What is the solution for y^{''}-12y^'+36y=0,y(-1)=2,y^'(-1)=5 ?

The solution for y^{''}-12y^'+36y=0,y(-1)=2,y^'(-1)=5 is y=-5e^{6+6t}-7e^{6+6t}t

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y^{''}-12y^'+36y=0,y(-1)=2,y^'(-1)=5

y^{^{''}} - 12 y^{^{'}} + 36 y = 0, y (- 1) = 2, y^{^{'}} (- 1) = 5

y = - 5 e^{6 + 6 t} - 7 e^{6 + 6 t} t

Solution steps

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

Solve linear ODE:

y = - 5 e^{6 + 6 t} - 7 e^{6 + 6 t} t

y = - 5 e^{6 + 6 t} - 7 e^{6 + 6 t} t

4 y^{^{''}} - 24 y^{^{'}} + 37 y = 0, y (0) = - 9

y^{^{''}} = - 4 y, y (0) = 1, y^{^{'}} (0) = 0

8 y^{^{''}} + 16 y^{^{'}} + 128 y = 0, y (0) = 5, y^{^{'}} (0) = 8

(d^{2} + d + 1) y = 0

4 y^{^{''}} - 4 y^{^{'}} - 3 y = 0, y (- 2) = e, y^{^{'}} (- 2) = - \frac{e}{2}

What is the solution for y^{''}-12y^'+36y=0,y(-1)=2,y^'(-1)=5 ?
The solution for y^{''}-12y^'+36y=0,y(-1)=2,y^'(-1)=5 is y=-5e^{6+6t}-7e^{6+6t}t