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

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

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

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

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

Solution steps

y^{''} (t) - \frac{1}{4} y = 0

Solve linear ODE:

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

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

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

y^{^{''}} - 4 y^{^{'}} + 4 y = 0, y (0) = 5, y^{^{'}} = 2

y^{^{''}} + 16 y^{^{'}} + 64 y = 0, y (0) = 5, y^{^{'}} (0) = 0

x^{^{''}} - 16 x = 0

y^{^{''}} - 18 y^{^{'}} + 81 y = 0, y (0) = 0, y^{^{'}} (0) = 5

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