What is the solution for y^{''}-12y^'+36y=0,y(0)=4,y^'(0)=15.451 ?

The solution for y^{''}-12y^'+36y=0,y(0)=4,y^'(0)=15.451 is y=4e^{6t}-8.549e^{6t}t

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y^{''}-12y^'+36y=0,y(0)=4,y^'(0)=15.451

y^{^{''}} - 12 y^{^{'}} + 36 y = 0, y (0) = 4, y^{^{'}} (0) = 15.451

y = 4 e^{6 t} - 8.549 e^{6 t} t

Solution steps

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

Solve linear ODE:

y = 4 e^{6 t} - 8.549 e^{6 t} t

y = 4 e^{6 t} - 8.549 e^{6 t} t

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

y^{^{''}} + 18 y^{^{'}} = 0, y (0) = 18

\frac{d ^{2} f}{d x ^{2}} = f (x)

(0 + 1) y^{^{''}} + (7 + 1) y^{^{'}} - (6 + 1) y = 0

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

What is the solution for y^{''}-12y^'+36y=0,y(0)=4,y^'(0)=15.451 ?
The solution for y^{''}-12y^'+36y=0,y(0)=4,y^'(0)=15.451 is y=4e^{6t}-8.549e^{6t}t