What is the solution for y^{''}-12y^'+36y=0,y(0)=9,y^'(0)=5 ?

The solution for y^{''}-12y^'+36y=0,y(0)=9,y^'(0)=5 is y=9e^{6t}-49e^{6t}t

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y^{''}-12y^'+36y=0,y(0)=9,y^'(0)=5

y^{^{''}} - 12 y^{^{'}} + 36 y = 0, y (0) = 9, y^{^{'}} (0) = 5

y = 9 e^{6 t} - 49 e^{6 t} t

Solution steps

y^{''} (t) - 12 y^{'} (t) + 36 y = 0

Solve linear ODE:

y = 9 e^{6 t} - 49 e^{6 t} t

y = 9 e^{6 t} - 49 e^{6 t} t

7 y^{^{''}} - 8 y^{^{'}} + y = 0, y (0) = 7, y^{^{'}} (0) = \frac{1}{7}

y^{^{''}} - 4 y^{^{'}} + 3 y = 0, y (0) = \frac{3}{5}, y^{^{'}} (0) = \frac{1}{5}

y^{^{''}} + 16 y = 0, y (0) = sin (k t)

2 y^{^{''}} = 3 y

y^{^{''}} + 2 y^{^{'}} - 3 y = 0, y^{^{'}} (0) = 0

What is the solution for y^{''}-12y^'+36y=0,y(0)=9,y^'(0)=5 ?
The solution for y^{''}-12y^'+36y=0,y(0)=9,y^'(0)=5 is y=9e^{6t}-49e^{6t}t