What is the solution for y^{''}-y=0,y(ln(2))=5,y(0)=1 ?

The solution for y^{''}-y=0,y(ln(2))=5,y(0)=1 is y=3e^t-2e^{-t}

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y^{''}-y=0,y(ln(2))=5,y(0)=1

y^{^{''}} - y = 0, y (ln (2)) = 5, y (0) = 1

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d ^{2} x}{d t ^{2}} + 4 \frac{d x}{d t} + 2 x = 0, x (0) = 0

y^{^{''''}} + 2 y^{^{'''}} + 2 y^{^{''}} + 2 y^{^{'}} + y = 0

2 \cdot y^{^{''}} + 8 \cdot y^{^{'}} + 8 y = 0

y^{^{''}} - 6 y^{^{'}} + 9 y = 0, y (0) = 0, y^{^{'}} (0) = 1

c^{2} \cdot f^{^{''}} (t) = a f^{^{''}} (t) - b f (t)

What is the solution for y^{''}-y=0,y(ln(2))=5,y(0)=1 ?
The solution for y^{''}-y=0,y(ln(2))=5,y(0)=1 is y=3e^t-2e^{-t}