What is the solution for 36y^{''}-132y^'+121y=0,y(0)=5,y^'(0)=5 ?

The solution for 36y^{''}-132y^'+121y=0,y(0)=5,y^'(0)=5 is y=5e^{(11t)/6}-(25e^{(11t)/6}t)/(6)

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36y^{''}-132y^'+121y=0,y(0)=5,y^'(0)=5

36 y^{^{''}} - 132 y^{^{'}} + 121 y = 0, y (0) = 5, y^{^{'}} (0) = 5

y = 5 e^{\frac{11 t}{6}} - \frac{25 e ^{\frac{11 t}{6}} t}{6}

Solution steps

36 y^{''} (t) - 132 y^{'} (t) + 121 y = 0

Solve linear ODE:

y = 5 e^{\frac{11 t}{6}} - \frac{25 e ^{\frac{11 t}{6}} t}{6}

y = 5 e^{\frac{11 t}{6}} - \frac{25 e ^{\frac{11 t}{6}} t}{6}

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

625 y^{^{''}} - 9 y = 0, y (0) = 9, y^{^{'}} (0) = 8

\frac{5}{8} \frac{d ^{2} x}{d t ^{2}} + 40 x = 0

y^{^{''''}} - 12 y^{^{''}} + 36 y = 0

y^{^{''}} + 1 y^{^{'}} + 0 y = 0

What is the solution for 36y^{''}-132y^'+121y=0,y(0)=5,y^'(0)=5 ?
The solution for 36y^{''}-132y^'+121y=0,y(0)=5,y^'(0)=5 is y=5e^{(11t)/6}-(25e^{(11t)/6}t)/(6)