What is the solution for y^{''}+2y^'-15y=0,y(0)=1 ?

The solution for y^{''}+2y^'-15y=0,y(0)=1 is y=(1-c_{2})e^{3t}+c_{2}e^{-5t}

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y^{''}+2y^'-15y=0,y(0)=1

y^{^{''}} + 2 y^{^{'}} - 15 y = 0, y (0) = 1

y = (1 - c_{2}) e^{3 t} + c_{2} e^{- 5 t}

Solution steps

y^{''} (t) + 2 y^{'} (t) - 15 y = 0

Solve linear ODE:

y = (1 - c_{2}) e^{3 t} + c_{2} e^{- 5 t}

y = (1 - c_{2}) e^{3 t} + c_{2} e^{- 5 t}

5 y^{^{''}} + 20 y = 0, y (\frac{π}{4}) = 10, y (0) = 15

2 y^{^{''}} + 6 y^{^{'}} = 0

y^{^{''}} + c^{2} y = 0, y (0) = a, y^{^{'}} (0) = 0

x^{^{''}} + 10 x^{^{'}} + 29 x = 0, x^{^{'}} (0) = 6

y^{^{''}} + 4 y = 0, y (0) = 1, y^{^{'}} (0) = 2

What is the solution for y^{''}+2y^'-15y=0,y(0)=1 ?
The solution for y^{''}+2y^'-15y=0,y(0)=1 is y=(1-c_{2})e^{3t}+c_{2}e^{-5t}