What is the solution for y^{''}-12y^'+36y=0,y(0)=3,y^'(0)= 39/2 ?

The solution for y^{''}-12y^'+36y=0,y(0)=3,y^'(0)= 39/2 is y=3e^{6t}+(3e^{6t}t)/2

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y^{''}-12y^'+36y=0,y(0)=3,y^'(0)= 39/2

y^{^{''}} - 12 y^{^{'}} + 36 y = 0, y (0) = 3, y^{^{'}} (0) = \frac{39}{2}

y = 3 e^{6 t} + \frac{3 e ^{6 t} t}{2}

Solution steps

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

Solve linear ODE:

y = 3 e^{6 t} + \frac{3 e ^{6 t} t}{2}

y = 3 e^{6 t} + \frac{3 e ^{6 t} t}{2}

y^{^{''}} + 18 y^{^{'}} + 80 y = 0, y (0) = 9, y^{^{'}} (0) = - 76

(D^{3} - 3 D) y = 0

(d - 5 d^{2} + 4 d - 20) x = 0

25 x^{^{''}} + 50 x^{^{'}} + 169 x = 0

45 y^{^{''}} + 180 y = 0

What is the solution for y^{''}-12y^'+36y=0,y(0)=3,y^'(0)= 39/2 ?
The solution for y^{''}-12y^'+36y=0,y(0)=3,y^'(0)= 39/2 is y=3e^{6t}+(3e^{6t}t)/2