What is the solution for y^{''}-12y^'+36y=0,y(0)=e^{6x} ?

The solution for y^{''}-12y^'+36y=0,y(0)=e^{6x} is y=c_{1}e^{6x}+c_{2}xe^{6x}

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y^{''}-12y^'+36y=0,y(0)=e^{6x}

y^{^{''}} - 12 y^{^{'}} + 36 y = 0, y (0) = e^{6 x}

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

Solution steps

y^{''} (x) - 12 y^{'} (x) + 36 y = 0

Solve linear ODE:

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

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

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

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

y^{^{'''}} + 2 y^{^{''}} - 63 y^{^{'}} = 0

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

0.05 q^{^{''}} + 2 q^{^{'}} + \frac{1}{0.01} q = 0

What is the solution for y^{''}-12y^'+36y=0,y(0)=e^{6x} ?
The solution for y^{''}-12y^'+36y=0,y(0)=e^{6x} is y=c_{1}e^{6x}+c_{2}xe^{6x}