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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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

y^{^{''}} - 6 y^{^{'}} + 9 y = e^{3 x} ln (x^{2} + 1)

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

y^{^{''}} - 2 y^{^{'}} + y = 2 x + x^{2} + sin (3 x)

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

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