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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

2^{60}

y^{^{'}} = \frac{2 x y - y ^{4}}{3 x ^{2}}

f (x) = x (x - 10)

5 y^{^{''}} + 32 y = 0, y (0) = 8, y^{^{'}} (0) = 5

\int \frac{1}{2} x cos (x^{2}) d x

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