What is the solution for y^{''}-16y=0,y(0)=e^{4x},y(0)=9,y^'(0)=3 ?

The solution for y^{''}-16y=0,y(0)=e^{4x},y(0)=9,y^'(0)=3 is y=c_{1}e^{4x}+c_{2}e^{-4x}

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y^{''}-16y=0,y(0)=e^{4x},y(0)=9,y^'(0)=3

y^{^{''}} - 16 y = 0, y (0) = e^{4 x}, y (0) = 9, y^{^{'}} (0) = 3

y = c_{1} e^{4 x} + c_{2} e^{- 4 x}

Solution steps

y^{''} (x) - 16 y = 0

Solve linear ODE:

y = c_{1} e^{4 x} + c_{2} e^{- 4 x}

y = c_{1} e^{4 x} + c_{2} e^{- 4 x}

What is the solution for y^{''}-16y=0,y(0)=e^{4x},y(0)=9,y^'(0)=3 ?
The solution for y^{''}-16y=0,y(0)=e^{4x},y(0)=9,y^'(0)=3 is y=c_{1}e^{4x}+c_{2}e^{-4x}