What is the solution for y^{''}-2y^'=1,y(0)=2,y^'(0)=1 ?

The solution for y^{''}-2y^'=1,y(0)=2,y^'(0)=1 is y=(2-5/4)e^{2t}+5/4-t/2

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y^{''}-2y^'=1,y(0)=2,y^'(0)=1

y^{^{''}} - 2 y^{^{'}} = 1, y (0) = 2, y^{^{'}} (0) = 1

y = (2 - \frac{5}{4}) e^{2 t} + \frac{5}{4} - \frac{t}{2}

Solution steps

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

Solve linear ODE:

y = (2 - \frac{5}{4}) e^{2 t} + \frac{5}{4} - \frac{t}{2}

y = (2 - \frac{5}{4}) e^{2 t} + \frac{5}{4} - \frac{t}{2}

y^{^{''}} - y^{^{'}} + 14 y = 5 + e^{\frac{x}{2}}

y^{^{''}} - 16 y = 9.6 e^{4 x} + 30 e^{x}

y^{^{''}} - y = 3 e^{2 x}

y^{^{''}} + y^{^{'}} + y = 13 e^{- 2 t} (cos (t) + 2 sin (t))

y^{^{'''}} + 6 y = 7 y^{^{'}} + 4 e^{x}

What is the solution for y^{''}-2y^'=1,y(0)=2,y^'(0)=1 ?
The solution for y^{''}-2y^'=1,y(0)=2,y^'(0)=1 is y=(2-5/4)e^{2t}+5/4-t/2