What is the solution for y^{''}-2y^'-15y=0,y(0)=2,y^'(0)=6 ?

The solution for y^{''}-2y^'-15y=0,y(0)=2,y^'(0)=6 is y=(2-1/2)e^{5t}+1/2 e^{-3t}

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y^{''}-2y^'-15y=0,y(0)=2,y^'(0)=6

y^{^{''}} - 2 y^{^{'}} - 15 y = 0, y (0) = 2, y^{^{'}} (0) = 6

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 3 y^{^{'}} + a y = 0, y (0) = 1, y^{^{'}} (0) = - 7

y^{^{''}} - 5 y^{^{'}} - 3 y = 0

\frac{d ^{2} y}{d x ^{2}} = 5 y, y (0) = - 2, y (2) = 10

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

y^{^{''}} + 7 y^{^{'}} + 6 y = 0

What is the solution for y^{''}-2y^'-15y=0,y(0)=2,y^'(0)=6 ?
The solution for y^{''}-2y^'-15y=0,y(0)=2,y^'(0)=6 is y=(2-1/2)e^{5t}+1/2 e^{-3t}