What is the solution for 16y^{''}-72y^'+81y=0,y^'(0)=8 ?

The solution for 16y^{''}-72y^'+81y=0,y^'(0)=8 is y=(e^{(9t)/4}(32-4c_{2}))/(9)+c_{2}e^{(9t)/4}t

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16y^{''}-72y^'+81y=0,y^'(0)=8

16 y^{^{''}} - 72 y^{^{'}} + 81 y = 0, y^{^{'}} (0) = 8

y = \frac{e ^{\frac{9 t}{4}} ( 32 - 4 c _{2} )}{9} + c_{2} e^{\frac{9 t}{4}} t

Solution steps

16 y^{''} (t) - 72 y^{'} (t) + 81 y = 0

Solve linear ODE:

y = \frac{e ^{\frac{9 t}{4}} ( 32 - 4 c _{2} )}{9} + c_{2} e^{\frac{9 t}{4}} t

y = \frac{e ^{\frac{9 t}{4}} ( 32 - 4 c _{2} )}{9} + c_{2} e^{\frac{9 t}{4}} t

x^{^{''}} + 16 x = 0, x^{^{'}} (0) = 5

y^{^{'''}} - 3 y^{^{''}} + 2 y^{^{'}} = 0

y^{^{''''}} - 6 y^{^{'''}} + 17 y^{^{''}} - 28 y^{^{'}} + 20 y = 0

y^{^{''}} + 9 y = 0, y (0) = 6, y^{^{'}} (0) = - 4

y^{^{''}} + 9 y = 0, y (0) = 6, y^{^{'}} (0) = - 5

What is the solution for 16y^{''}-72y^'+81y=0,y^'(0)=8 ?
The solution for 16y^{''}-72y^'+81y=0,y^'(0)=8 is y=(e^{(9t)/4}(32-4c_{2}))/(9)+c_{2}e^{(9t)/4}t