What is the solution for y^{''}-y^'-30y=0,y(0)=x,y^'(0)=5 ?

The solution for y^{''}-y^'-30y=0,y(0)=x,y^'(0)=5 is y=c_{1}e^{6x}+c_{2}e^{-5x}

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y^{''}-y^'-30y=0,y(0)=x,y^'(0)=5

y^{^{''}} - y^{^{'}} - 30 y = 0, y (0) = x, y^{^{'}} (0) = 5

y = c_{1} e^{6 x} + c_{2} e^{- 5 x}

Solution steps

y^{''} (x) - y^{'} (x) - 30 y = 0

Solve linear ODE:

y = c_{1} e^{6 x} + c_{2} e^{- 5 x}

y = c_{1} e^{6 x} + c_{2} e^{- 5 x}

x^{^{''}} = a x

- 3 y^{^{''}} + \frac{1}{12} y = 0

\frac{1}{8} y^{^{''}} + 4 y^{^{'}} + 28 y = 0, y (0) = 0, y^{^{'}} (0) = 8

\frac{d ^{4} y}{d x ^{4}} - 2 \frac{d ^{2} y}{d x ^{2}} - 8 y = 0

3 y^{^{''}} + 12 y = 0, y (0) = 8, y^{^{'}} (0) = 6

What is the solution for y^{''}-y^'-30y=0,y(0)=x,y^'(0)=5 ?
The solution for y^{''}-y^'-30y=0,y(0)=x,y^'(0)=5 is y=c_{1}e^{6x}+c_{2}e^{-5x}