What is the solution for 4y^{''}-36y^'+81y=0 ?

The solution for 4y^{''}-36y^'+81y=0 is y=c_{1}e^{(9t)/2}+c_{2}e^{(9t)/2}t

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4y^{''}-36y^'+81y=0

4 y^{^{''}} - 36 y^{^{'}} + 81 y = 0

y = c_{1} e^{\frac{9 t}{2}} + c_{2} e^{\frac{9 t}{2}} t

Solution steps

4 y^{''} (t) - 36 y^{'} (t) + 81 y = 0

Solve linear ODE:

y = c_{1} e^{\frac{9 t}{2}} + c_{2} e^{\frac{9 t}{2}} t

y = c_{1} e^{\frac{9 t}{2}} + c_{2} e^{\frac{9 t}{2}} t

4 y^{^{''}} - 4 y^{^{'}} + 37 y = 0, y (0) = 3, y^{^{'}} (0) = \frac{3}{2}

y^{^{''}} + 4 y^{^{'}} + y 8 = 0

y^{^{''}} = 8 y^{^{'}} - 15 y, y (0) = 0, y^{^{'}} (0) = - 2

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

0.1 y^{^{''}} + 0.4 y^{^{'}} + 2 y = 0

What is the solution for 4y^{''}-36y^'+81y=0 ?
The solution for 4y^{''}-36y^'+81y=0 is y=c_{1}e^{(9t)/2}+c_{2}e^{(9t)/2}t