What is the solution for 2y^{''}-8y=0,y(-2)=2,y^'(-2)=1 ?

The solution for 2y^{''}-8y=0,y(-2)=2,y^'(-2)=1 is y=(2e^4-(3e^4)/4)e^{2t}+(3e^{-2t-4})/4

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2y^{''}-8y=0,y(-2)=2,y^'(-2)=1

2 y^{^{''}} - 8 y = 0, y (- 2) = 2, y^{^{'}} (- 2) = 1

y = (2 e^{4} - \frac{3 e ^{4}}{4}) e^{2 t} + \frac{3 e ^{- 2 t - 4}}{4}

Solution steps

2 y^{''} (t) - 8 y = 0

Solve linear ODE:

y = (2 e^{4} - \frac{3 e ^{4}}{4}) e^{2 t} + \frac{3 e ^{- 2 t - 4}}{4}

y = (2 e^{4} - \frac{3 e ^{4}}{4}) e^{2 t} + \frac{3 e ^{- 2 t - 4}}{4}

y^{^{''}} + 3 y = sin (t)

f (x) = ln (x + x^{2} - 1)

\int (6 x - 5) d x

\frac{d}{d x} (lo g_{e} (cos (x)))

\frac{d}{d x} (\frac{1 + 2 x}{( 3 x - 1 ) ^{2}})

What is the solution for 2y^{''}-8y=0,y(-2)=2,y^'(-2)=1 ?
The solution for 2y^{''}-8y=0,y(-2)=2,y^'(-2)=1 is y=(2e^4-(3e^4)/4)e^{2t}+(3e^{-2t-4})/4