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

The solution for y^{''}-12y^'+36y=0,y(0)=4,y^'(0)=0 is y=4e^{6t}-24e^{6t}t

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

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

y = 4 e^{6 t} - 24 e^{6 t} t

Solution steps

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

Solve linear ODE:

y = 4 e^{6 t} - 24 e^{6 t} t

y = 4 e^{6 t} - 24 e^{6 t} t

\frac{d}{d x} (cos^{7} (x))

y^{^{'}} = \frac{x ^{2} y + y ^{3}}{x ^{3}}

\frac{\partial}{\partial y} (- sin (y))

\int (1 - \frac{x}{a})^{3} d x

y^{^{''}} + 20 y^{^{'}} + y = 12

What is the solution for y^{''}-12y^'+36y=0,y(0)=4,y^'(0)=0 ?
The solution for y^{''}-12y^'+36y=0,y(0)=4,y^'(0)=0 is y=4e^{6t}-24e^{6t}t