What is the solution for 25y^{''}-120y^'+144y=0 ?

The solution for 25y^{''}-120y^'+144y=0 is y=c_{1}e^{(12t)/5}+c_{2}e^{(12t)/5}t

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25y^{''}-120y^'+144y=0

25 y^{^{''}} - 120 y^{^{'}} + 144 y = 0

y = c_{1} e^{\frac{12 t}{5}} + c_{2} e^{\frac{12 t}{5}} t

Solution steps

25 y^{''} (t) - 120 y^{'} (t) + 144 y = 0

Solve linear ODE:

y = c_{1} e^{\frac{12 t}{5}} + c_{2} e^{\frac{12 t}{5}} t

y = c_{1} e^{\frac{12 t}{5}} + c_{2} e^{\frac{12 t}{5}} t

y^{^{'''}} - 7 y^{^{''}} + 16 y^{^{'}} - 12 y = 0

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

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

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

x^{^{''}} + a x^{^{'}} + b x = 0

What is the solution for 25y^{''}-120y^'+144y=0 ?
The solution for 25y^{''}-120y^'+144y=0 is y=c_{1}e^{(12t)/5}+c_{2}e^{(12t)/5}t