What is the solution for 4y^{''}-y=1,y^'(0)= 1/2 ,y(0)=0 ?

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

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4y^{''}-y=1,y^'(0)= 1/2 ,y(0)=0

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

y = e^{\frac{t}{2}} - 1

Solution steps

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

Solve linear ODE:

y = e^{\frac{t}{2}} - 1

y = e^{\frac{t}{2}} - 1

y^{^{''}} + 6 y^{^{'}} + 9 y = \frac{- 10 e ^{- 3 t}}{t ^{2} + 1}

y^{^{''}} - y^{^{'}} - 20 = 0

\frac{d ^{2} y}{d x ^{2}} - 9 y = e^{3 x} cos (2 x) + e^{2 x}

y^{^{''}} + 9 y = cos (t)

y^{^{''}} + 4 y^{^{'}} + 5 y = e^{3 x}

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