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

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

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

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

y = \frac{3}{2} e^{t} - \frac{3}{2} e^{- t}

Solution steps

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

Solve linear ODE:

y = \frac{3}{2} e^{t} - \frac{3}{2} e^{- t}

y = \frac{3}{2} e^{t} - \frac{3}{2} e^{- t}

y^{^{''}} + 49 y = 0, y (π) = 0, y (0) = 0

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

\frac{d ^{2} y}{d x ^{2}} + 5 y = 0

4 y^{^{''}} + 20 y^{^{'}} + 169 y = 0, y (0) = 3, y^{^{'}} (0) = 1

4 y^{^{''}} - 10 y^{^{'}} = 0

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