What is the solution for y^{''}+4y^'-5y=0,y(0)=2,y(0)=26 ?

The solution for y^{''}+4y^'-5y=0,y(0)=2,y(0)=26 is y=(2-c_{2})e^t+c_{2}e^{-5t}

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y^{''}+4y^'-5y=0,y(0)=2,y(0)=26

y^{^{''}} + 4 y^{^{'}} - 5 y = 0, y (0) = 2, y (0) = 26

y = (2 - c_{2}) e^{t} + c_{2} e^{- 5 t}

Solution steps

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

Solve linear ODE:

y = (2 - c_{2}) e^{t} + c_{2} e^{- 5 t}

y = (2 - c_{2}) e^{t} + c_{2} e^{- 5 t}

What is the solution for y^{''}+4y^'-5y=0,y(0)=2,y(0)=26 ?
The solution for y^{''}+4y^'-5y=0,y(0)=2,y(0)=26 is y=(2-c_{2})e^t+c_{2}e^{-5t}