What is the solution for y^{''}-12y^'+36y=0,y(0)=-2,y^'(0)=3 ?

The solution for y^{''}-12y^'+36y=0,y(0)=-2,y^'(0)=3 is y=-2e^{6t}+15e^{6t}t

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y^{''}-12y^'+36y=0,y(0)=-2,y^'(0)=3

y^{^{''}} - 12 y^{^{'}} + 36 y = 0, y (0) = - 2, y^{^{'}} (0) = 3

y = - 2 e^{6 t} + 15 e^{6 t} t

Solution steps

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

Solve linear ODE:

y = - 2 e^{6 t} + 15 e^{6 t} t

y = - 2 e^{6 t} + 15 e^{6 t} t

\int_{0}^{1} \frac{2}{t ^{2} + 1} d t

\frac{e ^{x}}{4 x ^{2} + 3}

f (x) = 2 π r^{2} + \frac{1000}{r}

y = \frac{6}{x}, (36, \frac{6}{6})

\frac{d y}{d x} = - 2 y + e^{x}

What is the solution for y^{''}-12y^'+36y=0,y(0)=-2,y^'(0)=3 ?
The solution for y^{''}-12y^'+36y=0,y(0)=-2,y^'(0)=3 is y=-2e^{6t}+15e^{6t}t