What is the solution for y^{''}-16y=0,y(0)=3,y^'(0)=5 ?

The solution for y^{''}-16y=0,y(0)=3,y^'(0)=5 is y=(3-7/8)e^{4t}+7/8 e^{-4t}

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y^{''}-16y=0,y(0)=3,y^'(0)=5

y^{^{''}} - 16 y = 0, y (0) = 3, y^{^{'}} (0) = 5

y = (3 - \frac{7}{8}) e^{4 t} + \frac{7}{8} e^{- 4 t}

Solution steps

y^{''} (t) - 16 y = 0

Solve linear ODE:

y = (3 - \frac{7}{8}) e^{4 t} + \frac{7}{8} e^{- 4 t}

y = (3 - \frac{7}{8}) e^{4 t} + \frac{7}{8} e^{- 4 t}

\frac{d ^{4} y}{d x ^{4}} - \frac{d ^{2} y}{d x ^{2}} + 4 y = 0

\frac{d ^{2} y}{d x ^{2}} = 0.15 y, y (0) = 240, y (10) = 150

y^{^{''}} - 4 y^{^{'}} + 13 y = 0, y (0) = - 2, y^{^{'}} (0) = - 16

y = 10 y^{^{''}}

y^{^{''}} - 2 y^{^{'}} - 6 y = 0

What is the solution for y^{''}-16y=0,y(0)=3,y^'(0)=5 ?
The solution for y^{''}-16y=0,y(0)=3,y^'(0)=5 is y=(3-7/8)e^{4t}+7/8 e^{-4t}