What is the solution for y^{''}+2y^'-15y=0,y(0)=x,y^'(0)=20 ?

The solution for y^{''}+2y^'-15y=0,y(0)=x,y^'(0)=20 is y=c_{1}e^{3x}+c_{2}e^{-5x}

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y^{''}+2y^'-15y=0,y(0)=x,y^'(0)=20

y^{^{''}} + 2 y^{^{'}} - 15 y = 0, y (0) = x, y^{^{'}} (0) = 20

y = c_{1} e^{3 x} + c_{2} e^{- 5 x}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{3 x} + c_{2} e^{- 5 x}

y = c_{1} e^{3 x} + c_{2} e^{- 5 x}

3 y^{^{''}} + 2 y^{^{'}} + 1 y = 0

\frac{y ^{^{''}}}{8} + 3 y^{^{'}} + 4 y = 0, y (0) = 0, y^{^{'}} (0) = - 5

y^{^{''}} + 4 y = 0, y (0) = 6, y^{^{''}} (0) = - 4

4 y^{^{''}} - 12 y^{^{'}} + 73 y = 0

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

What is the solution for y^{''}+2y^'-15y=0,y(0)=x,y^'(0)=20 ?
The solution for y^{''}+2y^'-15y=0,y(0)=x,y^'(0)=20 is y=c_{1}e^{3x}+c_{2}e^{-5x}