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

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

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

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

y = (3 - c_{2}) e^{t} + c_{2} e^{- t}

Solution steps

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

Solve linear ODE:

y = (3 - c_{2}) e^{t} + c_{2} e^{- t}

y = (3 - c_{2}) e^{t} + c_{2} e^{- t}

2 e^{x} + e^{2} x

3 + e^{- 0.1 t} cos (4 t)

\int 7 x cos (6 x) d x

\int \frac{3 t - 2}{t + 1} d t

\int \frac{3 e ^{- x} + 2 e ^{3 x}}{e ^{2 x}} d x

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