sin (x) - cos (x)

x^{^{''}} - 6 x^{^{'}} + 9 x = 2 t e^{3 t}

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

y^{^{''}} + 2 y^{^{'}} + y = \frac{e ^{- x}}{1 - x ^{2}}

y^{^{''}} - 5 y^{^{'}} + 4 y = x, y (0) = 0, y^{^{'}} (0) = 0

What is the solution for 2y^{''}-y^'-y=t+1,y(0)=1,y^'(0)=0 ?
The solution for 2y^{''}-y^'-y=t+1,y(0)=1,y^'(0)=0 is y=-t+e^t