What is the solution for y^{''}+y^'-2y=2t,y(0)=0,y^'(0)=4 ?

The solution for y^{''}+y^'-2y=2t,y(0)=0,y^'(0)=4 is y=(3/2+1/2)e^t-3/2 e^{-2t}-t-1/2

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y^{''}+y^'-2y=2t,y(0)=0,y^'(0)=4

y^{^{''}} + y^{^{'}} - 2 y = 2 t, y (0) = 0, y^{^{'}} (0) = 4

y = (\frac{3}{2} + \frac{1}{2}) e^{t} - \frac{3}{2} e^{- 2 t} - t - \frac{1}{2}

Solution steps

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

Solve linear ODE:

y = (\frac{3}{2} + \frac{1}{2}) e^{t} - \frac{3}{2} e^{- 2 t} - t - \frac{1}{2}

y = (\frac{3}{2} + \frac{1}{2}) e^{t} - \frac{3}{2} e^{- 2 t} - t - \frac{1}{2}

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x \to \infty lim ((1 + \frac{5}{x})^{\frac{x}{7}})

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What is the solution for y^{''}+y^'-2y=2t,y(0)=0,y^'(0)=4 ?
The solution for y^{''}+y^'-2y=2t,y(0)=0,y^'(0)=4 is y=(3/2+1/2)e^t-3/2 e^{-2t}-t-1/2