What is the solution for y^{''}+4y^'-5y=0,y(ln(2))=3,y(0)=33 ?

The solution for y^{''}+4y^'-5y=0,y(ln(2))=3,y(0)=33 is y=e^t+32e^{-5t}

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y^{''}+4y^'-5y=0,y(ln(2))=3,y(0)=33

y^{^{''}} + 4 y^{^{'}} - 5 y = 0, y (ln (2)) = 3, y (0) = 33

y = e^{t} + 32 e^{- 5 t}

Solution steps

y^{''} (t) + 4 y^{'} (t) - 5 y = 0

Solve linear ODE:

y = e^{t} + 32 e^{- 5 t}

y = e^{t} + 32 e^{- 5 t}

y^{^{'''}} - 4 y^{^{''}} - 45 y^{^{'}} = 0

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

y^{^{'''}} - 5 y^{^{''}} + 5 y^{^{'}} + 11 y = 0

36 y^{^{''}} + 108 y^{^{'}} + 97 y = 0, y^{^{'}} (1) = 3, y (1) = 4

25 y^{^{''}} - 80 y^{^{'}} + 64 y = 0

What is the solution for y^{''}+4y^'-5y=0,y(ln(2))=3,y(0)=33 ?
The solution for y^{''}+4y^'-5y=0,y(ln(2))=3,y(0)=33 is y=e^t+32e^{-5t}