What is the solution for 25y^{''}-20y^'+4y=0 ?

The solution for 25y^{''}-20y^'+4y=0 is y=c_{1}e^{(2t)/5}+c_{2}e^{(2t)/5}t

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25y^{''}-20y^'+4y=0

25 y^{^{''}} - 20 y^{^{'}} + 4 y = 0

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

Solution steps

25 y^{''} (t) - 20 y^{'} (t) + 4 y = 0

Solve linear ODE:

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

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

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What is the solution for 25y^{''}-20y^'+4y=0 ?
The solution for 25y^{''}-20y^'+4y=0 is y=c_{1}e^{(2t)/5}+c_{2}e^{(2t)/5}t